Chapter 10. Sequences and Series of Functions

For the materials to be discussed below, we will need to introduce the extended real number system [�1;C1]:Definitions. (1) Let [�1;C1] denote the set R[ f�1;C1g: We extend the ordering of R to [�1;C1] bydefining �1 < x < C1 for every x 2 R: (Since �1;C1 62 R; they are not numbers! We treat �1;C1 assymbols.)

Let S be a nonempty subset of [�1;C1]: Then C1 is an upper bound of S in [�1;C1] and �1 is a lowerbound of S in [�1;C1]: (So every nonempty set S has at least one upper and one lower bound.) Define sup S to bethe least upper bound of S in [�1;C1] and inf S to be the greatest lower bound of S in [�1;C1]: (Note that fornonempty S; sup S and inf S always exist in [�1;C1]:)(2) For x 2 R[ fC1g; c > 0; let x C .C1/ D C1 D .C1/ C x; c.C1/ D C1 D .C1/c; c.�1/ D �1 D.�1/c: Similarly, for x 2 R[f�1g; c < 0; let xC.�1/ D �1 D .�1/Cx; c.�1/ D C1 D .�1/c; c.C1/ D�1 D .C1/c: Also, j C 1j D C1 D j �1j:(3) For x1; x2; x3; : : : 2 [�1;C1];we say lim

n!1 xn D C1 iff for every real number r > 0; there exists a K 2 N such

that n � K implies xn > r: Similarly, we say limn!1 xn D �1 iff for every real number r > 0; there exists a K 2 N

such that n � K implies xn < �r:Below x1; x2; x3; : : :will denote a sequence of real numbers. If lim

n!1 xn exists, then we can see where the sequence

is converging. If the limit does not exist, we still would like to get some information on the distributionof the sequence.

Recall a subsequence of x1; x2; x3; x4; : : : is a sequence xn1 ; xn2 ; xn3 ; xn4 ; : : : ;where the indices n1; n2; n3; n4; : : :is a strictly increasing sequence of positive integers. For example, taking ni D 3i; we get the subsequencex3; x6; x9; x12; : : : : Taking ni D i2;we get the subsequence x1; x4; x9; x16; : : : : Taking ni to be the i-th prime number,we get the subsequence x2; x3; x5; x7; x11; : : : : For x1; x2; x3; : : : ; consider the set of subsequential limitsL D n

z 2 [�1;C1] : z D limk!1 xnk for some subsequence xn1 ; xn2 ; xn3; : : : of x1; x2; x3; : : :o:

If the sequence is bounded in R; then by the Bolzano-Weierstrass theorem, there is at least one subsequence that haslimit, i.e. L 6D ;: If the sequence is unbounded inR; then either�1 orC1 is in L; so again L 6D ;: Therefore supLand infL always exist in [�1;C1]:Definitions. Define the limit superior (or upper limit) of x1; x2; x3; : : : to be supL and the limit inferior (or lower limit)of x1; x2; x3; : : : to be infL: For limit superior, the notations are limsup

n!1 xn or limn!1 xn : For limit inferior, the notations

are liminfn!1 xn or lim

n!1 xn : (Briefly, the limit superior is the largest subsequential limit and the limit inferior is the least

subsequential limit.)

Remarks. (1) The limit superior and limit inferior of x1; x2; x3; x4; : : : always exist in [�1;C1]: We also haveliminfn!1 xn D infL � supL D limsup

n!1 xn :(2) If fx1; x2; x3; x4; : : :g is not bounded above in R; then C1 2 L; so limsup

n!1 xn D supL D C1: Similarly, iffx1; x2; x3; x4; : : :g is not bounded below inR; then liminfn!1 xn D �1:

(3) limn!1 xn D z 2 [�1;C1] () L D fzg () limsup

n!1 xn D z D liminfn!1 xn :

(4) limsupn!1 .�xn/ D sup.�L/ D � infL D � liminf

n!1 xn : For c > 0; limsupn!1 .cxn/ D sup.cL/ D c supL D c limsup

n!1 xn

and liminfn!1 .cxn/ D c liminf

n!1 xn : For every c 2 R; limsupn!1 .c C xn/ D sup.c C L/ D cC supL D c C limsup

n!1 xn and

liminfn!1 .c C xn/ D cC liminf

n!1 xn :78

Theorem. For x1; x2; x3; x4; : : : ; define

Mk D supfxk; xkC1; xkC2; xkC3; : : :g and mk D inffxk; xkC1; xkC2; xkC3; : : :g:(1) We have M1 � M2 � M3 � M4 � : : : ; m1 � m2 � m3 � m4 � : : : ;(2) limsup

n!1 xn D limk!1 Mk; liminf

n!1 xn D limk!1mk and both are limits of some subsequences of x1; x2; x3; : : : :

Proof. (1) Since Mk � xkC1; xkC2; xkC3; : : : ; so Mk � supfxkC1; xkC2; xkC3; : : :g D MkC1: The inequalities for mk’sare similar.

(2) As both the Mk and the mk sequences are monotone, their limits exist in [�1;C1]: Let M D limk!1 Mk: We have

to show supL D M: For every z 2 L; z D limj!1 xnj � lim

j!1 Mnj D M: So supL � M:Conversely, if M1 D C1; then x1; x2; x3; : : : is not bounded above in R and so supL D C1 D M: Oth-

erwise, M1 2 R: By the supremum property, since M1 D supfx1; x2; x3; : : :g; there exists some xn1 such thatM1 � 1 < xn1 � M1: Next, consider M1Cn1 D supfx1Cn1 ; x2Cn1; x3Cn1; : : :g D supfxk : k > n1g: By the supremum

property, there is n2 > n1 such that M1Cn1 � 1

2< xn2 � M1Cn1 : Keep on repeating this, we get strictly increasing

n j’s such that M1Cnj � 1j C 1

< xnjC1 � M1Cnj : Then xn1 ; xn2; xn3 ; : : : is a subsequence. By the sandwich theorem,

limj!1 xnj D lim

j!1 M1Cnj D M 2 L: Then M � supL: Therefore, supL D M: The limit inferior case is similar.

Corollary. If xn � yn; then limsupn!1 xn � limsup

n!1 yn and liminfn!1 xn � liminf

n!1 yn:Proof. This follows from observing that the Mk’s and the mk’s for xn’s are less than or equal to those for yn’s.

Examples. (1) Since limn!1 1C n

3C 4nD 1

4; so by remark (3), limsup

n!1 1C n

3C 4nD 1

4and liminf

n!1 1C n

3C 4nD 1

4:

(2) Let xn D �e�n if n is odden if n is even,

then 0 < xn < C1: So, the limits of the convergent subsequences are in [0;C1];i.e. L � [0;C1]: Since lim

n!1 x2nC1 D 0 2 L and limn!1 x2n D C1 2 L; so by definition, liminf

n!1 xn D infL D 0 and

limsupn!1 xn D supL D C1: (Alternatively, by part (2) of the theorem,

Mk D �supfe�k; ekC1; e�.kC2/; ekC3; : : :g if k is oddsupfek; e�.kC1/; ekC2; e�.kC3/; : : :g if k is even

D C1 H) limsupn!1 xn D lim

k!1 Mk D C1;mk D �

inffe�k; ekC1; e�.kC2/; ekC3; : : :g if k is oddinffek ; e�.kC1/; ekC2; e�.kC3/; : : :g if k is even

D 0 H) liminfn!1 xn D lim

k!1mk D 0:/(3) Consider the sequence �4; 0; 1;�3 1

2 ; 0; 12 ;�3 1

3 ; 0; 13 ;�3 1

4 ; : : : defined by xn D 8<:��3C 1jC1

�if n D 3 j C 1

0 if n D 3 j C 21

jC1 if n D 3 j C 3:

Observe x1 D �4 < x4 D �312< x7 D �3

13< � � � < 0 D x2 D x5 D x8 D � � � < � � � < x9 D 1

3< x6 D 1

2< x3 D 1:

Then the Mk sequence is 1; 1; 1; 12 ; 1

2 ; 12 ; 1

3 ; 13 ; 1

3 ; : : : ; 1jC1 ; 1

jC1 ; 1jC1 ; : : : : So by part (2) of the theorem, limsup

n!1 xnD limk!1 Mk D 0: Similarly, the mk sequence is �4;�3 1

2 ;�3 12 ;�3 1

2 ;�3 13 ;�3 1

3 ;�3 13 ; : : : ;�.3 C 1

jC1/;�.3 C 1jC1/;�.3 C 1

jC1/; : : : : Again by part (2) of the theorem, liminfn!1 xn D lim

k!1mk D �3:79

The following are theorems involving limit superior and limit inferior.

Theorem. For a sequence a1; a2; a3; a4; : : : of real numbers with an 6D 0 for all n; we have

liminfn!1 ����anC1

an

���� � liminfn!1 n

pjanj � limsupn!1 n

pjanj � limsupn!1 ����anC1

an

���� :(So, if lim

n!1 ����anC1

an

���� D L; then limn!1 n

pjanj D L : However, if limn!1 n

pjanj exists, then limn!1 ����anC1

an

���� may not exist.)

Proof. We will prove the rightmost inequality only. (The middle inequality follows from remark (1) and the proof

of leftmost inequality is similar to that of the rightmost inequality.) Let M D limsupn!1 ����anC1

an

���� : The case M D C1 is

trivial. So suppose M < C1: For every real number r > M; since

M D limk!1 Mk; where Mk D sup

�����akC1

ak

���� ; ����akC2

akC1

���� ; : : :� ;there is Mk such that Mk < r: Then

����akC1

ak

���� ; ����akC2

akC1

���� ; : : : < r: So for n > k; ����an

ak

���� D ����akC1

ak

���� � ����akC2

akC1

���� � � � ���� an

an�1

���� < rn�k ;which implies n

pjanj < r npjakjr�k : Since lim

n!1 np

c D 1 for every positive constant c; so

limsupn!1 n

pjanj � limsupn!1 r n

pjakjr�k D limn!1 r n

pjakjr�k D r:Taking limit as r ! MC; we get limsup

n!1 npjanj � M; which is the rightmost inequality.

Remarks. The theorem tells us that if the convergence of a series can be detected by the ratio test, then the convergenceof the series can also be detected by the root test, but not vice versa.

Theorem (Strong Form of Root Test). For a sequence a1; a2; a3; a4; : : : of real numbers,

(1) if limsupn!1 n

pjanj < 1; then16

nD1an converges absolutely.

(2) If limsupn!1 n

pjanj > 1; then16

nD1an diverges.

Proof. Let M D limsupn!1 n

pjanj: If M < 1; then choose r so that M < r < 1: Since

M D limk!1 Mk; where Mk D supf k

pjakj; kC1pjakC1j; : : :g;

there exists some Mk such that Mk < r: Then kpjak j; kC1

pjakC1j; : : : < r: So janj < rn for n � k: By the comparisontest, the series will converge absolutely.

If M > 1; then since M is the limit of some subsequence of npjanj; there are infinitely many n

pjanj > 1: Thenthere are infinitely many janj > 1: So lim

n!1 an 6D 0: By term test, the series diverges.

Theorem (Strong Form of Ratio Test). For a sequence a1; a2; a3; a4; : : : of real numbers with an 6D 0 for all n;(1) if limsup

n!1 ����anC1

an

���� < 1; then16

nD1an converges absolutely.

(2) If liminfn!1 ����anC1

an

���� > 1; then16

nD1an diverges.

Proof. This follow from the theorem above and the strong form of the root test.

80

Pointwise Convergence

Question. How do calculators compute sin x; cos x; ex; x y; ln x; : : :?For �1 < r < 1; 1

1� rD 1C r C r2 C r3 C � � � : Integration suggests� ln.1� r/ D r C r2

2C r3

3C r4

4C � � � :

Setting x D 1 � r suggests ln x D .x � 1/C .x � 1/22

� .x � 1/33

C � � � : As �1 < r < 1 is required, this formula

may only work for 0 < x D 1� r < 2: Note the formula is consisted of monomial terms, which may be computed byhand in principle. Now the formula is a series of functions in x : What is the meaning of convergence if the terms arefunctions?

Definitions. We say a sequence of functions Sn: E ! Rconverges pointwise on a set E to a function S: E ! Riff forevery x 2 E , lim

n!1 Sn.x/ D S.x/ 2 R(i.e. for every x 2 E , for every " > 0, there exists K 2 N (K depends on x and") such that n � K ) jSn.x/ � S.x/j < ".) In that case, S.x/ is said to be the pointwise limit of the sequence Sn.x/:Similarly, a series of functions

16kD1

fk .x/ converges pointwise on E to a function S : E ! R iff for every x 2 E ,

limn!1. f1.x/ C f2.x/ C : : : C fn.x// D S.x/, i.e. the sequence of partial sums Sn.x/ D f1.x/ C � � � C fn.x/ converges

pointwise on E to S.x/.Examples. (1) Show that the sequence Sn.x/ D .1� x2/n converge pointwise on [0; 1] to some function S.x/:

Solution. For every x 2 [0; 1], limn!1 Sn.x/ D lim

n!1.1 � x2/n D n0 if 0 < x � 11 if x D 0

. So Sn.x/ converges pointwise

on [0; 1] to the function S.x/ D n0 if 0 < x � 11 if x D 0

.

(2) Show that the series16

kD1ekx cos x converge pointwise on .�1; 0/; but not on .�1; 1]:

Solution. Recall the geometric series16

kD1rk converges (to

r

1� r) if and only if r 2 .�1; 1/:Now r D ex 2 .�1; 1/

if and only if x < 0: So16

kD1ekx cos x converges pointwise on .�1; 0/ to S.x/ D ex cos x

1� ex. Since

16kD1

ekx cos x

diverges when x D 0; it does not converge pointwise on .�1; 1]:Definition. A power series is a function f .x/ of the form a0 C a1.x � c/ C a2.x � c/2 C : : : D 16

kD0ak.x � c/k;where

c, a0, a1, a2, : : : are numbers and c is called the center of the power series. Its domain (of convergence) is the set ofnumbers, where the power series converges pointwise.

Power series are very important. For example, mathematical tables were built by using series such as

ex D 1C x C x2

2!C x3

3!C � � � D 16

kD0

xk

k!for �1 < x < C1

cos x D 1� x2

2!C x4

4!� x6

6!C � � � D 16

kD0

.�1/k x2k.2k/! for �1 < x < C1sin x D x � x3

3!C x5

5!� x7

7!C � � � D 16

kD0

.�1/k x2kC1.2k C 1/! for �1 < x < C1:Remarks. These series also converge for every complex numbers x by ratio test. If wet set x D i�; then we can checkthat ei� D cos � C i sin �: In particular, ei� D �1: Then ei� C 1 D 0; which is called Euler’s formula. It is the mostbeautiful formula in mathematics because it links up the five most important constants 1; 0; �; i; e in one equation.

Domain Theorem for Power Series. The domain of a power series f .x/ D 16kD0

ak.x � c/k is a nonempty interval

81

with c as midpoint. (The half-length of the interval is called the radius of convergence of the power series, commonlydenoted by R.) We have

R D 1

limsupk!1 k

pjakj :In fact, the power series converges absolutely on the open interval .c � R; c C R/; but it may or may not converge ateither endpoint.

Proof. This follows by applying the strong form of the root test to16

kD0ak.x � c/k :

Remarks. Although we have a formula for the radius of convergence, it is often easier to compute the radius using theratio test (especially when the coefficients involve factorials).

Examples. (1) Find the domain of f .x/ D 16kD0

xk

k!.

Solution. By the ratio test, since limk!1 ���� xkC1.k C 1/! k!

xk

���� D limk!1 jxj

k C 1D 0 < 1; the series converges for every x : The

domain of f .x/ isRD .�1;C1/: Here c D 0 and R D 1:(2) Find the domain of f .x/ D 16

kD0k!.x � �/k :

Solution. By the ratio test, since limk!1 ���� .k C 1/!.x � �/kC1

k!.x � �/k ���� D limk!1.k C 1/jx � � j D �

0 if x D �;1 if x 6D �; the series

converges only for x D �: The domain of f .x/ is f�g D [�; �]: Here c D � and R D 0:(3) Find the domain of f .x/ D 16

kD1

.�1/k .x � 25/kk

.

Solution. By the ratio test, since limk!1 ���� .�1/kC1.x � 25/kC1

k C 1k.�1/k.x � 25/k ���� D lim

k!1 jx � 25j k

k C 1D jx � 25j;

the series converges absolutely for jx � 25j < 1 (i.e. 24 < x < 26) and diverges for jx � 25j > 1 (i.e. x < 24

or x > 26/: When x D 24, the series becomes16

kD1

1k

which diverges by p-test. When x D 26, the series becomes16kD1

.�1/kk

which converges by the alternating series test. The domain of f .x/ is the interval .24; 26]:Here c D 25

and R D 1.

Question. How do we expand functions into series?

If f .x/ D a0 C a1x C a2x2 C a3x3 C � � � C anxn C � � � ; then formal computation yields a0 D f .0/; a1 Df 0.0/; a2 D f 00.0/=2; a3 D f 000.0/=6; : : : ; an D f .n/.0/=n!; : : : : In case the function is not defined at 0 (for example,log x), we can try to expand the function at any point c; where f is infinitely differentiable. Similarly, we can get thecoefficients an in terms of f .n/.c/: So for an arbitrary (infinitely differentiable) function f .x/ defined about c; we canattempt to form the power series

f .c/ C f 0.c/.x � c/ C f 00.c/2

.x � c/2 C f 000.c/6

.x � c/3 C � � � C f .n/.c/n!

.x � c/n C � � � D 16kD0

f .k/.c/k!

.x � c/k;which is called the Taylor series of f about c. (For the Taylor series of f about c D 0; it is traditionally called theMaclaurin series of f:)Remarks. Unfortunately, the Taylor series of a function does not always equal to the function away from the center.

For example, the function f .x/ D �e�1=x2

if x 6D 00 if x D 0

can be shown to be infinitely differentiable and f .n/.0/ D 0

82

for every n 2 N: So the Taylor series of f .x/ about c D 0 is16

kD00xk D 0; i.e. the Taylor series is the zero function.

Therefore, the Taylor series of f .x/ equals f .x/ only at the center c D 0:Nevertheless, the following theorems provide sufficient conditions for many common functions to equal their

Taylor series.

Taylor Series Theorem. If f : .a; b/ ! R is infinitely differentiable , c 2 .a; b/ and there are constants M; � > 0

such that j f .n/.x/j � �Mn for every x 2 .a; b/, n 2 N, then f .x/ D 16kD0

f .k/.c/k!

.x � c/k for every x 2 .a; b/:Proof. Apply Taylor’s theorem. Note jRn.x/j � �Mn

n!jx � cjn D � yn

n!; where y D Mjx � cj: By example (1),

16nD0

yn

n!

converges. By term test, limn!1 yn

n!D 0: By sandwich theorem, lim

n!1 Rn.x/ D 0:Examples.(1) For f .x/ D sin x , f .n/.x/ D � .�1/k cos x if n D 2k C 1.�1/k sin x if n D 2k

. So j f .n/.x/j � 1 D 1 �1n for every x 2 R:Taking c D 0; by the Taylor series theorem, sin x D 16

kD0

.�1/k x2kC1.2k C 1/! for all x 2 R.

Remarks. For 0 � x � �2; jR18.x/j � jxj18

18!� .�=2/18

18!< 6� 10�13: So x � x3

3!C � � � C x17

17!can be used to compute

sin x to 10 decimal places.

(2) For f .x/ D ex on .�w;w/, f .n/.x/ D ex : So j f .n/.x/j D ex � ew D ew � 1n for every x 2 .�w;w/: Taking

c D 0; by the Taylor series theorem, ex D 16kD0

xk

k!for all x 2 .�w;w/. Since w is arbitrary, ex D 16

kD0

xk

k!for all

x 2 R.

Bernstein’s Theorem. Suppose f and all its derivatives are nonnegative on a closed and bounded interval [c; cC r]:For x 2 [c; cC r/; we have f .x/ D 16

kD0

f .k/.c/k!

.x � c/k :Proof. By substitution, we may assume c D 0: The case x D 0 is clear. Suppose x 2 .0; r/: Apply Taylor’s

formula with integral remainder Rn.x/ D 1.n � 1/! Z x

0.x � t/n�1 f .n/.t/ dt : The conclusion follows if we can show

0 � Rn.x/ � �x

r

�nf .r/; since the expression on the right has limit 0 as n !1: Substituting t D x.1 � u/; we get

Rn.x/ D xn.n � 1/! Z 1

0un�1 f .n/�x.1� u/� du: Since f .nC1/ � 0 implies f .n/ increasing, we get f .n/�x.1 � u/� �

f .n/�r.1 � u/� for all u 2 [0; 1]: HenceRn.x/

xnis increasing. So 0 � Rn.x/

xn� Rn.r/

rn� f .r/

rn; where Rn.r/ � f .r/

is because f .r/ D n�16kD0

f .k/.0/k!

rk C Rn.r/: Multiplying all terms of the inequalities by xn; we get the result.

Next we will extend the binomial theorem to cover all real exponents.

Binomial Theorem. For a 2 R; if �1 < x < 1; then.1 C x/a D 1C ax C a.a � 1/2!

x2 C a.a � 1/.a � 2/3!

x3 C � � � D 1C 16kD1

a.a � 1/ � � � .a � k C 1/k!

xk :Proof. Let f .x/ D .1 C x/a; then f .n/.x/ D a.a � 1/ � � � .a � n C 1/.1 C x/a�n : By Taylor’s formula with inte-

gral remainder, Rn.x/ D 1.n � 1/! Z x

0.x � t/n�1 f .n/.t/ dt D a.a � 1/ � � � .a � n C 1/.n � 1/! Z x

0

�x � t

1 C t

�n�1.1 C t/a�1 dt :83

For x 2 .�1; 1/; the function g.t/ D x � t

1 C thas derivative g0.t/ D �1� x.1 C t/2 < 0: On the closed interval with end-

points 0 and x; g takes values between 0 and x : So there jg.t/j � jg.0/j D jxj: Let k D ���Z x

0.1 C t/a�1 dt

���; thenjRn.x/j � ja.a � 1/ � � � .a � n C 1/kxn�1j.n � 1/!| {z }call this bn

: Since limn!1 bnC1

bnD lim

n!1 ja � njjxjn

D jxj < 1; by ratio test,16

nD1bn con-

verges. By term test, limn!1 bn D 0: Then lim

n!1 Rn.x/ D 0 by the sandwich theorem.

Here are a few more common Taylor series. (Note the series only converge on a small interval to the functions.)

They are obtained by taking the cases a D �1; a D �1 (with x replaced by x2) and a D �1

2(with x replaced by�x2)

of the binomial theorem and integrating term-by-term. These can be justified by applying Abel’s limit theorem and theintegration theorem for uniformly convergence to be covered soon. Alternatively, they can be justified by integratingthe Taylor’s formula with integral remainder for the corresponding binomial functions and using the bn estimates aboveto squeeze the new remainder to 0.

ln.1 C x/ D x � x2

2C x3

3� x4

4C � � � D 16

kD1

.�1/k�1xk

kfor � 1 < x � 1

tan�1 x D x � x3

3C x5

5� x7

7C � � � D 16

kD0

.�1/k x2kC1

2k C 1for � 1 � x � 1

sin�1 x D x C 1

2

x3

3C 1 � 3

2 � 4

x5

5C � � � D x C 16

kD1

1 � 3 � 5 � � � .2k � 1/2 � 4 � 6 � � � .2k/ x2kC1

2k C 1for � 1 � x � 1

Power series are not only good for computing the values of functions, but they can be used to expand otherfunctions into series for limit, differentiation and integration purposes. Here we will practice expanding functions intoseries of functions first.

Examples. (1) For x 2 .0;1/; x x D ex ln x D 16kD0

.x ln x/kk!

:(2) For x 2 .�1; 1/;we have�1 < �x2 < 1 and so

1p1� x2

D .1 C .�x2//�1=2 D 1C 1

2x2 C 3

8x4 C 5

16x6 C � � � :

(3) For x > 0; we have 0 < e�x < 1: Using the geometric series formulaa

1� rD a C ar C ar2 C ar3 C � � � with

a D x3e�x; r D e�x; we havex3

ex � 1D x3e�x

1� e�xD x3e�x C x3e�2x C x3e�3x C � � � D 16

kD1x3e�kx :

Uniform Convergence

Power series for sine and cosine allowed us to compute values of these functions. As another application of power

series, we will find the sum of certain convergent series of numbers. Consider the series 1� 12C 1

3� 1

4C � � � ;which

converges by the alternating series test. Here is a “clever” way to find the sum. Let f .x/ D x � x2

2C x3

3� x4

4C � � �:

(We can check that this power series converges pointwise on .�1; 1]:) The sum of the alternating series above is f .1/:To find this value, we write f .1/ D f .1/ � f .0/ D Z 1

0f 0.x/ dx : Now, one may try to differentiate term-by-term to

get f 0.x/ D 1� x C x2 � x3 C x4 � � � � D 1

1C x: So f .1/ D Z 1

0f 0.x/ dx D Z 1

0

1

1C xdx D ln 2:

84

Let’s look at the argument more carefully. First, we see that every term of f .x/ is differentiable, but does thisimply f .x/ is differentiable? Next suppose f .x/ is differentable. Can we simply differentiate term-by-term? Note

f .x/ D 16kD1

.�1/k�1xk

kand f 0.x/ D d

dx

16kD1

.�1/k�1xk

k; but the assumed derivative is

1� x C x2 � x3 � � � � D 16kD1.�1/k�1xk�1 D 16

kD1

d

dx

� .�1/k�1xk

k

�:If f .x/ has finitely many terms, term-by-term differentiation is justified by mathematical induction. However, forinfinitely many terms, term-by-term differentiation needs to be justified.

Questions. (1) If Sn.x/ converges pointwise on an interval to S.x/ and each Sn.x/ is continuous (or differentiable), isit necessary that S.x/ will also be continuous (or differentiable)?

(2) If gk.x/ are continuous, differentiable, integrable on an interval [a; b] for k D 1; 2; 3; : : :, then

limx!x0

n6kD1

gk.x/ D n6kD1

limx!x0

gk.x/; d

dx

n6kD1

gk.x/ D n6kD1

d

dxgk.x/; Z b

a

n6kD1

gk.x/ dx D n6kD1

Z b

agk.x/ dx

for every positive integer n: If n is replaced by the symbol 1; will the equations still be true always?

Examples.(1) Continuous functions Sn.x/ D xn converge pointwise on [0; 1] to S.x/ D n0 if 0 � x < 11 if x D 1

. However,

S.x/ is not continuous on [0; 1]: This gives a negative answer to the first question above.

Let g1.x/ D S1.x/ and gk.x/ D Sk.x/ � Sk�1.x/ for k > 1: Then

limx!1

16kD1

gk.x/ D limx!1

limn!1 Sn.x/ D lim

x!1S.x/ D 0; but

16kD1

limx!1

gk.x/ D limn!1 lim

x!1Sn.x/ D lim

n!1 1 D 1:This gives a negative answer to the second question above.

(2) Consider the function S.x/ D 16kD0

x2.1C x2/k D x2 C x2

1C x2C x2.1 C x2/2 C : : : : Now S.0/ D 0: For x 6D 0,

S.x/ D x2

�1C 1

1C x2C 1.1C x2/2 C 1.1 C x2/3 C : : :� D x2 1

1� 11Cx2

D 1C x2:Thus S.x/ D 16

kD0

x2.1 C x2/k converges pointwise on R to�

0 if x D 01C x2 if x 6D 0

. Although every term gk.x/ Dx2.1C x2/k is differentiable on Rwith g0k.x/ D 2x � 2.k � 1/x3.1C x2/k�1

so that the partial sum Sn.x/ D n6kD0

gk.x/ is

differentiable, but S.x/ D limn!1 Sn.x/ is not differentiable at x D 0: This gives a negative answer to the first

question above.

Note g0k.0/ D 0 and S0n.0/ D 0: So at x D 0;d

dx

16kD0

gk.x/ D d

dxlim

n!1 Sn.x/ D d

dxS.x/ does not exist, but

16kD0

d

dxgk.x/ D lim

n!1 d

dxSn.x/ D 0:

This gives a negative answer to the second question above.

(3) Since Q\ [0; 1] is countable, its elements can be arranged into a sequence r1; r2; r3; : : : ; where each element

appears exactly one time. The functions Sn.x/ D n1 if x D r1; r2; : : : ; rn

0 otherwiseare integrable on [0; 1] and converge

85

pointwise on [0; 1] to S.x/ D n1 if x 2 Q\ [0; 1]0 otherwise

;which is not integrable on [0; 1]:Again this gives a negative

answer to the first question above.

ny = Sn(x)

0 1/ 2/n n 1

Next, the continuous (hence integrable) functions

Sn.x/ D 8<: n2x if 0 � x < 1=n�n2x C 2n if 1=n � x < 2=n0 if 2=n � x � 1

converge pointwise on [0; 1] to S.x/ D 0 because limn!1 Sn.0/ D 0 and

for x 2 .0; 1]; Sn.x/ D 0 when x > 2n, n > 2

xso that lim

n!1 Sn.x/ D 0: Define g1.x/ D S1.x/ and gk.x/ DSk.x/ � Sk�1.x/ for k > 1; we have16kD1

Z 1

0gk.x/ dx D lim

n!1 Z 1

0Sn.x/ dx D lim

n!1 1 D 1; butZ 1

0

16kD1

gk.x/ dx D Z 1

0lim

n!1 Sn.x/ dx D Z 1

00 dx D 0:

Again this give a negative answer to the second question above.

Now recall that continuity, differentiability and integration are concepts involving limits:

g.x0/ D limx!x0

g.x/; g0.x0/ D limh!0

g.x0 C h/� g.x0/h

; Z b

ag.x/ dx D lim

n!1 n6kD1

g

�a C b � a

nk

�b � a

n:

In general, for S.x/ D limn!1 Sn.x/ or g.x/ D 16

kD1gk.x/, as the above examples showed, we have

limx!x0

limn!1 Sn.x/ 6D lim

n!1 limx!x0

Sn.x/; limx!x0

16kD1

gk.x/ 6D 16kD1

limx!x0

gk.x/Id

dxlim

n!1 Sn.x/ 6D limn!1 d

dxSn.x/; d

dx

16kD1

gk.x/ 6D 16kD1

d

dxgk.x/IZ b

alim

n!1 Sn.x/ dx 6D limn!1 Z b

aSn.x/ dx ; Z b

a

16kD1

gk.x/ dx 6D 16kD1

Z b

agk.x/ dx :

Question. Are there special conditions which ensure an interchange of limit operations is correct?

Notations. For a function f : E ! R; the quantity supfj f .x/j : x 2 Eg is often denoted by k f kE or k f k1 and iscalled the sup-norm of f on E :

S(x) + ε

S(x) − ε

S (x)

Sn(x)( )

E

Definitions. A sequence of functions Sn: E ! Rconverges uniformlyon E to a function S: E ! R iff

limn!1 kSn � SkE D lim

n!1.supfjSn.x/ � S.x/j : x 2 Eg/ D 0

(i.e. for every " > 0, there exists K 2 N (K depends only on ") suchthat n � K ) jSn.x/ � S.x/j < " for every x 2 E :)

A series16

kD1gk.x/ converges uniformly on E to S.x/ iff its partial

sum sequence Sn.x/ D n6kD1

gk.x/ converges uniformly on E to S.x/.Theorem. If Sn.x/ converges uniformly on E to S.x/, then Sn.x/ converges pointwise on E to S.x/. Similarly, if16kD1

gk.x/ converges uniformly on E; then16

kD1gk.x/ converges pointwise on E :

86

Proof. For all x 2 E; jSn.x/� S.x/j � kSn� SkE : So limn!1 kSn � SkE D 0 implies lim

n!1 jSn.x/ � S.x/j D 0 for every

x 2 E; which implies limn!1 Sn.x/ D S.x/:

Uniform convergence is a sufficient condition that allows interchange of operations, as will be seen in the followingtheorems. The proofs will be given in Appendix 4.

Continuity Theorem for Uniform Convergence. If every Sn.x/ is continuousat c 2 E and Sn.x/ converges uniformlyon E to S.x/; then S.x/ is continuous at c; i.e.

limx!c

limn!1 Sn.x/ D lim

x!cS.x/ D S.c/ D lim

n!1 Sn.c/ D limn!1 lim

x!cSn.x/:

Similarly, if every gk.x/ is continuous at c 2 E and16

kD1gk.x/ converges uniformly on E; then

16kD1

gk.x/ is continuous

at c; i.e.

limx!c

16kD1

gk.x/ D 16kD1

gk.c/ D 16kD1

limx!c

gk.x/:Integration Theorem for Uniform Convergence. If every Sn.x/ is integrable on a bounded interval [a; b] and Sn.x/converges uniformly on [a; b] to S.x/; then S.x/ is also integrable on [a; b] andZ b

a

�lim

n!1 Sn.x/� dx D Z b

aS.x/ dx D lim

n!1 Z b

aSn.x/ dx :

Similarly, if every gk.x/ is integrable on [a; b] and16

kD1gk.x/ converges uniformly on [a; b]; then

16kD1

gk.x/ is integrable

on [a; b] and Z b

a

� 16kD1

gk.x/� dx D 16kD1

Z b

agk.x/ dx :

At this point we expect a similar theorem may hold for differentiation. However, this is not the case as thefollowing example shows.

Example. Consider the function Sn.x/ which equals jxj outside In D h�1

n; 1

n

iand on In;we take the graph of Sn.x/

to be the quarter circle with center at�0; 2

n

�and radius

p2

n: Then the functions Sn.x/ are differentiable on R and

supfjSn.x/ � jxjj : x 2 Rg D Sn.0/ D 2�p2

n; which has limit 0. So the functions Sn.x/ converge uniformly onR tojxj; which is not differentiable.

Even if a sequence of differentiable functions converge uniformly on R to a differentiable function, limit and

differentiation may not be interchanged! For example, the functions Sn.x/ D x

1C nx2converge uniformly on R to

S.x/ D 0 because S0n.x/ � S0.x/ D 1� nx2.1 C nx2/2 implies supfjSn.x/ � S.x/j : x 2 Rg D Sn. 1pn/ D 1

2p

n; which has

limit 0. However, at x D 0; d

dxlim

n!1 Sn.x/ D d

dx0 D 0; but at x D 0; lim

n!1 d

dxSn.x/ D lim

n!1 1 D 1:Differentiation Theorem for Uniform Convergence. Let a; b 2 Rwith a < b: If every Sn.x/ is differentiable on.a; b/; S0n.x/ converges uniformly on .a; b/ to T .x/ and Sn.x0/ converges for at least one x0 2 .a; b/; then Sn.x/converges uniformly on .a; b/ to a differentiable function S.x/ with

d

dx

�lim

n!1 Sn.x/� D d

dxS.x/ D T .x/ D lim

n!1� d

dxSn.x/� :

87

Similarly, if every gk.x/ is differentiable on .a; b/; 16kD1

g0k.x/ converges uniformly on .a; b/ and16

kD1gk.x0/ converges

for at least one x0 2 .a; b/; then16

kD1gk.x/ converges uniformly on .a; b/ to a differentiable function with

d

dx

16kD1

gk.x/ D 16kD1

d

dxgk.x/:

Remarks. For unbounded intervals, in the differentiation theorem for uniform convergence, Sn.x/ may not convergeuniformly on the interval to the pointwise limit function S.x/:To see this,consider Sn : R! Rdefined by Sn.x/ D x=n:Then S0n.x/ D 1=n converges uniformly on RD .�1;C1/ to T .x/ D 0 and Sn.0/ D 0 converges to 0. However,Sn.x/ converges only pointwise onR to S.x/ D 0; but not uniformly because kSn � Sk.�1;C1/ D C1: Neverthelessthe other conclusions, namely the pointwise limit function S.x/ exists, is differentiable and differentiation may bedone term-by-term at every point on the interval are true. This is because for every w on the unbounded interval wecan restrict the unbounded interval to an open bounded interval containingw and x0 to get these conclusions.

Because of the above theorems, in order to interchange operations, we now need to know how to check uniformconvergence for sequence and series of functions. For sequences of functions, we will introduce the L-test below, andfor series, the M-test.

Theorem (L-test). If Sn.x/ converges pointwise on E to S.x/ and there are constants Ln such that jSn.x/�S.x/j � Ln

for every x 2 E and limn!1 Ln D 0; then Sn.x/ converges uniformly on E to S.x/:

Proof. Since jSn.x/ � S.x/j � Ln for every x 2 E; so kSn � SkE D supfjSn.x/ � S.x/j : x 2 Eg � Ln : Sincelim

n!1 Ln D 0; by sandwich theorem, limn!1 kSn � SkE D 0:

Examples. (1) Show Sn.x/ D sin nxpn

converges uniformly onR.

Solution. Note that the pointwise limit function on R is S.x/ D limn!1 sin nxp

nD 0. For every x 2 R;jSn.x/ � S.x/j D jSn.x/j � 1p

nand lim

n!1 1pnD 0: By L-test, Sn.x/ converges uniformly onR to S.x/:

(2) Show Sn.x/ D e� cos2.1=x/=n converges uniformly on .0; 1/:Solution. The pointwise limit function on .0; 1/ is S.x/ D lim

n!1 e� cos2.1=x/=n D 1: For w � 0; by the

mean value theorem, je�w � 1j D j.�e�c/.w � 0/j � jwj for some c between w and 0: For every x 2 .0; 1/;jSn.x/ � S.x/j D je� cos2.1=x/=n � 1j � ���cos2.1=x/n

��� � Ln D 1

n: Since lim

n!1 Ln D 0; by the L-test, Sn.x/ con-

verges uniformly on .0; 1/:(3) Show Sn.x/ D xn converges uniformly on [0; t ] for t < 1, but Sn.x/ does not converge uniformly on [0; 1].

(1, 1)

]0 1

Solution. The pointwise limit function on [0; t ] is S.x/ D 0. For x 2 [0; t ];jSn.x/ � S.x/j � Ln D t n: Now limn!1 Ln D lim

n!1 t n D 0 because t < 1: By the

L-test, Sn.x/ converges uniformly on [0; t ]:Next suppose Sn.x/ converges uniformly on [0; 1]: Since every Sn.x/ is con-

tinuous on [0; 1]; by the continuity theorem for uniform convergence, the limitfunction S.x/ will also be continuous on [0; 1]: However the pointwise limit func-

tion is S.x/ D n0 if x 6D 11 if x D 1

; which is not continuous at x D 1; a contradiction.

So Sn.x/ does not converge uniformly on [0; 1]:88

Theorem (Weierstrass M-test). If for k D 1, 2, 3, : : :, there are constants Mk such that jgk.x/j � Mk for every x 2 E

and16

kD1Mk converges, then

16kD1

gk.x/ converges uniformly on E.

Proof. The n-th partial sum is Sn.x/ D n6kD1

gk.x/ and the sum is S.x/ D 16kD1

gk.x/; which converges absolutely for

every x 2 E by comparing with16

kD1Mk: We have jSn.x/ � S.x/j � 16

kDnC1jgk.x/j � 16

kDnC1Mk D Ln for every x 2 E :

Since16

kD1Mk converges, so lim

n!1 Ln D limn!1� 16kD1

Mk � n6kD1

Mk� D 0: By L-test,

16kD1

gk.x/ converges uniformly on E :Examples. (1)

16kD1

sin kx

k2converges uniformly onR since for all x 2 R; ���� sin kx

k2

���� � 1k2D Mk and

16kD1

1k2

converges.

(2) Show16

kD1

�ln x

x

�k

converges uniformly on [1;1/.Solution. By calculus, max

x2[1;1/ ���� ln x

x

���� D maxx2[1;1/ ln x

xD 1

e. So

������ ln x

x

�k����� � �

1e

�k D Mk for every x 2 [1;1/.Since

16kD1

Mk converges by the geometric series test, Weierstrass M-test gives us the result.

(3) (Here is another application of power series!) FindZ 1

0x x dx to five decimal places.

Solution. For x 2 .0; 1]; x x D ex ln x : (As x ! 0C; x ln x D .ln x/= 1x ! 0 by l’Hopital’s rule. So at x D 0;we set

x ln x D 0 and x x D 1 to be continuous. Then x x is integrable on [0; 1]:) Since ew D 16kD0

wk

k!for every w 2 R;we

have x x D ex ln x D 16kD0

.x ln x/kk!

: By calculus, maxx2[0;1]

jx ln xj D maxx2[0;1]

.�x ln x/ D 1

e: So

���� .x ln x/kk!

���� � Mk D 1

k!ek

and16

kD0Mk D 16

kD0

1

k!ekD e1=e: By Weierstrass M-test,

16kD0

.x ln x/kk!

converges uniformly on [0; 1] to x x : By the

integration theorem for uniform convergence,Z 1

0x x dx D Z 1

0

16kD0

.x ln x/kk!

dx D 16kD0

Z 1

0

.x ln x/kk!

dx D 16kD0

.�1/kk!.k C 1/kC1

Z 10

uke�u du| {z }Dk!

D 16kD0

.�1/k.k C 1/kC1;

where we substituted u D �.k C 1/ ln x and integrated by parts k times in the computations. Taking 10 terms, we

findZ 1

0x x dx � 0:78343:

(4) In this example, we will show the integration theorem for uniform convergence may not hold for improperRiemann integrals. Let

Sn.x/ D (0 if x 2 [0; n/[ [3n;1/.x � n/=n2 if x 2 [n; 2n/�.x � 3n/=n2 if x 2 [2n; 3n/ :The graph of Sn.x/ on [0;1/ is consisted of the segments from .0; 0/ to .n; 0/, from .n; 0/ to .2n; 1=n/; from.2n; 1=n/ to .3n; 0/ and the part of the positive x-axis starting at .3n; 0/: Since 0 � Sn.x/ � 1

nfor every x � 0;

so the pointwise limit of Sn.x/ is S.x/ D limn!1 Sn.x/ D 0: Since lim

n!1 kSn � Sk[0;1/ D limn!1 1

nD 0; by definition

Sn.x/ converges uniformly on [0;1/ to S.x/ D 0: NowZ 10

limn!1 Sn.x/ dx D Z 1

0S.x/ dx D 0; but lim

n!1 Z 10

Sn.x/ dx D limn!1 1 D 1:

89

(5) FindZ 1

0

x3

ex � 1dx to five decimal places. (The exact answer is known to be

�4

15:)

Solution. (Note as x ! 0C; the integrand tends to 0. So at 0, the integrand is set to 0 to be continuous.)

For x 2 .0;1/; 0 < e�x < 1: Sox3

ex � 1D x3e�x

1� e�xD x3e�x C x3e�2x C x3e�3x C � � � D 16

kD1x3e�kx for all

x 2 .0;1/. We have Z 10

x3

ex � 1dx D limw!1Z w

0

x3

ex � 1dx D limw!1 Z w

0

16kD1

x3e�kx dx :By calculus, for every x 2 [0; w]; jx3e�kx j � �3

k

�3

e�. 3k /k D 27

e3k3D Mk: Since

16kD1

Mk D 27

e3

16kD1

1

k3converges

by p-test, Weierstrass M-test implies16

kD1x2e�kx converges uniformly on [0; w]:Applying the integration theorem

for uniform convergence, we should getZ 10

x3

ex � 1dx D limw!1 Z w

0

16kD1

x3e�kx dx D limw!1 16kD1

Z w0

x3e�kx dx :Next substitute w D 1=t for t > 0; and define gk.t/ D Z 1=t

0x3e�kx dx : The value gk.0/ is defined to be

limt!0C gk.t/ D Z 1

0x3e�kx dx D 6

k4; obtained by integrating by parts three times. Observe

16kD1

gk.t/ converges

uniformly on [0;1/: (We check the M-test. Since x3e�kx � 0;jgk.t/j D Z 1=t

0x3e�kx dx � Z 1

0x3e�kx dx D 6

k4for every t 2 [0;1/:

Finally16

kD1

6

k4converges by p-test.) By the continuity theorem for uniform convergence,

Z 10

x3

ex � 1dx equals

limw!1 16kD1

Z w0

x3e�kx dx D limt!0C 16

kD1gk.t/ D 16

kD1lim

t!0C gk.t/ D 16kD1

Z 10

x3e�kx dx D 16kD1

6k4D �4

15� 6:49394:

Other than the L-test and the M-test, there are a few useful facts for checking uniform convergence.

Facts. (1) (Subset Property) If Sn.x/ converges uniformly on A to S.x/ and E � A; then Sn.x/ converges uniformlyon E to S.x/ because 0 � kSn � SkE � kSn � SkA ! 0:(2) (Union Property) If Sn.x/ converges uniformly on A and B to S.x/; then Sn.x/ converges uniformly on A [ B toS.x/ because 0 � kSn � SkA[B � kSn � SkA C kSn � SkB ! 0:(3) (Bounded Multiplier Property) If Sn.x/ converges uniformly on E to S.x/ and g.x/ is a bounded function on E;then g.x/Sn.x/ converges uniformly on E to g.x/S.x/ because 0 � kgSn � gSkE � kgkEkSn � SkE ! 0:(4) (Substitution Property) If Sn.x/ converges uniformly on E to S.x/ and h : E0 ! E; then Sn

�h.x/� converges

uniformly on E0 to S�h.x/� because E 0 D fh.x/ : x 2 E0g � E implies 0 � kSn � h � S � hkE0 D kSn � SkE0 �kSn � SkE ! 0: Briefly, this means we can do substitution for uniform convergent sequences and series and still

remain uniformly convergent.

Question. Are there special cases where pointwise convergence of nice functions implies uniform convergence?

For closed and bounded intervals, the following two theorems provide such special situations.

90

Dini’s Theorem. If continuous functions Sn converge pointwise on a closed and bounded interval [a; b] to a continuousfunction S and for every x 2 [a; b]; Sn.x/ is an increasing sequence of real numbers, then the convergence is uniformon [a; b]:

For series, if each term gk.x/ is nonnegative, continuous on [a; b] and16

kD1gk.x/ converges pointwise on [a; b]

to a continuous function S.x/; then the convergence is uniform on [a; b]: The statement is also true if increasing isreplaced by decreasing or nonnegative is replaced by nonpositive.

Proof. The continuous function fn D S � Sn converges pointwise on [a; b] to 0 and for every x 2 [a; b]; fn.x/decreases to 0. Since kSn � Sk[a;b] D k fnk[a;b]; it suffices to show fn converges uniformly on [a; b] to 0. Supposethis is false, then there is " > 0 such that for every K ; there is n � K and xn 2 [a; b] with j fn.xn/j � ": ByBolzano-Weierstrass, xn has a subsequence xnk converging to some w 2 [a; b]:

Now for every positive integer n; consider those nk � n: Then fn.xnk / � fnk .xnk / D j fnk .xnk /j � ": Since fn iscontinuous and xnk ! w; by the sequential continuity theorem, fn.w/ D lim

k!1 fn.xnk / � ": Then 0 D limn!1 fn.w/ � ";

a contradiction to fn.w/ converging to 0 as n tends to infinity.

For the second statement, let Sn D g1 C � � � C gn and apply the first statement. For the third statement, multiplyevery function by �1 and apply the first two statements.

Remarks. Dini’s Theorem is false on intervals without an endpoint. For example, Sn.x/ D �xn converges pointwiseon [0; 1/ to S.x/ D 0 and for every x 2 [0; 1/; Sn.x/ is increasing, but lim

n!1 kSn � Sk[0;1/ D 1 so that the convergence

is not uniform on [0; 1/:Abel’s Limit Theorem. If a power series

16kD0

ak.x � c/k converges pointwise on a closed and bounded interval [u; v]

to S.x/, then16

kD0ak.x�c/k converges uniformly on [u; v] to S.x/. (By the continuity theorem for uniform convergence,

S.x/ is continuous on [u; v]: Hence,16

kD0ak.u � c/k D S.u/ D lim

x!uC S.x/ and16

kD0ak.v � c/k D S.v/ D lim

x!v� S.x/:/Remarks. Abel’s Limit Theorem is false on intervals without an endpoint. For example,

16kD0

xk converges pointwise

on [0; 1/ to S.x/ D 11� x

; but kSn � Sk[0;1/ D supf��� xnC1

1� x

��� : x 2 [0; 1/g D 1 for every n so that the convergence is

not uniform on [0; 1/:For a proof of Abel’s limit theorem, please see Appendix 4. This theorem has a useful consequence in differenti-

ating power series.

Theorem (Differentiation of Power Series). If a power series16

kD0ak.x � c/k converges pointwise on .c� R; cC R/

to S.x/ where the radius of convergence R > 0; then S.x/ is differentiable there and S0.x/ D 16kD1

kak .x � c/k�1 for

x 2 .c � R; c C R/.Proof. The radius of convergence of S.x/ D 16

kD0ak.x � c/k is R D 1

limsupk!1 k

pjakj by the domain theorem for power

series. For the derived series16

kD1kak .x � c/k�1 , its radius of convergence is

1

limsupk!1 k

pjkak j D 1

limsupk!1 k

pjak j D R; since kp

k ! 1 as k !1:So both series converges pointwise on the interval .c � R; c C R/. For x 2 .c � R; c C R/; take " > 0 so that[x � "; x C "] � .c � R; c C R/: Abel’s limit theorem implies both series converge uniformly on [x � "; x C "] (andhence also on .x � "; x C "/ by the subset property). Applying the differentiation theorem for uniform convergence

on .x � "; x C "/; we get S0.x/ D 16kD1

kak.x � c/k�1:91

Examples. (1) (Here is yet another application of power series!) Find L D limx!0

x2ex5 � sin.x2/x6 � ex7 C 1

:Solution. Recall ew D 1C w C w2

2!C � � � and sinw D w � w3

3!C � � � for �1 < w < C1: Hence,

L D limx!0

x2.1C x5 C � � �/ � .x2 � x6

3! C � � �/x6 � .1C x7 C � � �/ C 1

D limx!0

16 x6 C x7 C � � �x6 � x7 C � � � D lim

x!0

16 C x C � � �1� x C � � � :

Since1

6C x C � � � D ��

x2ex5 � sin.x2/�=x6 if x 6D 01=6 if x D 0

and 1� x C � � � D � .x6 � ex7 C 1/=x6 if x 6D 01 if x D 0

;both power series converge pointwise on [�1; 1]; hence both are continuous at 0 by Abel’s limit theorem. So

limx!0

.16C x C � � �/ D 1

6and lim

x!0.1 � x C � � �/ D 1: Therefore, L D 1

6:

(2) Show 1� 12C 1

3� 1

4C � � � D 16

kD1

.�1/k�1

kD ln 2 and justify the steps clearly.

Solution. Step 1. The series converges by the alternating series test since 1 � 12� 1

3� � � � and lim

k!1 1kD 0:

Step 2. Define f .x/ D 16kD1

.�1/k�1xk

k: Then f .1/ converges by step 1. We have

limk!1 ��� .�1/k xkC1

k C 1

k.�1/k�1xk

��� D limk!1 kjxj

k C 1D jxj < 1 if and only if � 1 < x < 1:

By the ratio test and step 1, we find that the domain of f is .�1; 1]: Applying differentiation of power series on.�1; 1/;f 0.x/ D 16

kD1.�1/k�1xk�1 D 1� x C x2 � x3 C � � � D 1

1C x) f .x/ D f .0/ C Z x

0f 0.t/ dt D ln.1 C x/:

So f .x/ D x � x2

2C x3

3� x4

4C � � � D �

ln.1C x/ if �1 < x < 11� 1

2 C 13 � 1

4 C : : : if x D 1.

Step 3. Using Abel’s limit theorem on [0; 1]; 1� 12 C 1

3 � 14 C : : : D f .1/ D lim

x!1� f .x/ D limx!1� ln.1 C x/ D ln 2:

Remarks. (1) The series converges too slowly to ln 2. (For example, by the hundredth term, which is 1100 D 0:01, it

still affects the second decimal place of ln 2.) To get ln 2, it is better that we consider for x 2 .�1; 1/;ln�1C x

1� x

� D ln.1Cx/�ln.1�x/ D �x� x2

2C x3

3� x4

4C� � �����x� x2

2� x3

3� x4

4�� � �� D 2

�xC x3

3C x5

5C x7

7C� � ��

and set x D 13 to get ln 2 D ln

�1C .1=3/1� .1=3/� D 2

�13C .1=3/3

3C .1=3/5

5C .1=3/7

7C � � �� D 0:6931 : : : ; which is cor-

rect to four decimal places already by the fourth term.

(2) Differentiation of power series may be false if .c � R; c C R/ is replaced by .c � R; c C R]; [c � R; c C R/ or

[c � R; c C R]: We saw16

kD1

.�1/k�1xk

kD x � x2

2C x3

3� � � � converges pointwise on .�1; 1] to f .x/ D ln.1 C x/;

but the derived series16

kD1.�1/k�1xk�1 D 1� x C x2 � � � � converges pointwise only on .�1; 1/ and diverges at x D 1

even though f 0.1/ D 1

2exist.

92

(3) Example (2) above showed that16

kD1

.�1/k�1xk

kconverges pointwise on [0; 1] to ln.1 C x/: So by Abel’s limit

theorem, the convergence is uniform on [0; 1]: However, if you try to check this by Weierstrass’ M-test, then it will

lead to Mk D 1

k; but

16kD1

Mk D 16kD1

1

kdiverges!

(3) Find the value of 1� 12 � 1

3 C 14 C 1

5 � 16 � 1

7 C : : :.Solution. Step 1. The given series 1� 1

2� 1

3C 1

4C 1

5� 1

6� 1

7C : : : D �1� 1

2

�� �1

3� 1

4

�C �1

5� 1

6

�� : : :D 16kD1.�1/kC1

�1

2k � 1� 1

2k

� D 16kD1

.�1/kC1.2k � 1/.2k/ converges by the alternating series test.

Step 2. Define f .x/ D 16kD1

.�1/kC1x2k.2k � 1/.2k/ : Then f .1/ D f .�1/ converges by step 1. We have

limk!1��� .�1/kC2x2kC2.2k C 1/.2k C 2/ .2k � 1/.2k/.�1/kC1x2k

��� D limk!1 .2k � 1/.2k/x2.2k C 1/.2k C 2/ D x2 < 1 if and only if � 1 < x < 1:

By ratio test and step 1, it has domain [�1; 1]: Applying differentiation of power series twice on .�1; 1/;f 0.x/ D 16

kD1

.�1/kC1x2k�1

2k � 1D x � x3

3C x5

5� : : : and f 00.x/ D 1� x2 C x4 � : : : :

(Note the power series for f 0.x/ also converge at x D 1 by the alternating series test.) So for �1 < x < 1, we have

f 00.x/ D 1

1C x2) f 0.x/ D f 0.0/ C Z x

0f 00.t/ dt D tan�1 x) f .x/ D f .0/ C Z x

0f 0.t/ dt D Z x

0tan�1 t dt D x tan�1 x � 1

2ln.1C x2/:

Step 3. Applying Abel’s limit theorem on [0; 1];we get

1� 1

2� 1

3C 1

4C 1

5� 1

6� 1

7C : : : D f .1/ D lim

x!1� f .x/ D �4� 1

2ln 2

1� 1

3C 1

5� 1

7C 1

9� : : : D f 0.1/ D lim

x!1� f 0.x/ D limx!1� tan�1 x D tan�1 1 D �

4:

Remarks. This series converges too slowly. To compute� , one uses the relation tan.�C�/ D tan� C tan �1� tan� tan � : Taking

x D tan�; y D tan �; we get tan�1 x C tan�1 y D � C � D tan�1

�tan� C tan �

1� tan� tan �� D tan�1

�x C y

1� x y

� : Then�4D tan�1 1 D tan�1 1

2C tan�1 1

3D 2 tan�1 1

3C tan�1 1

7D : : : D 12 tan�1 1

18C 8 tan�1 1

57� 5 tan�1 1

239:

So using tan�1 x D f 0.x/ D x � x3

3C x5

5� � � � for �1 < x < 1 from above, we get� D 48 tan�1 1

18C 32 tan�1 1

57� 20 tan�1 1

239D 48

�118� 1

3 � 183C 1

5 � 185� : : :� C 32

�157� 1

3 � 573C 1

5 � 575� : : :� � 20

�1

239� 1

3 � 2393C : : :�D 3:1415926 : : : ;

93

which is correct to seven decimal places with these eight terms.

Appendix 1: Space-Filling Curves

Do all curves have zero area? Our intuition will think so, but our intuition is wrong. In 1890, Peano showedthere is a continuous curve in R2 that passes through every point of the unit square [0; 1] � [0; 1]: We will presentI. J. Schoenberg’s 1938 example of such a curve, which is easier to understand.

On [0; 2]; define g.t/ D 8><>: 0 if t 2 [0; 1=3][ [5=3; 2]3t � 1 if t 2 [1=3; 2=3]1 if t 2 [2=3; 4=3]�3t C 5 if t 2 [4=3; 5=3]

: Extend g to Rby g.t C 2/ D g.t/: Next define

f .t/ D �x.t/; y.t/�; where x.t/ D 16

nD1

g.32n�2t/2n

; y.t/ D 16nD1

g.32n�1t/2n

: Since g is continuous onR; ��� g.3 j t/2n

��� � 12n

onRand16

nD1

1

2nD 1; using the Weierstrass M-test and the continuity theorem, we see that x.t/ and y.t/ are continuous

onR: We will show S D f f .t/ : t 2 [0; 1]g is the unit square [0; 1]� [0; 1]:Since 0 � x.t/; y.t/ � 1; S � [0; 1] � [0; 1]: Next, for .a; b/ 2 [0; 1] � [0; 1]; write a and b in base 2 as

a D .0:a1a2a3 � � �/2 D 16nD1

an

2nand b D .0:b1b2b3 � � �/3 D 16

nD1

bn

2n; where an; bn D 0 or 1. Let c D 2.0:a1b1a2b2 � � �/3D 2

16nD1

cn

3n; where c2m�1 D am and c2m D bm: Since 2

16nD1

13nD 1; so c 2 [0; 1]:We will show f .c/ D �

x.c/; y.c/� D.a; b/:We will show g.3m�1c/ D cm for m D 1; 2; 3; : : : : (Then g.32n�2c/ D c2n�1 D an; g.32n�1c/ D c2n D bn and

we will get x.c/ D a; y.c/ D b:) Now 3m�1c D 2.c1c2 � � � cm�1:cm cmC1cmC2 � � �/3 D 2m�16kD1

ck3m�1�k C 216

kDm

ck

3k�mC1:

Note the first term is an even integer and the second term dm is in [0; 1]; since 216

kDm

1

3k�mC1D 1: Further,

dm D 216

kDm

ck

3k�mC1

�� 261kDmC1 1=3k�mC1 D 1=3 if cm D 0� 2=3 if cm D 1

:So if cm D 0; then dm 2 [0; 1=3] and if cm D 1; then dm 2 [2=3; 1]: Since g.t C N/ D g.t/ for any even integer N ; so

g.3m�1c/ D g.dm/ D �0 if cm D 01 if cm D 1

D cm :Appendix 2: Everywhere Continuous, But Nowhere Differentiable Functions

Must every continuous function be differentiable at some place? Our intuition will think so, but our intuition iswrong again. In 1872, Weierstrass proved the following result.

Theorem. There is a function continuous onRwhich is not differentiable at every x 2 R:Proof. On [0; 1]; define f0.x/ D �

x if 0 � x < 1=21� x if 1=2 � x < 1

: Extend f0 toRby f0.x C 1/ D f0.x/: Then f0 has roots

only at the integers. Let fk .x/ D f0.2k x/2k

and f .x/ D 16kD0

fk .x/: Since each fk is continuous on R; j fk.x/j � 12k

onRand16

kD0

1

2kconverges, by Weierstrass M-test and the continuity theorem, f is a continuous function onR:

Next we will show f is not differentiable onR: Since f; like f0; is also periodic with period 1, it is enough to showf is not differentiable at x 2 [0; 1/:Note that f 0k .x/ D �1 if 2kC1x is not an integer. Suppose f is differentiable at some

94

x 2 [0; 1/: If x D .0:d1d2d3 : : :/2; then let an D .0:d1d2 : : : dn/2 and bn D .0:d1d2 : : : dn111 : : :/2: So x 2 [an; bn/ andan increases to x while bn decreases to x : As n !1;

f .bn/ � f .an/bn � an

� f 0.x/ D � f .bn/ � f .x/bn � an

C f .x/ � f .an/bn � an

�� f 0.x/� bn � x

bn � anC x � an

bn � an

�D f .bn/� f .x/bn � an

� f 0.x/ bn � x

bn � anC f .x/ � f .an/

bn � an� f 0.x/ x � an

bn � anD � f .bn/ � f .x/bn � x

� f 0.x/| {z }! f 0.x/� f 0.x/D0

�� bn � x

bn � an| {z }�1

�C � f .x/ � f .an/x � an

� f 0.x/| {z }! f 0.x/� f 0.x/D0

�� x � an

bn � an| {z }�1

�! 0:So f 0.x/ D lim

n!1 f .bn/� f .an/bn � an

: For k D 0; 1; : : : ; n � 1; we have k C 1 � n: So the interval .2kC1an; 2kC1bn/ ��.d1d2 : : : dkC1/2; .d1d2 : : : dkC1:111 : : :/2�; which contains no integer. Hence, f 0k .x/ D �1 for x 2 .an; bn/: By the

mean value theorem, there is cn 2 .an ; bn/ such thatfk .bn/ � fk .an/

bn � anD f 0k .cn/ D �1: For k D n; nC 1; : : : ; 2kbn and

2kan are integers and so fk .bn/ D 0 D fk .an/: Then

f 0.x/ D limn!1 f .bn/ � f .an/

bn � anD lim

n!1 n�16kD0

fk .bn/ � fk .an/bn � an

D 16kD0�1;

which diverges by term test and gives a contradiction to f differentiable at x :Appendix 3: Weierstrass Approximation Theorem

For an infinitely differentiable function, such as cos x; we can compute the values of the function using its Taylorseries most of the time. However, if the function is only continuous, but not differentiable, then there is no Taylorseries. The following theorem asserts that for every continuous function f .x/ on a closed and bounded interval, thereexists a sequence of polynomials Pn.x/ converging uniformly on the interval to f .x/:Weierstrass Approximation Theorem. Let f : [0; 1] ! Rbe continuous. Forevery " > 0, there is a polynomial P.x/ such that j f .x/ � P.x/j < " for allx 2 [0; 1]. Taking " D 1=n;we get a sequence of polynomials Pn.x/ converginguniformly on [0; 1] to f .x/:

The theorem is also true for every closed and bounded interval [a; b] bysubstituting x D .t � a/=.b� a/ and compact subsets ofR(to be defined later).(Note for " D 10�12, P.x/ will agree with f .x/ to at least 10 decimal places forall x 2 [0; 1].)

f - ε

f +ε

0 1

Remarks. Combining the continuous, nowhere differentiable function with the Weierstrass approximation theorem,we can see that there is a sequence of polynomials converging uniformly on [0; 1] to a nowhere differentiable function.So the uniform limit of continuous functions can be very bad in terms of differentiabilty!

To prove the Weierstrass approximation theorem, we will introduce the Bernstein polynomials of a continuousfunction.

Definition. For each n 2 N, the n-th Bernstein polynomial of f .x/ is

fn.x/ D n6kD0

f� k

n

��n

k

�| {z }number

xk.1� x/n�k| {z }polynomial

; deg fn � n:95

Examples. Find the n-th Bernstein polynomial of f .x/ D 1; g.x/ D x and h.x/ D x2 respectively.

Solution. The binomial theorem asserts that .x C a/n D n6kD0

�n

k

�xkan�k : Differentiating x; then multiplying by

x

n;

and repeating these operations, we get .x C a/n�1x D n6kD0

k

n

�n

k

�xkan�k and

[.n � 1/.x C a/n�2x C .x C a/n�1]x

nD n6

kD0

k2

n2

�n

k

�xkan�k :

Setting a D 1� x; we get fn.x/ D n6kD0

�n

k

�xk.1 � x/n�k D 1; gn.x/ D n6

kD0

k

n

�n

k

�xk.1� x/n�k D x and

hn.x/ D n6kD0

k2

n2

�n

k

�xk.1 � x/n�k D .n � 1/x2 C x

n:

As n !1; we get fn.x/! f .x/; gn.x/! g.x/ and hn.x/! h.x/:Formula.

n6kD0

�x � k

n

�2�

n

k

�xk.1 � x/n�k D n6

kD0

�x2 � 2x

� k

n

�C �k

n

�2��

n

k

�xk.1� x/n�kD x2 � 2x2 C .n � 1/x2 C x

nD x � x2

n:

Proof of Weierstrass Approximation Theorem. For any " > 0, since f is uniformly continuous on [0; 1], there

is � > 0 such that for every x , w 2 [0; 1], jx � wj < � ) j f .x/ � f .w/j < "2

. By the extreme value theorem,j f .x/j � M for some M . Choose integer n > M"�2: Let P.x/ be the n-th Bernstein polynomial of f .

For every x 2 [0; 1];we have 0 � .x � 12/2 D x2 � x C 1

4; which implies x � x2

.†/� 14: Next define S D n

k: k D0; 1; : : : ; n and

���x � k

n

��� < �o. Note k 62 S implies1�2.x � k

n/2.�/�1: Then for every x 2 [0; 1];j f .x/ � P.x/j D ��� f .x/ n6

kD0

�n

k

�xk.1� x/n�k| {z }D1 by examples

� n6kD0

f . k

n/�n

k

�xk.1 � x/n�k

���D ���� n6kD0

�f .x/ � f . k

n/��n

k

�xk.1 � x/n�k

���� � n6kD0

���� f .x/ � f . k

n/���� �n

k

�xk.1 � x/n�k� 6

k2S

���� f .x/ � f . k

n/���� �n

k

�xk.1� x/n�k C 6

k 62S

���� f .x/ � f . k

n/���� �n

k

�xk.1 � x/n�k� 6

k2S

"2

�n

k

�xk.1 � x/n�k C 6

k 62S2M

�n

k

�xk.1 � x/n�k� "

2

n6kD1

�n

k

�xk.1� x/n�k C 2M 6

k 62S

1�2.x � k

n/2| {z }

by.�/ �n

k

�xk.1 � x/n�k� "

2C 2M

x � x2

n�2| {z }by formula

� "2C M

2n�2| {z }by.†/ < "

2C "

2D ":

96

Remarks. In analysis, if you want to prove something is true for continuous functions, you should first see if it is truefor polynomials. Very often the truth of the polynomial case implies the truth for the continuous case by Weierstrassapproximation theorem.

Appendix 4 : Proofs of Theorems

Proof of the Continuity Theorem for Uniform Convergence. For every " > 0; let "0 D "=3: Since Sn.x/ convergesuniformly on E to S.x/; there exists K 2 N such that n � K implies jSn.x/�S.x/j < "0 for every x 2 E : Since SK .x/is continuous at c; so for this "0; there exists � > 0 such that for every x 2 E; jx�cj < � implies jSK .x/�SK .c/j < "0:Then for every x 2 E; jx � cj < � impliesjS.x/ � S.c/j � jS.x/ � SK .x/j C jSK .x/ � SK .c/j C jSK .c/ � S.c/j < "0 C "0 C "0 D ":The statement for series of functions follows by considering the partial sum sequence.

Proof of the Integration Theorem for Uniform Convergence. Let "n D kS � Snk[a;b]: Then jS.x/ � Sn.x/j � "n;which is equivalent to Sn.x/ � "n � S.x/ � Sn.x/ C "n for all x 2 [a; b]: SoZ b

aSn.x/ dx � "n.b � a/ D .L/ Z b

a.Sn.x/ � "n/ dx � .L/ Z b

aS.x/ dx� .U / Z b

aS.x/ dx � .U / Z b

a.Sn.x/ C "n/ dx D Z b

aSn.x/ dx C "n.b � a/: .†/

Hence, 0 � .U / Z b

aS.x/ dx � .L/ Z b

aS.x/ dx � 2"n.b � a/: Since lim

n!1 "n D 0; by sandwich theorem, we get.U / Z b

aS.x/dx D .L/ Z b

aS.x/dx; i.e. S.x/ is integrable on [a; b]:By (†),

����Z b

aSn.x/ dx � Z b

aS.x/ dx

���� � "n.b � a/:By sandwich theorem again, lim

n!1 Z b

aSn.x/ dx D Z b

aS.x/ dx D Z b

alim

n!1 Sn.x/ dx :There is a Cauchy criterion for uniform convergence similar to the Cauchy criterion for a sequence of numbers.

Theorem (Uniform Cauchy Criterion). Sn.x/ converges uniformly on E if and only if for every " > 0, there exists

K 2 N such that m, n � K.�/) jSn.x/ � Sm.x/j < " for every x 2 E :

Proof. For the ‘if’ direction, the condition (*) implies Sn.x/ is a Cauchy sequence for every x 2 E . So the sequenceSn.x/ converges to some number S.x/ for every x 2 E . For every " > 0;by (*) again, there is K 2 Nsuch that m; n � Kimplies jSm.x/ � Sn.x/j < "=2 for every x 2 E; which implies jSn.x/ � S.x/j D lim

m!1 jSn.x/ � Sm.x/j � "=2 < "for every x 2 E :

For the ‘only if’ direction, given " > 0; there exists K such that n � K ) jSn.x/� S.x/j < "=2 for every x 2 E :Then m; n � K ) jSn.x/ � Sm.x/j � jSn.x/ � S.x/j C jS.x/ � Sm.x/j < "=2C "=2 D ":Proof of the Differentiation Theorem for Uniform Convergence. For " > 0; let "0 D "

2.b � a/ : Since fSn.x0/g is a

Cauchy sequence and S0n.x/ satisfies the uniform Cauchy criterion, there is K 2 N such that m; n � K impliesjSn.x0/ � Sm.x0/j < "2

and jS0n.t/ � S0m.t/j < "0 for every t 2 .a; b/: .�/Applying the mean value theorem to Sn � Sm; we get for every x; y 2 .a; b/; there is t between x and y such thatjSn.x/ � Sm.x/ � Sn.y/ C Sm.y/j D ����S0n.t/� S0m.t/�.x � y/��� � jx � yj"0 < .b � a/"0 D "

2: .��/

97

Then m; n � K implies jSn.x/ � Sm.x/j � jSn.x/ � Sm.x/ � Sn.x0/ C Sm.x0/j C jSn.x0/ � Sm.x0/j < " for everyx 2 .a; b/: So Sn.x/ satisfies the uniform Cauchy criterion and hence converges uniformly on .a; b/ to some functionS.x/:

Next, for every y 2 .a; b/; the function gn.x/ D 8<: Sn.x/ � Sn.y/x � y

if x 2 .a; b/ n fygS0n.y/ if x D y

is continuous at y because

limx!y

gn.x/ D gn.y/: By (*) and (**), jgn.x/� gm.x/j < "0 for every x 2 .a; b/: So gn.x/ satisfies the uniform Cauchy

criterion and hence converges uniformly on .a; b/: Let g.x/ D limn!1 gn.x/ D ( S.x/ � S.y/

x � yif x 2 .a; b/ n fyg

T .y/ if x D y: By

the continuity theorem for uniform convergence, g.x/ is continuous at y: So

d

dyS.y/ D lim

x!y

S.x/ � S.y/x � y

D limx!y

g.x/ D g.y/ D T .y/:In computing an integral, the method of integration by parts allows us to switch a difficult integral to an easier

integral. Similarly, in summing a series, there is a method called summation by parts, which allows us to switch aseries to another more manageable series.

Theorem (Summation by Parts). Let Aj D j6kDi

ak D ai C aiC1 C : : :C aj and1bk D bkC1 � bk.k C 1/� kD bkC1 � bk; then

m6kDi

akbk D Am bm � m�16kDi

Ak1bk :Proof. Note ai D Ai and ak D Ak � Ak�1 for k > i. So,

m6kDi

akbk D Ai bi C .AiC1 � Ai /biC1 C : : : C .Am � Am�1/bm D Am bm � Ai .biC1 � bi/ � : : : � Am�1.bm � bm�1/:Proof of Abel’s Limit Theorem. We first prove the special case c D 0 and [u; v] D [0; 1]: Since

16kD0

ak converges to

S.1/, the partal sum sequence sk D a1 C � � � C ak is a Cauchy sequence of numbers. So for every " > 0; there is K

such that j > n � K.�/) jsj � snj D ���� j6

kDnC1ak

���� < "2

. We now show16

kD0akxk converges uniformly on [0; 1] to S.x/ by

verifying the uniform Cauchy criterion. Let Sn.x/ D a0 C a1x C : : : C anxn and Aj D j6kDnC1

ak for j > n. For every

x 2 [0; 1], m > n > K , we apply summation by parts (where we take i D n C 1 in the formula) to getjSm.x/ � Sn.x/j D ���� m6kDnC1

ak xk

���� D ����Am xm � m�16kDnC1

Ak .xkC1 � xk/���� � jAm jxm C m�16kDnC1

jAk j.xk � xkC1/by .�/� "

2

�xm C .xnC1 � xnC2/ C : : : C .xm�1 � xm/� D "

2xnC1 < ":

For the general case, depending on c < u or u � c � v or v < c; we have [u; v] is a subset of [c; v] or [u; c] [ [c; v]or [u; c]; respectively. To show uniform convergence on [u; v]; it suffices to show uniform convergence on [c; v] and[u; c] by the subset and union property. If c < v; then let r D v � c and x 0 D h.x/ D .x � c/=r; a0k D akrk : We have

a0kx 0k D ak.x � c/k : So16

kD0ak.x � c/k converges pointwise on [c; v] implies

16kD0

a0kx 0k converges pointwise on [0; 1]:Hence

16kD0

a0k x 0k converges uniformly on [0; 1] by the argument above, which implies16

kD0ak.x � c/k converges uniformly

on [c; v] by substitution property. Similarly, if u < c; then letting r D c � u and x 0 D .c � x/=r; a0k D ak.�r/k ; wewill get uniform convergence on [u; c]:

98

Chapter 11. Set Operations, Images and Inverse Images

Usually, a set contains elements that are numbers, for exampleZD f: : : ;�1; 0; 1; : : :g: Later we will consider aset whose elements are sets, for example # D fQ;R; C ; : : :g: We will say a collection (or class) to mean a set of sets.

Definitions. (a) The union of one or more sets is the set whose elements belong to at least one of these sets. The union

of sets S1; S2; : : : ; Sn may be denoted by S1 [ S2 [ � � � [ Sn orn[

iD1

Si : If we have a collection # of sets, then we may

take the union of all sets in #; which will be denoted by[

S�2# S� (or[� S� or

[#).

(b) The intersection of one or more sets is the set whose elements belong to every one of these sets. As usual, we willuse \ to denote intersection. There are similar notations for intersection of one or more sets as those for union.

(c) The complement of B in A (denoted by A n B) is the set whose elements belong to A; but not to B:We have the following frequently used facts:

(1) If S� � T� for every �; then[� S� �[� T� and

\� S� �\� T�:(2) For S � R;Rn .Rn S/ D S: If A � B; then for every set C; C n B � C n A:(3) (Distributive Law) A \ �[� S�� D[� .A \ S�/ and A [ �\� S�� D\� .A [ S�/(4) (de Morgan’s Law) A n �[� S�� D\� .A n S�/ and A n �\� S�� D[� .A n S�/

Definitions. For a function f : A ! B and S � A and C � B; the set f .S/ D f f .x/ : x 2 Sg is the image of S underf and the set f �1.C/ D fx : f .x/ 2 Cg is the inverse image (or preimage) of C under f:Remarks. In the definitions above, y 2 f .S/ () y D f .x/ for some x 2 S: Also, x 2 f �1.C/ () f .x/ 2 C:

y

16

9

4

x3 4

1

-1-2

y

16

9

x3 4-3-4

y = x2

Examples. Let f : R! Rbe defined by f .x/ D x2: Then f .[�2;�1/ [ .3; 4]/ D fx2 : x 2 [�2;�1/ [ .3; 4]g D.1; 4] [ .9; 16] and f �1.[9; 16// D fx : x2 2 [9; 16/g D .�4;�3] [ [3; 4/:99

The following are frequently used facts concerning images and inverse images of a function f : A ! B:

(5) If U � V � A; then f .U / � f .V /: If X � Y � B; then f �1.X / � f �1.Y /:(6) U � f �1. f .U // and f . f �1.X // � X :(7) f

�[� S�� D[� f .S�/ f �1�[� S�� D[� f �1.S�/

(8) f�\� S�� �\� f .S�/ f �1

�\� S�/ D\� f �1.S�/(9) If X � A; then f .A n X / � f .A/ n f .X /: If Y � B; then f �1.B n Y / D f �1.B/ n f �1.Y /:

For the inclusion signs in facts (6), (8), (9), we now show there are examples where the two sides are not equal.Let f : R! Rbe defined by f .x/ D x2:(a) For fact (6), let U D .�1; 0]; then f �1. f .U // D f �1.[0;C1// D R 6D U: Let X D R; then f . f �1.X // D

f .R/ D [0;C1/ 6D X :(b) For fact (8), let S1 D .�1; 0] and S2 D [0;C1/; then f .S1 \ S2/ D f .f0g/ D f0g; but f .S1/ \ f .S2/ D

[0;C1/ \ [0;C1/ D [0;C1/: So, in this case, f .S1 \ S2/ 6D f .S1/ \ f .S2/:(c) For fact (9), let A D Rand X D .�1; 0]; then A n X D .0;C1/; f .A/ n f .X / D [0;C1/ n [0;C1/ D ;; but

f .A n X / D f ..0;C1// D .0;C1/: So, in this case, f .A/ n f .X / 6D f .A n X /:Next, we will demonstrate how to prove some of the facts and leave the rest to the readers.

Proof of A n �[� S�� D\� .A n S�/:x 2 A n �[� S�� () x 2 A and x 62[� S�() x 2 A and x 62 S� for every �() x 2 A n S� for every �() x 2\� .A n S�/

Proof of f . f �1.X // � X :y 2 f . f �1.X // () y D f .x/; where x 2 f �1.X /() y D f .x/; where f .x/ 2 XH) y 2 X

(Note in the last step, the (H direction is false if y cannot be written in the form f .x/:) In fact, f . f �1.X // D X ifand only if X � f .A/; where A is the domain of f:Proof of f �1�[� S�� D[� f �1.S�/:

x 2 f �1�[� S�� () f .x/ 2[� S�() f .x/ 2 S� for at least one �() x 2 f �1.S�/ for at least one �() x 2[� f �1.S�/100

Proof of f�\� S�� �\� f .S�/:

y 2 f�\� S�� () y D f .x/ for some x; where x 2\� S�() y D f .x/ for some x; where x 2 S� for every �H) y 2 f .S�/ for every �() y 2\� f .S�/

(Note in the third step, the (H direction is false in general because y 2 f .S�/ means y D f .x�/ for some x� 2 S�:This x� may not be the same for every �:)

101

Chapter 12. Lebesgue Measure and Integral on RThere are two unsatisfactory features with Riemann integration:

(1) The integrable functions must be continuous almost everywhere. In particular, functions that are discon-

tinuous everywhere, such as f .x/ D � 1 if x 2 Q0 if x 2 RnQ; are not integrable. These include many important

functions in probability, physics and mathematics.

(2) For functions difficult to integrate, we use series expansion and properties of uniform convergence to computethe integrals. Uniformly convergent sequences involving easily integrable functions are not always available.

These reasons are more than sufficient for the development of a more powerful integral with simpler conditions onterm-by-term integration. In 1902, such a theory was introduced by the French mathematician Henri Lebesgue. Thetheory can be developed as follow:

Step 1. Study sets that are built from intervals which can be measured.

Step 2. Measure the length of as many sets as possible using the sets in Step 1.

Step 3. Identify functions having the property that the area under their graphs can be approximated by rectangleswhose bases are sets in Step 2.

Step 4. Assign an integral to each of the functions in Step 3 and study the basic properties of this integral.

Step 5. Obtain powerful theorems for term-by-term integration applicable to a wide collection of sequences andseries of functions.

Below we will develop the Lebesgue theory of integration on the real line following these steps.

Step 1: Open, Closed and Compact Sets

InR; a set is bounded iff it is bounded above and bounded below. A set is unbounded iff it is not bounded.

The open intervals inRcan be divided into the following types:.a; b/;| {z }the bounded type

.a;C1/; .�1; a/; .�1;C1/| {z }the unbounded types

; ; D .a; a/| {z }the degenerate type

:Similarly, the closed intervals can be divided into bounded and unbounded types. Now observe that Rn [a; b] D.�1; a/ [ .b;C1/: This suggests that closed and open intervals are complement of each other.

Definitions. InR; (1) a set S is open iff it is the union of a collection of open intervals;

(2) a set W is closed iffRn W is an open set;

(3) a set K is compact iff K is a closed and bounded set.

Remark. SinceRn .Rn S/ D S; so a set S is open if and only ifRn S is closed.

Examples. (1) Every open interval is an open set. In particular,RD .�1;C1/ and ; are open sets.

102

(2) The set .�1; 1/ [ .2; 5/ [ .3; 7/ is an open set. Note the open intervals may overlap.

(3) For f .x/ D sin x; f �1..0;C1// D � � �[ .�2�;��/ [ .0; �/[ .2�; 3�/[ � � � D [n2Z�2n�; .2n C 1/�� is open.

(4) The intervals .�1; a]; [a;C1/;RD .�1;C1/ and ; are closed because their complementsRn .�1; a] D .a;C1/; Rn [a;C1/ D .�1; a/; Rn .�1;C1/ D ;; Rn ; D R are open.

(Exercise: Show thatRand ; are the only open and closed sets inR:)(5) (Every finite set is compact.) If x1 < x2 < � � � < xn�1 < xn; then the set fx1; x2; : : : ; xn�1; xng is closed becauseRn fx1; x2; : : : ; xn�1; xng D .�1; x1/ [ .x1; x2/ [ � � � [ .xn�1; xn/ [ .xn;C1/ is open:

The set is also compact because it is bounded above by xn and below by x1:(6) The set W D f0g [ f 1

n : n 2 Ng D f0; 1; 12 ; 1

3 ; 14 ; : : :g is closed becauseRn W D .�1; 0/ [ � � � [ .1

4; 1

3/ [ .1

3; 1

2/ [ .1

2; 1/ [ .1;C1/ is open:

W is compact because in addition to being closed, it is bounded above by 1 and bounded below by 0.

(7) Any closed and bounded interval [a; b] is compact. The Cantor set

K D [0; 1] n ��1

3; 2

3

� [ �1

9; 2

9

� [ �7

9; 8

9

� [ � 1

27; 2

27

� [ � 7

27; 8

27

� [ �19

27; 20

27

� [ �25

27; 26

27

� [ � � �| {z }call this set S

�(always removing the open middle thirds of the remaining intervals) is a closed and bounded set becauseRn K D .�1; 0/ [ S [ .1;C1/ is open and K is bounded above by 1 and bounded below by 0. Hence theCantor set is compact.

Note every element of an open interval is inside the interval and is not an end point.

Definitions. An element x of a set S is an interior point of S iff .x � r; x C r/ � S for some r > 0: The interior of S(denoted by Int.S/ or S� ) is fx 2 S : x is an interior point of Sg:Structure Theorem for Open Sets. For a nonempty set S inR; the following are equivalent.

(1) S is an open set.

(2) Every element of S is an interior point of S; i.e. Int.S/ D S:(3) S is a countable union of pairwise disjoint open intervals. (Here “pairwise disjoint” means the intersection of

every pair is the empty set. The pairwise disjoint open intervals are called components of S.)

Proof. (1) ) (2) Suppose S is open, then S D[� .a�; b�/: If x 2 S; then x 2 .a�; b�/ for at least one �: Let r be the

minimum of x �a� and b�� x; then r > 0 and .x � r; x C r/ � .a�; b�/ � S: So every x 2 S is an interior point of S:(2) ) (3) Suppose Int.S/ D S: If x 2 S; then .x � r; x C r/ � S for some r > 0: Now let Ix be the largest intervalcontaining x such that Ix � S: (Here Ix is an interval from mx D inffm : .m; x] � Sg to Mx D supfM : [x; M/ � Sg:)If a D mx or Mx is a real number, then a 62 S (otherwise .a� r 0; aC r 0/ � S for some r 0 > 0; then Ix [ .a� r 0; aC r 0/is a larger interval in S containing x :) So Ix is an open interval.

For x; y 2 S; if Ix \ Iy 6D ;; then Ix D Iy (otherwise Ix [ Iy is a larger interval in S containing x and y:) So S isthe union of these pairwise disjoint open intervals.

Finally there are only countably many of these intervals because in each of these intervals I;we can pick a rationalnumber rI in the interval I: These rational numbers form a subset ofQ: Since the assignment of rI to I is bijective, Sis the union of countably many pairwise disjoint open intervals.

103

(3) ) (1) This follows from the definition of open sets.

Definition. We define the length of a nonempty open set to be the sum of the lengths of the pairwise disjoint openintervals in part (3) above. Also, we denote the length of an open set S by �.S/: For the empty set, we define �.;/ D 0:Example. The set S D .0; 1/ [ .2; 4/ [ .3; 5/ is open. To express S as a countable union of pairwise disjoint openintervals, we write S D .0; 1/ [ .2; 5/: So the length �.S/ D .1� 0/C .5� 2/ D 4:

Concerning the lengths of open sets, we have the following useful facts.

Facts. (1) If In is a countable sequence of pairwise disjoint open intervals containing in .a; b/; then 6n�.In/ � b� a:

(If there are finitely many intervals, then we can prove by induction. The case of one interval is easy. When thereis more than one intervals, let Ij D .c; d/ be the interval with the least left endpoint. Then 6

n 6D j�.In/ � b� d by

inductive step since the other intervals are in .d; b/: Hence, 6n�.In/ � .b � d/ C .d � c/ D b � c � b � a: For the

infinite case, we take limit.)

(2) If U; V are open sets with U � V ; then �.U / � �.V /: Let the components of U and V be In and Jm respectively.For each In; take x 2 In : Then x 2 Jm for some m and Jm is the largest open interval in V containing x : SinceIn is also an interval containing x in V ; we get In � Jm : For each m; let Sm D fn : In � Jmg: By fact (1),�.V / D 6

m�.Jm/ � 6

m6

n2Sm

�.In/ D �.U /:Theorem (Topological Properties of Open Sets).

(1) ; andRare open sets.

(2) The union of any collection of open sets is an open set.

(3) The intersection of finitely many open sets is an open set.

Proof. Part (1) is clear. For part (2), if S� is open for a collection of �’s, then each S� D[i

.a�i; b�i/ for some open

intervals .a�i ; b�i/ and so[� S� D[� [

i

.a�i; b�i/ is open.

For part (3), if S1; : : : ; Sn are open and x 2 n\kD1

Sk; then for each k; there is a rk > 0 such that .x�rk ; xCrk/ � Sk:Let r be the minimum of r1; : : : ; rn : Then r > 0 and .x � r; x C r/ � n\

kD1

.x � rk ; x C rk / � n\kD1

Sk: So every element

ofn\

kD1

Sk is an interior point. By the structure theorem for open sets,n\

kD1

Sk is open.

Remarks. Part (3) above may be false if “finitely” is replaced by “infinitely.” For example, although each interval��1k; 1C 1

k

�is open,

1\kD1

��1k; 1C 1

k

� D [0; 1] is not an open set (as 0 or 1 is not an interior point of [0; 1]).

Terminology. For any set X (eg. X D R), a collection C of subsets of X satisfying the properties:

(1) ; and X are in C(2) the union of any collection of elements of C is in C(3) the intersection of finitely many elements of C is in C

is said to be a topology on X : By the topological properties of open sets, the class C D fS : S is an open set in Rg isa topology onR: This terminology will come up in later courses.

104

Theorem (Topological Properties of Closed Sets).

(1) ; andRare closed sets.

(2) The intersection of any collection of closed sets is a closed set.

(3) The union of finitely many closed sets is a closed set.

Proof. Part (1) is clear. For part (2), suppose W�’s are closed. By de Morgan’s law and part (2) of the topologicalproperties of open sets,Rn �\� W�� D[� .Rn W�/ is open, so

\� W� is closed. For part (3), suppose W1; : : : ;Wn are

closed. Again by de Morgan’s law and part (3) of the topological properties of open sets,Rn � n[kD1

Wk� D n\

kD1

.Rn Wk/is open, so

n[kD1

Wk is closed.

Remarks. Part (3) above may be false if “finitely” is replaced by “infinitely.” For example, although each interval�1k; 1� 1

k

�is closed,

1[kD3

�1k; 1� 1

k

� D .0; 1/ is not a closed set.

Corollary. If S is open and W is closed, then S n W D S \ .Rn W / is open and W n S D W \ .Rn S/ is closed.

Question. Is every closed set a countable union of pairwise disjoint closed intervals? (If so, then we can define thelength of a closed set as the sum of the lengths of these intervals.)

Unfortunately, as we will see later, the answer is no. So it is more difficult to define the length of a closed set.However, for compact sets, it is easier to define their lengths. For every compact set K ; since K is bounded, K � .a; b/for some open interval .a; b/: So there are always bounded open set S containing K : By the corollary, S n K is open.

Definition. The length of a compact set K is �.K / D �.S/ � ��S n K| {z }open

�;where S is any bounded open set containing K :Remarks. In the definition, any bounded open set S containing K gives the same answer. To see that, write

S D 1[iD1.ai ; bi/ with pairwise disjoint .ai ; bi/: If .ai ; bi/ \ K D ; for some i; then .ai ; bi/ will be in S and S n K and

hence �..ai ; bi // will be cancelled in �.S/ � �.S n K /: So we may as well assume every .ai ; bi / \ K 6D ;:Next, write .inf K ; sup K / n K D 1[

jD1.cj ; dj/ with pairwise disjoint .cj ; dj/: If .cj ; dj/ 6� S for some j; then since

cj ; dj 2 K � S; so cj 2 .am; bm/ and dj 2 .an; bn/ for some m; n and cj < bm � an < dj : Then S0 D S [ .am; bn/is open, S0 � S � K ; �.S0/ D �.S/ C jbm � anj and �.S0 n K / D �.S n K / C jbm � anj: So �.S0/ � �.S0 n K / D�.S/ � �.S n K /: Thus, we may assume S � ..inf K ; sup K / n K / [ K D [inf K ; sup K ]:

Finally we can cancel �.S \ .�1; inf K // and �.S \ .sup K ;C1// from both �.S/ and �.S n K / to arrive at�.K / D �..inf K ; sup K // � �..inf K ; sup K / n K /:Examples. (1) For every closed and bounded interval [x; y]; we have [x; y] � .x � 1; y C 1/ and so�.[x; y]/ D .y C 1/� .x � 1/� ��.x � 1; x/ [ .y; y C 1/� D .y C 1/ � .x � 1/� .1 C 1/ D y � x :

105

(2) Recall the Cantor set K is compact. Now K � .�1; 2/: So�.K / D �2� .�1/� � ��.�1; 2/ n K

�D 3� ��.�1; 0/ [ .1; 2/ [ �1

3; 2

3

� [ �1

9; 2

9

� [ �7

9; 8

9

�[ � 1

27; 2

27

� [ � 7

27; 8

27

� [ �19

27; 20

27

� [ �25

27; 26

27

� [ � � ��D 3� �1C 1C 13C 1

9C 1

9| {z }2 times

C 127C 1

27C 1

27C 1

27| {z }4 times

C � � ��D 3� �2C 16nD0

2n 1

3nC1

� D 3� 2C 13

1� 23

! D 0:Answer to Question. No! The Cantor set K is closed. Since its length is 0, the only closed intervals contained inK are single points fxg D [x; x]: So every countable union of these (single point) closed intervals is a countable set,which cannot be K as K is uncountable. To see K uncountable, consider the nonterminating base 3 representation ofx 2 [0; 1]: Let x D .0:a1a2a3 : : :/3 with all ai D 0; 1 or 2 and ai not eventually 0. Note

x 62 .13; 2

3/() a1 6D 1; x 62 .1

9; 2

9/ [ .4

9; 5

9/ [ .7

9; 8

9/() a2 6D 1; : : : :

Hence x 2 K if and only if all ai D 0 or 2: Then f .x/ D .a1; a2; a3; : : :/ is a bijection from K onto f0; 2g � f0; 2g �f0; 2g � � � � ; which is uncountable by example (10) of Chapter 3 and so is K :To define the length of a closed set, we view the set as a “limit” of compact sets.

Definition. We define the length of a closed set W to be�.W / D limx!C1 ��W \ [�x; x]| {z }

compact

�:Remark. As x increases, �.W \ [�x; x]/ increases. So the limit will always exist or it is C1: In particular, wecan compute it along any sequence fxng tending to C1: If W is compact, then for x > j sup W j; j inf W j; we haveW \ [�x; x] D W: So the length of W as a closed set is the same as the length of W as a compact set.

Examples. (1) For an unbounded closed interval [a;C1/; by definition,�.[a;C1// D limx!C1 �.[a;C1/ \ [�x; x]/ D lim

x!C1�.[a; x]/ D limx!C1.x � a/ D C1:

(2) Let W D 1[kD1

[k; k C 13k

] D [1; 113

] [ [2; 219

] [ [3; 3127

] [ � � �: Then W is closed because Rn W D .�1; 1/ [.1 13 ; 2/ [ .2 1

9 ; 3/ [ � � � is open. Now the length�.W / D limn!C1 �.W \ [�n; n]/ D lim

n!C1 n�16kD1

1

3kD 16

kD1

1

3kD 1

3

1� 13

D 1

2:

Step 2: Measurable Sets

Next we will try to measure arbitrary sets from inside and outside.

106

Definitions. Let B be a subset ofR:(1) Define the Lebesgue outer measure of B to be m�.B/ D inff�.S/ : S is open and B � Sg:(2) Define the Lebesgue inner measure of B to be m�.B/ D supf�.K / : K is compact and K � Bg:(3) In case B is bounded, we say B is (Lebesgue) measurable iff m�.B/ D m�.B/: In this case, we call this number

the Lebesgue measure of B and denote it by m.B/:(4) In case B is unbounded, we say B is (Lebesgue) measurable iff B \ [a; b] is measurable for every a � b: In this

case, the Lebesgue measure of B is

m.B/ D limx!C1m.B \ [�x; x]| {z }

bounded

/:(Note this equation also holds for bounded B since B \ [�x; x] D B for x > j sup Bj; j inf Bj:)

Remarks. Most sets we can imagine are measurable. However, there still exist some extraordinary sets, which arenot measurable (see appendix 2 to step 2). In fact, there is a nonmeasurable set in every set with positive Lebesguemeasure! All known descriptions of these nonmeasurable sets are complicated. Outside set theory, they almost neverarise in practice.

Theorem. For any subset B of R; 0 � m�.B/ � m�.B/: In particular, if m�.B/ D 0; then B is measurable andm.B/ D 0:Proof. For every open set S and compact set K such that K � B � S; T D S \ .inf K � 1; sup K C 1/ is a boundedopen subset of S containing K :We have �.K / D �.T /��.T nK / � �.T / � �.S/: This implies�.S/ is an upper boundof f�.K / : K is compact and K � Bg: So m�.B/ D supf�.K / : K is compact and K � Bg � �.S/: This impliesm�.B/ is a lower bound of f�.S/ : S is open and B � Sg: So m�.B/ � inff�.S/ : S is open and B � Sg D m�.B/:The second sentence of the theorem is clear.

Remarks. A set B such that m.B/ D 0 is called a set of measure 0 or a null set. (For example, empty set, finite sets,countable sets and the Cantor set K are sets of measure 0 as will be shown shortly.) These sets are regarded as smallsets in measure theoretical situations. If a property is true except on a set of measure 0, then we say the property istrue almost everywhere. (The abbreviation is a.e.) For example, if f .x/ D g.x/ except for x D 0; then f .x/ D g.x/a.e.. If hn.x/! h.x/ except for x in the Cantor set K ; then hn.x/! h.x/ a.e.. Also, Lebesgue’s theorem in Chapter9 asserts that a bounded function on [a; b] is Riemann integrable if and only if it is continuous almost everywhere.

Theorem. Every interval is measurable and the Lebesgue measure is the length of the interval. (Here an interval froma to b is defined to have length b� a:)Proof. (Bounded Cases) First we will take care of some trivial cases. Since ; is open and �.;/ D 0; it follows thatm�.;/ D 0: By the theorem above, ; is measurable and m.;/ D 0:

Next, if I D fag, then I � .a� "; aC "/ for every " > 0: It follows that m�.I / � 2" for all " > 0: So m�.I / D 0:By the theorem above, I is measurable and m.I / D 0:

For the nontrivial cases, let I be a bounded interval with endpoints a and b .a < b/: Since I � S" D .a�"; bC"/for all " > 0; it follows that m�.I / � b� a C 2" for all " > 0: So m�.I / � b� a: Similarly, using [a C "; b� "] � Ifor 0 < " < .b � a/=2; we will get b � a � m�.I /: By the theorem above, b � a � m�.I / � m�.I / � b � a: Som�.I / D b � a D m�.I /: Then I is measurable and m.I / D b � a:

(Unbounded Cases) Clearly, the intersection of every unbounded interval with [a; b] is a bounded interval, hencethe intersection is measurable. We will begin with R D .�1;C1/: We have m.R/ D lim

x!C1m.R\ [�x; x]/ Dlim

x!C1m.[�x; x]/ D limx!C1 2x D C1: Next, m..�1; a// D lim

x!C1m..�1; a/ \ [�x; x]/ D limx!C1m.[�x; a// D

limx!C1a C x D C1: Similarly, m..�1; a]/ D C1;m..a;C1// D C1 and m.[a;C1// D C1:

So in all cases, the Lebesgue measure of an interval is the length of the interval.

107

Remark. By the structure theorem for open sets, we also have m�.B/ D infn 16

jD1jaj � bj j : B � 1[

jD1

.aj; bj/o:Theorem (Properties of Outer Measure).

(a) m�.;/ D 0:(b) If A � B; then m�.A/ � m�.B/:(c) If A D 1[

kD1

Ak ; then m�.A/ � 16kD1

m�.Ak /:Proof. Part (a) is proved in the last theorem. Part (b) follows because B � S ) A � S and sof�.S/ : S is open and B � Sg � f�.S/ : S is open and A � Sg:Taking infimum, we get m�.A/ � m�.B/: For part (c), for every " > 0; by the remark above, there are open intervals.akj ; bkj / such that Ak � 1[

jD1

.akj ; bkj / and16

jD1jakj � bkj j < m�.Ak / C "

2k: Then

A � 1[kD1

1[jD1

.akj ; bkj/ and m�.A/ � 16kD1

16jD1jakj � bkj j � 16

kD1m�.Ak / C ":

The result then follows by the infinitesimal principle.

Remarks. (1) There are similar properties for inner measure with similar proofs.

(2) In advanced courses, Lebesgue measure on subsets of R is generalized to subsets of an arbitrary set X : An outermeasure on X is defined to be a function on the subsets of X satisfying the three properties of outer measure above.

In showing a set is measurable, there is a theorem due to Caratheodory in 1914 which is often very useful.

Caratheodory’s Theorem. A set S is measurable if and only if for every set X; m�.X / D m�.X \ S/ C m�.X n S/:Remarks. (1) Since X D .X \ S/[.X n S/; so by part (c) of the properties of outer measure, we always havem�.X / � m�.X \ S/C m�.X n S/: So to check measurability of a set, we have to prove only the reverse inequality.

(2) In advanced courses, this equivalent condition is used to define measurable sets. The advantages are that (i) wedo not need to consider bounded and unbounded sets separately, (ii) the condition does not involve inner measure and(iii) when outer measures are defined by the properties above, there is no need to base the theory on open subsets(i.e. topology).

The proofs of Caratheodory’s theorem and the following useful properties of measurable sets will be presented inappendix 1 to step 2.

Theorem (Properties of Measurable Sets).

(1) If S is open, then S is measurable and m.S/ D �.S/:(2) If K is compact, then K is measurable and m.K / D �.K /: If W is closed, then W is measurable and m.W / D�.W /:(3) If C is measurable, m.C/ D 0 and C0 � C; then C0 is measurable and m.C0/ D 0: Also, if A;C are measurable

and m.C/ D 0; then A [ C; A n C are measurable and m.A [ C/ D m.A/ D m.A n C/:(4) If Ak is measurable for every k 2 N; then

1[kD1

Ak is measurable ( i.e. countable union of measurable sets is

measurable) and1\

kD1

Ak is also measurable (i.e. countable intersection of measurable sets is measurable). If

108

A1; : : : ; An are measurable, thenn[

kD1

Ak is measurable (by taking AnC1 D AnC2 D � � � D ; in the statement

above) andn\

kD1

Ak is measurable (by taking AnC1 D AnC2 D � � � D R in the statement above).

(5) If A is measurable, thenRn A is measurable. If A; B are measurable, then B n A D B \ .Rn A/ is measurable.

Terminologies. For any set X (eg. X D R), a collection C of subsets of X satisfying the properties:

(1) X is in C(2) if A is in C; then X n A is in C(3) the union of countably many elements of C is in C

is said to be a � -algebra (or � -field) in X : By properties (1), (4), (5), the class C D fS : S is a measurable set in Rg isa � -algebra inR: These terminologies will come up in later courses.

When a set is known to be measurable, the natural problem then is to find its Lebesgue measure. The next fewfacts will be helpful.

Theorem (Computation Formulas for Lebesgue Measure).

(1) (Translation Invariance) If A is measurable, then x C A D fx C a : a 2 Ag; A C x D fa C x : a 2 Ag aremeasurable and m.x C A/ D m.A/ D m.A C x/ for every x 2 R:

(2) (Countable Subadditivity) If A1; A2; A3; : : : are measurable sets, then m� 1[

kD1

Ak� � 16

kD1m.Ak /: Also, if we take

AnC1 D AnC2 D � � � D ;; then we have m� n[

kD1

Ak� � n6

kD1m.Ak /:

Remark. This follows from part (c) of the properties of outer measure since m.Ak / D m�.Ak /:(3) (Countable Additivity) If A1; A2; A3; : : : are pairwise disjoint measurable sets, then m

� 1[kD1

Ak� D 16

kD1m.Ak /: (As

in countable subadditivity,1 may be replaced by n:)Remarks. As a consequence, if A � B and both are measurable, then m.A/ � m.B/ since m.B/ D m.A/ Cm.B n A/: Also, if A � B and m.A/ <1; then m.B n A/ D m.B/ �m.A/:

(4) (Monotone Set Theorem) Let A1; A2; A3; : : : be measurable. If A1 � A2 � � � � ; then m� 1[

kD1

Ak� D lim

k!1m.Ak /:However, if A1 � A2 � � � � and m.Aj / <1 for some j; then m

� 1\kD1

Ak� D lim

k!1m.Ak /:Remarks. The condition m.Aj / < 1 is necessary. To see this, consider taking the sets Ak D .k;C1/: Then

m� 1\

kD1

Ak� D m.;/ D 0; but lim

k!1 m.Ak / D 1:Following these useful facts, we will present some examples.

Examples. (1) Since the Cantor set K is compact, by property (2), m.K / D �.K / D 0:109

(2) Let B be countable, eg. B D N;Z;Q: We can write B D fx1; x2; x3; : : :g with no element repeated. Then

B D 1[kD1

fxkg is measurable by property (4). By countable additivity, m.B/ D m� 1[

kD1

fxkg� D 16kD1

m.fxk g/ D 0: In fact,

if m.Ak / D 0 for every k 2 N; then by countable subadditivity, m� 1[

kD1

Ak� � 16

kD1m.Ak / D 0; so m

� 1[kD1

Ak� D 0:

Thus, m.Q/ D 0: By the remark to countable additivity, m.RnQ/ D C1 � 0 D C1: Also, by property (3),m.[0; 1] \Q/ D 0 and so m.[0; 1] nQ/ D m.[0; 1]/ �m.[0; 1] \Q/ D 1� 0 D 1:(3) Let K be the Cantor set. Show that the set A D fx 2 R : x D y C z for some y 2 K ; z 2 Qg is measurable andm.A/ D 0:Solution. SinceQ is countable, writeQ as fr1; r2; r3; : : :g: Then A D 1[

nD1

.K C rn/: By property (2), K is measurable.

By translation invariance, K C rn is measurable and m.K C rn/ D m.K / D 0 for every n 2 N: By property (4), A ismeasurable. By example (2) above, m.A/ D 0:(4) Show that the set

A D fx 2 [0; 1] : x has a decimal representation 0:a1a2a3 : : : such that a1; a2; a3; : : : are oddgis measurable.

Solution. Let An D fx 2 [0; 1] : x has a decimal representation 0:a1a2a3 : : : such that a1; a2; : : : ; an are oddg: Then

A1 � A2 � A3 � � � � and A D 1\nD1

An : Let B D f.a1; a2; : : : ; an/ : ai D 1; 3; 5; 7; 9 for i D 1; 2; : : : ; ng: Then B has

5n elements and An D [.a1;:::;an/2B

[0:a1a2 : : : an000 : : : ; 0:a1a2 : : : an999 : : :] is the union of 5n pairwise disjoint closed

intervals, each of length1

10n: By property (4), An and A are measurable. Since m.A1/ � m.[0; 1]/ D 1; by montone

set theorem, m.A/ D limn!1 m.An / D lim

n!1 5n� 1

10n

� D 0:Appendix 1 to Step 2 : Proofs of Properties and Computation Formulas

We will begin by proving a few lemmas for Caratheodory’s theorem.

Lemma 1. If K is compact and O is open, then �.K / � �.K n O/ C �.O/: Also, if K � O; then �.O/ D�.K / C �.O n K /:Proof. Let P be a bounded open set containing K : By definition, �.K / D �.P/ � �.P n K /: Note

P n .K n O/ D P n �K \ .Rn O/� D .P n K / [ �P n .Rn O/� D .P n K / [ .P \ O/:Now the sets P n K ; P \ O are open. So ��P n .K n O/� � ��.P n K / [ .P \ O/� � �.P n K /C �.P \ O/: Hence�.K n O/ C �.O/ D �.P/ � ��P n .K n O/� C �.O/� �.P/ � �.P n K / � �.P \ O/ C �.O/D �.K / � �.P \ O/ C �.O/ � �.K /:Let O D 1[

kD1

.ak; bk/ with .ak; bk/ pairwise disjoint. If K � O; then let inf K 2 .ai ; bi /; sup K 2 .aj ; bj/ and

O 0 D .ai ; bj/ \ O: Now O n [ai; bj] and O 0 n K are disjoint open sets and their union is O n K (as ai ; bj 62 O). So�.O/ D �.O 0/ C �.O n [ai; bj]/ D �.O 0 n K /C �.K / C �.O n [ai; bj]/ D �.K / C �.O n K /:110

Lemma 2. For every pair of disjoint A; B � R; we have m�.A [ B/ � m�.A/ C m�.B/ � m�.A [ B/:Proof. Let O be open and B � O: Let K be compact and K � A[ B: By lemma 1, �.K / � �.K n O/C �.O/: SinceK n O is compact in A; �.K / � m�.A/ C �.O/: Taking supremum over such K and infimum over such O; we getm�.A [ B/ � m�.A/ Cm�.B/:

For the inequality on the right, the case m�.A [ B/ D 1 is clear. If m�.A [ B/ < 1; then let O be open andA[ B � O: For every compact K � A; by lemma 1, �.O/ D �.K /C�.O n K /: Since O n K is open and B � O n Kby disjointness, we have m�.B/ � �.O n K /: Then �.K / C m�.B/ � �.O/: Taking supremum over such K andinfimum over such O; we have m�.A/ Cm�.B/ � m�.A [ B/:Lemma 3. If U is open with �.U / <1 and S is unbounded measurable, then m�.U \ S/ � m�.U \ S/:Proof. Let U D 1[

iD1

.ai ; bi/with .ai; bi/pairwise disjoint. Since�.U / <1; there is n 2 Nsuch that16

iDnC1jai � bi j < ":

Let W D n[iD1

.ai ; bi/: Then

m�.U \ S/ � m�.W \ S/ Cm��.U n W / \ S� � m�.W \ S/C m�.U n W /� m�.W \ S/ C 16

iDnC1jai � bi j < n6

iD1m�.[ai ; bi] \ S/ C ":

For i D 1; 2; : : : ; n; let "i D min� jai � bi j

2; "

4n

�: Since [ai C "i ; bi � "i ] \ S is measurable, there is a compact

set Ki � [ai C "i; bi � "i ] \ S such that m�.[ai C "i ; bi � "i] \ S/ � "2n

< �.Ki /: Since Ki � .ai ; bi/; the sets

K1; K2; : : : ; Kn are pairwise disjoint compact. So K D K1 [ K2 [ � � � [ Kn is a compact set in U \ S andn6

iD1m.[ai C "i; bi � "i ] \ S/� "

2< �.K /: Then

m�.U \ S/� " < n6iD1

m.[ai ; bi] \ S/ � n6iD1

�m.[ai C "i; bi � "i ] \ S/C 2"i

�� n6iD1

m.[ai C "i ; bi � "i] \ S/ C "2� �.K / C " � m�.U \ S/ C ":

Since " > 0 is arbitrary, we have m�.U \ S/ � m�.U \ S/:Lemma 4. For every open V ; we have m�.V / D �.V / D m�.V /:Proof. That m�.V / D �.V / is clear. Let V D 1[

iD1

.ai ; bi/ with .ai ; bi/ pairwise disjoint. For n D 1; 2; 3; : : : ; let"i;n D jai � bi j2n

; then Kn D n[iD1

[ai C "i;n ; bi � "i;n] is compact and Kn � V : We have

limn!1 �.Kn/ D lim

n!1.1� 1

n/ n6

iD1jai � bi j D �.V /:

So m�.V / D �.V /:Caratheodory’s Theorem. A set S is measurable if and only if for every set X; m�.X / D m�.X \ S/ C m�.X n S/:Proof. For the if direction, let X D [a; b]; then ja � bj D m�.[a; b] \ S/ C m�.[a; b] n S/: By lemma 2, ja � bj Dm�.[a; b]/ � m�.[a; b]\ S/Cm�.[a; b] n S/: These imply m�.[a; b]\ S/ � m�.[a; b]\ S/: As the reverse inequalityis clear, we get m�.[a; b] \ S/ D m�.[a; b]\ S/: This yields S is measurable in the bounded and unbounded cases.

For the only-if direction, the case m�.X / D 1 is easy since property (c) of outer measure implies m�.X / �m�.X \ S/ Cm�.X n S/ � 1: So we may reduce to the case m�.X / <1:

111

Assume S satisfies m�.U \ S/ � m�.U \ S/ for all open U with �.U / < 1: Then for every " > 0; there existsan open U � X such that �.U / � m�.X / C " <1: Now

m�.X / C " � �.U / D m�.U / .by definition of U and lemma 4/� m�.U \ S/C m�.U n S/ � m�.U \ S/C m�.U n S/ .by lemma 2 and assumption/� m�.X \ S/C m�.X n S/ � m�.X / .by U � X; properties (b) and (c) of outer measure/:Letting "! 0; we get m�.X / � m�.X \ S/C m�.X n S/ � m�.X /: Hence, m�.X / D m�.X \ S/C m�.X n S/:

Now every open set S satisfies the italicized assumption above by lemma 4 since U \ S is open. Also, bylemma 3, every unbounded measurable set S satisfies the assumption. In the remaining case, S is not open andis bounded measurable. Since U is open, by lemma 4, U satisfies the assumption as S: So for X D S; we havem�.S/ D m�.S \U /Cm�.S nU / by the open case as U is open. Also, by lemma 2, m�.S/ � m�.S \U /Cm�.S nU /:Since S is bounded measurable, m�.S/ D m�.S/ < 1; so the last two sentences imply m�.U \ S/ � m�.U \ S/:Thus the remaing case also satisfies the assumption and we are done.

We will use Caratheodory’s theorem to prove the following theorems, which yield properties (4), (5) of measurablesets and countable additivity.

Theorem. If A and B are measurable, thenRn A; A \ B; A [ B are measurable.

Proof. Let A0 D RnA: Since X nA D X\.RnA/ D X\A0; so the Caratheodory condition is symmetric in A and A0; soA measurable implies A0 D Rn A measurable. For A\B; since A is measurable, so m�.X / D m�.X \ A/Cm� .X n A/:Since B is measurable,

m�.X \ A/ D m�..X \ A/ \ B/C m�..X \ A/ n B/; m�.X n A/ D m�..X n A/ \ B/ C m�..X n A/ n B/:Since X n .A \ B/ D �.X \ A/ n B/� [ �.X n A/ \ B

� [ �.X n A/ n B�; we get by part (c) of the properties of outer

measure thatm�.X n .A \ B// � m�..X \ A/ n B/ Cm�..X n A/ \ B/ C m�..X n A/ n B/:

Combining with the equations above, we get m�.X / � m�.X \ A \ B/ C m�.X n .A \ B//: Since X D .X \ A \B/ [ .X n .A \ B//; the reverse inequality is true by part (c) of the properties of outer measure. By Caratheodory’stheorem, A \ B is measurable.

Finally,Rn .A [ B/ D .Rn A/ \ .Rn B/ is measurable and hence A [ B is measurable.

Theorem. Let A1; A2; : : : be pairwise disjoint measurable sets and A D 1[kD1

Ak ; then for every set S; we have

m�.S/ D 16kD1

m�.S \ Ak / Cm�.S n A/ D m�.S \ A/ C m�.S n A/:Proof. Let Bn D A1 [ � � � [ An : Since An is measurable, by Caratheodory’s theorem,

m�.S \ Bn/ D m�..S \ Bn/ \ An/ C m�..S \ Bn/ n An / D m�.S \ An/C m�.S \ Bn�1/:By induction, this gives m�.S \ Bn/ D n6

kD1m�.S \ Ak /: Since Bn is measurable by the last theorem and S n A �

S n Bn; so m�.S/ D m�.S \ Bn/C m�.S n Bn/ � n6kD1

m�.S \ Ak / Cm�.S n A/: Letting n ! 1; we get m�.S/ �16kD1

m�.S \ Ak / Cm�.S n A/ � m�.S \ A/ Cm�.S n A/ � m�.S/; where the last two inequalities are by part (c) of

the properties of outer measure because S \ A D 1[kD1

.S \ Ak / and S D .S \ A/ [ .S n A/: Therefore, we have the

desired equation.

112

Remarks. (1) By Caratheodory’s theorem, this implies A D 1[kD1

Ak is measurable. (In case, the Ak ’s are not pairwise

disjoint, write A as Ai [ .A2 n A1/[ �A3 n .A1 [ A2/�[ � � � ; then A will be measurable by the pairwise disjoint case.)

Using de Morgan’s law, we see that1\

kD1

Ak is also measurable. Hence property (4) of measurable sets is true. Taking

S D A; the equation yields countable additivity.

(2) If A1; A2; A3; : : : are measurable (not necessarily pairwise disjoint), then countable subadditivity follows by writing1[kD1

Ak as A1 [ .A2 n A1/ [ �A3 n .A1 [ A2/� [ � � �; which is the union of pairwise disjoint measurable sets so that

m�� 1[kD1

Ak� D m�.A1/Cm�.A2 n A1/Cm��A3 n .A1 [ A2/�C� � � � m�.A1/Cm�.A2/Cm�.A3/C� � � D 16

kD1m�.Ak /:

(3) Monotone set theorem follows from countable additivity by taking B1 D A1 and Bn D An n An�1 .n > 1/ in thefirst statement and Bj D Aj ; Bn D Aj n An .n > j / in the second statement.

(4) Translation invariance can be explained as follow. First note that ��.x C ai; x C bi /� D jai � bi j D ��.ai ; bi/�:Hence, for S open, �.x C S/ D �.S/ and so m�.x C A/ D m�.A/ for every set A: Suppose A is measurable so that forevery set X; m�.X / D m�.X \ A/ Cm�.X n A/: If x 2 R; then for every set Y; let X D �x C Y so that

m��Y \ .x C A/� Cm��Y n .x C A/� D m��.x C X / \ .x C A/� C m��.x C X / n .x C A/�D m��x C .X \ A/� C m�.x C .X n A/�D m�.X \ A/ C m�.X n A/D m�.X / D m�.Y /:(5) Since every open interval .a; b/ is measurable and m

�.a; b/� D ja�bj D ��.a; b/�; so every open set S D[iD1

.ai ; bi /(with .ai ; bi / pairwise disjoint) is a measurable set and m.S/ D 16

iD1m�.ai ; bi /� D 16

iD1jai � bi j D �.S/ by countable

additivity. Hence property (1) is true. Since the complement of closed and compact sets are open, closed andcompact sets are measurable. For a compact set K ; if S is a bounded open set containing K ; then S n K is open and�.S/ D m.S/ D m.K / Cm.S n K / D m.K /C �.S n K / so that m.K / D �.S/� �.S n K / D �.K /: Hence for closedset W; by monotone set theorem, m.W / D lim

n!1m.W \ [�n; n]/ D limn!1 �.W \ [�n; n]/ D �.W /:

(6) The first statement of property (3) of measurable sets follows from the first theorem of step 2,part (b) of the propertiesof outer measure. For the second statement, if A;C are measurable and m.C/ D 0; then m.C n A/ D 0 D m.A \ C/and by countable additivity, m.A [ C/ D m.A/ Cm.C n A/ D m.A/ D m.A \C/ C m.A n C/ D m.A n C/:

Appendix 2 to Step 2 : Proof of the Existence of Nonmeasurable Sets

Next we will show the existence of nonmeasurable sets. First we will recall the concept of an equivalent relation.

Definition. An equivalence relation R on a set E is a subset R of E � E such that

(a) (reflexive property) for every x 2 E; .x; x/ 2 R;(b) (symmetric property) if .x; y/ 2 R; then .y; x/ 2 R;(c) (transitive property) if .x; y/; .y; z/ 2 R; then .x; z/ 2 R:

We write x � y if .x; y/ 2 R: For each x 2 E; let [x] D fy : x � yg: This is called the equivalence classcontaining x : Note that every x 2 [x] by (a) so that

[x2E

[x] D E : If x � y; then [x] D [y] because by (b) and (c),

z 2 [x] () z � x () z � y () z 2 [y]: If x 6� y; then [x] \ [y] D ; because assuming z 2 [x] \ [y] will

113

lead to x � z and z � y;which imply x � y; a contradiction. So every pair of equivalence classes are either the sameor disjoint. Therefore, R partitions the set E into mutually disjoint equivalence classes.

Below we will need the Axiom of Choice which states that if fS� : � 2 �g is a nonempty collection of nonemptydisjoint subsets of a set X; then there exists a set V � X containing exactly one element from each set S�:Theorem. Every measurable set A with m.A/ > 0 has a nonmeasurable subset.

Proof. If A is unbounded, then m.A/ D limx!C1m.A \ [�x; x]/ > 0 implies m.A \ [�x; x]/ > 0 for some x : It is

enough to show this A \ [�x; x] has a nonmeasurable subset. Replacing A by A \ [�x; x]; we may assume A isbounded.

For x; y 2 R; define .x; y/ 2 R if y � x 2 Q: We can easily check that � is an equivalence relation on R andso we have a partitionRD [�2� S�; where S�’s are disjoint sets such that x; y are in the same S� if and only if x � y:SinceQ is countable, each S� is countable. By the axiom of choice, there is a set V containing exactly one element x�from each S�:

For each r 2 Q; let V C r D fx� C r : x� 2 V g; thenRD fx� C r : x� 2 V ; r 2 Qg D [r2QV C r:

If y 2 .V C i/ \ .V C j / for some i; j 2 Q; then there exists x�; x� 2 V such that y D x� C i D x� C j so thatx� � x� D i � j 2 Q: By the definition of V ; this implies x� D x� and so i D j: Hence .V C i/ \ .V C j / D ; fori 6D j:

Next we will show m�.V C i/ D 0 for every i 2 Q: Let T D Q \ .�1; 1/: For every compact K � V C i;the sets K C r with r 2 T are disjoint as K C r � V C i C r: Since

[r2T

.K C r/ � .inf K � 1; sup K C 1/; we

have m�[

r2T

.K C r/� <1: Now m�[

r2T

.K C r/� D 6r2T

m.K C r/ D 6r2T

m.K / implies m.K / D 0 for every compact

K � V : Therefore, m�.V C i/ D 0: Note

A D R\ A D �[r2Q.V C r/� \ A D [

r2Q�.V C r/ \ A�:

Since m�.A/ D m.A/ > 0; by property (c) of outer measure, at least one of the set .V C r/ \ A has nonzero outermeasure, say m��.V C i/\ A

� > 0: Then m��.V C i/\ A� � m�.V C i/ D 0 < m��.V C i/\ A

�implies .V C i/\ A

is a nonmeasurable subset of A:Remarks. In the case A D R;we see that one of the V Cr is nonmeasurable and hence every V Cr 0 D V CrC .r 0�r/is also nonmeasurable. Then RD [

r2QV C r is measurable and the V C r’s are pairwise disjoint, but none of them is

measurable.

Step 3: Measurable Functions

Consider the function f : [0; 1] ! R defined by f .x/ D � 3 if x 2 Q2 if x 2 RnQ: On [0; 1]; every lower Riemann

sum equals 2 because every nonempty interval contains irrational numbers and every upper Riemann sum equals 3because every nonempty interval contains rational numbers. So the function is not Riemann integrable.

However, if instead of using the usual rectangles to approximate the area under the graph, we use rectangles whosebases are measurable sets, then the area under the graph can be split into two such rectangles .[0; 1]\Q/� [0; 3] and.[0; 1] nQ/ � [0; 2]: If the area of such a rectangle is defined by length of base times height, then the area under thegraph of f would be 0� 3C .1� 0/� 2 D 2: Lebesgue made use of this idea to extend integration to more functions.

114

More generally, let f : [a; b] ! [0; k] be a function. If E1; E2; : : : ; En are pairwise disjoint measurable sets suchthat E1 [ E2 [ � � �[ En D [a; b]; then the collection Q D fE1; E2; : : : ; Eng is called a measurable partition of [a; b]:Let m j D inff f .x/ : x 2 E j g and Mj D supf f .x/ : x 2 E jg for j D 1; 2; : : : ; n: The lower Lebesgue sum for f with

respect to Q is L. f; Q/ D n6jD1

m j m.E j /: The upper Lebesgue sum for f with respect to Q is U. f; Q/ D n6jD1

Mjm.E j /:(The cases E j’s are intervals lead to lower and upper Riemann sums.)

Question: What kind of functions will the supremum of their lower Lebesgue sums equal the infimum of their upperLebesgue sums?

Definition. Let A be a measurable set. We say f : A ! R is a (Lebesgue) measurable function iff f �1..a; b// is ameasurable set for every a; b 2 R:Lebesgue Integral Theorem. Let f be a nonnegative valued bounded function on [x0; x1]: We have f : [x0; x1] ! Ris measurable if and only if

supfL. f; Q/ : Q is a measurable partition of [x0; x1]g D inffU. f; Q/ : Q is a measurable partition of [x0; x1]g:(This common value is called the Lebesgue integral of f on [x0; x1] and is denoted by

Z[x0;x1]

f dm:)Remarks. The proof will be presented in the appendix 1 to step 3. This theorem tells us that the bounded nonnegativevalued measurable functions on [x0; x1] have the property that the area under their graphs can be approximated byrectangles whose bases are measurable sets. Our goal now is to know as many measurable functions as possible.

Theorem. The following are equivalent.

(1) f : A ! R is measurable

(2) f �1.[a; b// is measurable for every a; b 2 R(3) f �1.[a; b]/ is measurable for every a; b 2 R(4) f �1..a; b]/ is measurable for every a; b 2 R(5) f �1..�1; b]/ is measurable for every b 2 R(6) f �1..�1; b// is measurable for every b 2 R(7) f �1.[a;C1// is measurable for every a 2 R(8) f �1..a;C1// is measurable for every a 2 R:

Proof. (1) ) (2) We have [a; b/ D 1\nD1

.a � 1

n; b/: So

f �1�[a; b/� D f �1

1\nD1

�a � 1

n; b�! D 1\

nD1

f �1�.a � 1

n; b/�;

which is measurable by (1).

(2) ) (3) We have f �1.[a; b]/ D 1\nD1

f �1.[a; b C 1n//; which is measurable by (2).

(3) ) (4) We have f �1..a; b]/ D 1[nD1

f �1.[a C 1

n; b]/; which is measurable by (3).

(4) ) (5) We have f �1..�1; b]/ D 1[nD1

f �1..�n; b]/; which is measurable by (4).

115

(5) ) (6) We have f �1..�1; b// D 1[nD1

f �1..�1; b � 1

n]/; which is measurable by (5).

(6) ) (7) We have f �1.[a;C1// D f �1.Rn .�1; a// D A n f �1..�1; a//; which is measurable by (6).

(7) ) (8) We have f �1..a;C1// D 1[nD1

f �1.[a C 1

n;C1//; which is measurable by (7).

(8) ) (1) We have

f �1..a; b// D f �1�.a;C1/ n 1\nD1

.b � 1

n;C1/� D f �1..a;C1// n 1\

nD1

f �1..b � 1

n;C1//;

which is measurable by (8).

Examples. (1) Let S be a set. The functionXS.x/ D �1 if x 2 S0 if x 62 S

is called the characteristic function of S: (In proba-

bility, it is called the indicator function of S and is denoted by 1S.x/:) Note thatX�1S .[a;C1// D (R if a � 0

S if 0 < a � 1; if 1 < a:

So if S is a measurable set, then XS is a measurable function.

(2) Every monotone function f onRis measurable. Just check all sets f �1..a;C1// are intervals, hence measurable.We leave this as an exercise.

(3) Every continuous function on a measurable domain is measurable by the following theorem.

Definitions. Let U � A � R: We say U is open in A if and only if U D V \ A for some open set V :Topological Continuity Theorem. Let A; B � R: The following are equivalent.

(1) f : A ! B is continuous.

(2) For every open set S; f �1.S/ is open in A; i.e. f �1.S/ D A \ V for some open set V :(3) For every bounded open interval I , f �1.I / is open in A; i.e. f �1.I / D A \ V for some open set V :

Proof. (1) ) (2) Suppose f is continuous. Let S be an open set and a 2 f �1.S/: Then f .a/ 2 S D Int.S/: So thereexists

�f .a/ � "; f .a/ C "� � S for some " > 0: Since f is continuous at a; for this " > 0; there exists �a > 0 such

that x 2 A and jx � aj < �a ) j f .x/ � f .a/j < ": Then

x 2 A \ .a � �a; a C �a/ ) f .x/ 2 � f .a/ � "; f .a/ C "� � S ) x 2 f �1.S/:Hence, A\.a��a ; aC�a/ � f �1.S/:Letting V D [

a2 f �1.S/.a � �a; a C �a/;we have V is open and f �1.S/ D [a2 f �1.S/fag� A \ V D [

a2 f �1.S/�A \ .a � �a; a C �a/� � f �1.S/: Therefore, f �1.S/ D A \ V :(2) ) (3) Since a bounded open interval I is an open set, so f �1.I / is open in A by (2).

(3)) (1) Let a 2 A: For " > 0; consider the bounded open interval I D �f .a/�"; f .a/C"�:By (3), f �1.I / D A\V

for some open V : Since f .a/ 2 I; so a 2 f �1.I / � V : Then there exists � > 0 such that .a � �; a C �/ � V : Thisimplies

x 2 A; jx � aj < � H) x 2 A \ .a � �; a C �/ � A \ V D f �1.I /H) f .x/ 2 I D �f .a/ � "; f .a/ C "�H) j f .x/ � f .a/j < ":

Therefore, f is continuous at every a 2 A:116

Remarks. (1) In the topological continuity theorem, if all the words “open” were replaced by “closed”, the statementswill remain true. This follows from W D Rn S implies f �1.W / D f �1.Rn S/ D f �1.R/ n f �1.S/ D A n f �1.S/ DA \ .Rn V /:(2) Note the topological continuity theorem tells us that to define a continuous function, all we need to know is whichsets are open in the domain, i.e. a topology. In a course on topology where we deal with abstract spaces, there may notbe a distance between points. Continuous functions can still be defined using open sets on the spaces.

Next we will present basic properties of measurable functions. The purpose is to decide which functions aremeasurable without having to use the definition. We remind the readers that for measurable set A and functionsf; g : A ! R; we say f D g almost everywhere iff f .x/ D g.x/ except for x in a set of measure 0. In that case, wedenote it by f D g a.e.

Theorem (Properties of Measurable Functions). Let A be a measurable set.

(1) If f : A ! R is measurable, f .A/ � B and g : B ! R is continuous, then g � f : A ! R is measurable.

(2) If f : A ! R is measurable, c 2 R and n 2 N, then cf; j f j; f n; e f ; cos f; sin f; 1

f(provided f .x/ 6D 0 for all

x 2 A) and np

f (provided f .x/ � 0) are measurable functions.

(3) If f1; f2 : A ! R are measurable, then f1 C f2; f1 � f2; f1 f2; f1

f2(provided f2.x/ 6D 0 ), max. f1; f2/ and

min. f1; f2/ are measurable functions.

(4) Let f1; f2; f3; : : : : A ! Rbe a sequence of measurable functions. For x 2 A; define the following functions.sup fn/.x/ D supf f1.x/; f2.x/; f3.x/; : : :g; .inf fn/.x/ D inff f1.x/; f2.x/; f3.x/; : : :g;.limsupn!1 fn/.x/ D limsup

n!1 fn.x/; .liminfn!1 fn/.x/ D liminf

n!1 fn.x/; . limn!1 fn/.x/ D lim

n!1 fn.x/:If each function .sup fn /.x/; .inf fn/.x/; .limsup

n!1 fn /.x/; .liminfn!1 fn/.x/; . lim

n!1 fn/.x/ exists everywhere on A; then

each of the functions sup fn ; inf fn ; limsupn1 fn ; liminf

n!1 fn ; limn!1 fn is measurable.

(5) Let f; g : A ! Rbe two functions such that f D g a.e.. If f is measurable, then g is measurable.

(6) If f : A ! R is a measurable function, then the function F : R! Rdefined by

F.x/ D �f .x/ if x 2 A0 if x 2 Rn A

D . fXA/.x/ is measurable.

Remarks. (1) Property (4) remains true if “exist everywhere on A” is replaced by “exist almost everywhere on A.”To see this, let E be the set where sup fn exists on A: Then m.A n E/ D 0: So E D A n .A n E/ is measurable.Now sup fnXE exists everywhere on A and hence it is measurable. Since sup fn D sup fnXE a.e. on A; so sup fn ismeasurable by property (5) if we consider it as sup fnXE : The other functions are similar.

(2) In place of measurable functions, if we consider Riemann integrable functions, then properties (1), (2) and (3) aretrue and properties (4) and (5) are false.

For property (4), the limit of Riemann integrable functions may not be Riemann integrable when the limit exists.For example, letQ\ [0; 1] D fr1; r2; r3; : : :g and fn D Xfr1;r2;:::;rn g : [0; 1] ! [0;C1/ is Riemann integrable becauseit is discontinuous at only finitely many points. However, lim

n!1 fn D XQ\[0;1] : [0; 1] ! [0;C1/ is not Riemann

integrable as in the beginning of step 3.

For Lebesgue integration, the limit function is measurable (hence can be integrated) whenever the limit exists bythe property (4) above. This is an advantage of Lebesgue integration over Riemann integration.

117

For property (5), the function f D 0 is Riemann integrable on every interval A D [a; b] and f D XQ\[a;b] a.e.,but XQ\[a;b] is not Riemann integrable.

Examples. (1) Show that f : R! Rdefined by f .x/ D jexXQ.x/ C max.cos x; 12/j is a measurable function.

Solution. Let f1.x/ D ex ; f2.x/ D XQ.x/; f3.x/ D cos x; f4.x/ D 12 : SinceQ is measurable, f2 is measurable. Since

f1; f3; f4 are continuous, they are measurable by the topological continuity theorem. Then f1 f2 and max. f3; f4/ aremeasurable. Hence f1 f2 Cmax. f3; f4/ is measurable. Therefore, f D j f1 f2 C max. f3; f4/j is measurable.

(2) Show that f : R! Rdefined by f .x/ D 16kD1

X[k;C1/.x/2k

is measurable.

Solution. The series converges pointwise on Rby comparing with16

kD1

1

2k: Now each term of the series is a constant

times a characteristic function of an interval, hence it is measurable. The k-th partial sum function fk is measurablebecause it is a sum of finitely many measurable functions. Since f D lim

k21 fk ; f is measurable.

(3) Show that f : [�2�; 3�] ! Rdefined by f .x/ D ( cos x if x 2 [�2�; 0/sin x if x 2 [0; 2�]�3 otherwise

is measurable.

Solution. We will express f using characteristic functions. Since [�2�; 3�] D [�2�; 0/ [ [0; 2�] [ .2�; 3�]; weconsider using X[�2�;0/;X[0;2� ];X.2�;3� ]: Now.cos x/X[�2�;0/.x/ C .sin x/X[0;2� ].x/ � 3X.2�;3� ].x/ D ( cos x C 0C 0 if x 2 [�2�; 0/

0C sin x C 0 if x 2 [0; 2�]0C 0� 3 if x 2 .2�; 3�]

D f .x/:Since cos x; sin x;�3 are continuous, hence measurable, and X[�2�;0/;X[0;2� ];X.2�;3� ] are measurable, so sums andproducts of these functions are measurable. Therefore, f is measurable.

(4) Show that if m.A/ D 0; then every function f : A ! R is measurable.

Solution. Since m.A/ D 0 and f �1..a; b// � A; so f �1..a; b// is measurable for every a; b 2 R: Therefore, f ismeasurable.

(5) Show that if f : R! R is continuous, then g : R! Rdefined by g.x/ D Z x

0f .t/ dt is a measurable function.

Solution. By the fundamental theorem of calculus, g0.x/ D f .x/: Since g is differentiable, g is continuous. Therefore,g is measurable by the topological continuity theorem.

Remarks. This example shows the integral of a continuous function is measurable. If f is only Riemann integrable(not necessarily continuous) on every interval [a; b]; then the function g defined above will still be continuous, hencemeasurable.

What about derivatives? As an exercise, show that if f : R! R is differentiable, then f 0 will be measurable.Note f 0 need not be continuous!

(6) Give an example of a nonmeasurable function.

Solution. To get such a function, take a nonmeasurable set S: Then the function XS is nonmeasurable becauseX�1S .[1; 1]/ D S is nonmeasurable.

118

Appendix 1 to Step 3 : Proofs of Properties and Theorems

Proof of Property (1). By the topological continuity theorem, for every a; b 2 R; g�1�.a; b/� D B \ S for some open

set S: By the structure theorem for open sets, S D 1[nD1

.an; bn/: For every a; b 2 R;x 2 .g � f /�1�.a; b/�() .g � f /.x/ D g. f .x// 2 .a; b/() f .x/ 2 g�1�.a; b/� D B \ S() f .x/ 2 S D 1[

nD1

.an; bn/() x 2 f �1� 1[nD1

.an; bn/� D 1[nD1

f �1�.an; bn/�:So .g � f /�1

�.a; b/� D 1[nD1

f �1�.an; bn/�; which is measurable by the fact that f is measurable and property (4) of

measurable sets.

Proof of Property (2). Just take g.x/ to be cx; jxj; xn; ex; cos x; sin x; 1x and n

px; respectively in property (1) of

measurable functions.

Proof of Property (3). First we will show f1 C f2 is measurable. For every a 2 R;x 2 . f1 C f2/�1�.a;C1/�() f1.x/ C f2.x/ 2 .a;C1/() f1.x/ C f2.x/ > a() f1.x/ > a � f2.x/() there exists q 2 Q such that f1.x/ > q > a � f2.x/() there exists q 2 Q such that f1.x/ 2 .q;C1/ and f2.x/ 2 .a � q;C1/() for some q 2 Q; x 2 f �1

1

�.q;C1/� \ f �12

�.a � q;C1/�() x 2 [q2Q� f �1

1

�.q;C1/� \ f �12

�.a � q;C1/�� :Therefore, . f1 C f2/�1�.a;C1/� D [

q2Q� f �11

�.q;C1/�| {z }measurable

\ f �12

�.a � q;C1/�| {z }measurable

�is measurable by the fact that f1; f2 are measurable and property (4) of measurable sets.

Next � f2 D .�1/ f2 is measurable by the property (2) of measurable functions. So f1 � f2 D f1 C .� f2/ ismeasurable by the first part. Next f1 f2 D 1

4

�. f1 C f2/2 � . f1 � f2/2� is measurable by the property (2) of measurable

functions and the first two parts. Thenf1

f2D f1

� 1f2

�is measurable by the property (2) of measurable functions and

the third part.

Finally,max. f1; f2/C min. f1; f2/ D f1 C f2

max. f1; f2/� min. f1; f2/ D j f1 � f2jH) max. f1; f2/ D 1

2. f1 C f2 C j f1 � f2j/ and min. f1; f2/ D 1

2. f1 C f2 � j f1 � f2j/

are measurable.

119

Proof of Property (4). First we will show sup fn is measurable. For every b 2 R;x 2 .sup fn/�1..�1; b]/ () .sup fn/.x/ � b() fn.x/ � b for n D 1; 2; 3 : : :() fn.x/ 2 .�1; b] for n D 1; 2; 3; : : :() x 2 1\

nD1

f �1n ..�1; b]/:

So .sup fn /�1..�1; b]/ D 1\nD1

f �1n ..�1; b]/; which is measurable by the fact that fn ’s are measurable and property

(4) of measurable sets. So sup fn is measurable. Next inf fn D � sup.� fn/ implies inf fn is measurable.

By the first part, Mk.x/ D supf fk .x/; fkC1.x/; fkC2.x/; : : :g is measurable for every k: Since M1.x/ � M2.x/ �M3.x/ � : : : ; limsup

n!1 fn D limk!1 Mk D inf Mk is measurable. Similarly, liminf

n!1 fn D limk!1mk D sup mk is mea-

surable, where mk .x/ D inff fk .x/; fkC1.x/; fkC2.x/; : : :g: Finally, if limn!1 fn.x/ exists for every x 2 A; then

limn!1 fn D limsup

n!1 fn is measurable.

Proof of Property (5). Let E D fx 2 A : f .x/ 6D g.x/g; then m.E/ D 0 and B D A n E is measurable. If x 2 B;then f .x/ D g.x/: For every a 2 R; the set S D fx 2 E : g.x/ > ag is a subset of E : So S is measurable. Now

g�1..a;C1// D fx 2 A : g.x/ > ag D fx 2 B : g.x/ > ag [ fx 2 E : g.x/ > agD fx 2 B : f .x/ > ag [ S D .B \ f �1..a;C1/// [ S

is measurable. Therefore, g is measurable.

Proof of Property (6). F is measurable because

F�1..a; b// D �f �1..a; b// if 0 62 .a; b/f �1..a; b// [ .Rn A/ if 0 2 .a; b/

is measurable for every a; b 2 R:Next we will prove the Lebesgue integral theorem. We first make some remarks.

Remarks. (1) For a bounded function f , L. f; Q/ � U. f; Q/ because

m j D inff f .x/ : x 2 E jg � supf f .x/ : x 2 E jg D Mj :(2) If P D fE1; � � � ; Emg and Q D fF1; � � � ; Fng are measurable partitions of [a; b]; then R D fEi \ Fj : i D

1; : : : ;mI j D 1; : : : ; ng is also a measurable partition of [a; b] by the distributive law. This is a commonrefinement of P and Q and has the property that L. f; P/ � L. f; R/ � U. f; R/ � U. f; Q/: The proof is thesame as the proof of the refinement theorem for Riemann integral, just replace partitions by measurable partitions.

(3) (Integral Criterion) The Lebesgue integral of a bounded function f on [x0; x1] exists, that is

supfL. f; Q/ : Q is a measurable partition of [x0; x1]g D inffU. f; Q/ : Q is a measurable partition of [x0; x1]gif and only if for every " > 0; there is a measurable partition Q of [x0; x1] such that U. f; Q/�L. f; Q/ < ": Theproof is the same as the proof of the integral criterion for Riemann integral, just replacing partitions by measurablepartitions.

120

Proof of the Lebesgue Integral Theorem. Suppose f : [x0; x1] ! [0; M/ is measurable. For every " > 0; take

positive integer n > .x1 � x0/M" : For j D 1; 2; : : : ; n; let Sj D f �1�� . j � 1/M

n; j M

n

��: Then Q D fS1; � � � ; Sng is

a measurable partition of [x0; x1]: Since. j � 1/M

n� f .x/ < j M

nfor all x 2 Sj; we getU. f; Q/ � L. f; Q/ � n6

jD1

M

nm.Sj / D .x1 � x0/M

n< ":

By the integral criterion, the Lebesgue integral of f on [x0; x1] exists.

Conversely, suppose the Lebesgue integral of f on [x0; x1] exists. For every n 2 N; by the integral criterion, there

is a measurable partition Qn D fE1; � � � ; Ekg of [x0; x1] such that U. f; Qn/� L. f; Qn/ < 1

n: Let mi D inff f .x/ :

x 2 Ei g and Mi D supf f .x/ : x 2 Ei g; then gn.x/ D k6iD1

miXEi .x/ and hn.x/ D k6iD1

MiXEi .x/ are measurable and

gn.x/ � f .x/ � hn.x/ for all x 2 [x0; x1]:Also g D sup gn and h D inf hn are measurable and g � f � h on [x0; x1]:We will show g D h a.e., that is S D fx 2 [x0; x1] : h.x/ � g.x/ > 0g is of measure 0. This will imply g D f a.e.,which implies f is measurable by property (5).

Note S D 1[jD1

Sj; where Sj D fx 2 [x0; x1] : h.x/ � g.x/ > 1jg: Let Tn; j D fx 2 [x0; x1] : hn.x/ � gn.x/ > 1

jg:

Since gn.x/ � g.x/ � h.x/ � hn.x/; so Sj � Tn; j for all n; j 2 N: For x 2 Ei\Tn; j ; Mi �mi D hn.x/ � gn.x/ > 1

j:

Sincek[

iD1

.Ei \ Tn; j / D [x0; x1] \ Tn; j D Tn; j and the Ei ’s are pairwise disjoint,

m.Tn; j /j

D k6iD1

1

jm.Ei \ Tn; j / � k6

iD1.Mi � mi /m.Ei \ Tn; j / � k6

iD1.Mi �mi /m.Ei / D U. f; Qn/ � L. f; Qn/ < 1

n:

Since Sj � Tn; j for all n; so m.Sj/ � m.Tn; j / < j

nfor all n:As n !1;we get m.Sj/ D 0: So m.S/ D m

� 1[jD1

Sj

� D 0:Appendix 2 to Step 3 : Cantor’s Function

Define a sequence of functions fn : [0; 1] ! Rby

f1.x/ D x and fnC1.x/ D ( fn.3x/=2 if x 2 [0; 1=3]1=2 if x 2 .1=3; 2=3/1=2C fn.3x � 2/=2 if x 2 [2=3; 1]

for n D 1; 2; 3; : : : :By mathematical induction, we can easily checked that fn .0/ D 0; fn.1/ D 1; fn is continuous on [0; 1] andk fnC1 � fnk[0;1] D 1

2n3for n D 1; 2; 3; : : : :

Consider the series f1.x/ C 16kD1. fkC1.x/ � fk .x// D lim

n!1� f1.x/ C n�16kD1

�fkC1.x/ � fk .x/�� D lim

n!1 fn.x/: Sincej fkC1.x/ � fk .x/j � Mk D 1

2k3for every x 2 [0; 1] and

16kD1

Mk D 1

3by the geometric series test, so the series

converges uniformly on [0; 1] by the M-test. Therefore, f .x/ D limn!1 fn.x/ is a continuous function on [0; 1] by

the continuity theorem of uniform convergence. By mathematical induction, we can also checked that every fn isan increasing function, hence f is also an increasing function. For a graph of this function, please visit the websitehttp://facweb.stvincent.edu/academics/mathematics/Cantorm/cantorm.html .

121

This function f is called Cantor’s function or the Cantor-Lebesgue function or Lebesgue’s singular function orDevil’s Staircase. The last terminology is because the graph of the function is like a staircase with irregular steps. Thisfunction exhibits many phenomena about Lebesgue measures and integrals.

Facts. (1) We will show the formulaZ b

af 0.x/ dx D f .b/� f .a/ does not always hold. This does not contradict the

fundamental theorem of calculus because f is not differentiable everywhere on [a; b]:To see this, let K be the Cantor set. The function f is constant on each open interval in [0; 1] n K . The constant

values of f on the intervals .13; 2

3/; .1

9; 2

9/; .7

9; 8

9/; . 1

27; 2

27/; . 7

27; 8

27/; .19

27; 20

27/; .25

27; 26

27/; : : : are respectively

1

2; 1

4; 3

4; 1

8; 3

8; 5

8; 7

8; : : :: So f 0.x/ D 0 for every x 2 [0; 1] n K : Since m.K / D 0; we may say f is differentiable

almost everywhere on [0; 1] and f 0 D 0 a.e. on [0; 1]: (Such a function f with f 0 D 0 a.e. is called a singularfunction.) In particular, f 0 is measurable on [0; 1]:Now, the surprising fact is that

0 D Z 1

0f 0.x/ dx 6D f .1/ � f .0/ D 1� 0 D 1:

This does not violate the fundamental theorem of calculus because f is not differentiable everywhere on [0; 1]:(2) We will show continuous bijections do not necessarily take measurable sets to measurable sets and compositionsof two measurable functions may not be measurable.

We begin with the function g.x/ D f .x/ C x on [0; 1]: Then g is a continuous and strictly increasing bijectionfrom [0; 1] to [0; 2]: By the continuous inverse theorem, g has a continuous inverse function h : [0; 2] ! [0; 1]: Sincef is constant on every open interval .a; b/ � [0; 1] n K ; f .a/ D f .b/ and so .g.a/; g.b// D . f .a/ C a; f .b/ C b/has length ja � bj: It follows that �.[0; 1] n K / D 1 implies �.g.[0; 1] n K // D 1: Then m.g.K // D 1 > 0: Therefore,g.K / contains a nonmeasurable set V . Now W D h.V / � K and �.K / D 0 implies W is measurable. So we see gis a continuous bijection from a measurable set W onto a nonmeasurable set V : In particular, h D g�1 is continuous(hence measurable) and W is a measurable set, but h�1.W / D V is nonmeasurable!

Furthermore, consider the function XW � h : [0; 2] ! R If x 2 V ; then h.x/ 2 W and so XW � h.x/ D 1;but if x 2 [0; 2] n V ; then h.x/ 62 W and so XW � h.x/ D 0: This implies XW � h D XV : Since XW is measurable,h is continuous and XV is nonmeasurable, this provides an example of a composition of measurable function with acontinuous (hence measurable) function resulting in a nonmeasurable function!

(3) (American Mathematical Monthly, Problem 6378) We will show that the set of all compositions � � �; where� : [0; 1] ! Rand � : R! Rare measurable, is the set of all functions : [0; 1] ! R:To see this, we will start with the fact that the Cantor function f : [0; 1] ! [0; 1] is surjective as it is continuous

on [0; 1] with f .0/ D 0 and f .1/ D 1: So we can find a right inverse function e : [0; 1] ! [0; 1] of f by takinge.x/ D sup f �1.fxg/ for example. So f � e is the identity function on [0; 1]: (Basically e.0/ D 0 and for everyx 2 .0; 1]; write the nonterminating base 2 representation 0:a1a2a3 : : : of x and we get e.x/ equals to the base 3number 2.0:a1a2a3 : : :/) This implies e.[0; 1]/ � K . From f increasing, it follows that e is strictly increasing (for ifa < b; then assuming e.a/ � e.b/; we will get a D f .e.a// � f .e.b// D b; a contradiction, so e.a/ < e.b/). Hence,e is measurable.

Now for every function : [0; 1] ! R; take � D e : [0; 1] ! e.[0; 1]/ and� .t/ D � . f .t// if t 2 e.[0; 1]/0 if t 2 Rn e.[0; 1]/ :

Then � : R! R is equal to 0 almost everywhere onRbecause e.[0; 1]/ � K . So � is measurable. Finally, for everyx 2 [0; 1];we have �.x/ D e.x/ 2 e.[0; 1]/ and so � � �.x/ D . f .e.x/// D .x/:

122

Step 4: Lebesgue Integral and its Properties

Let f : [a; b] ! [0;C1/ be a measurable function. The Lebesgue integral theorem is one approach to define theLebesgue integral for this type of function on [a; b] in case it is bounded above. For unbounded nonnegative valuedfunction, we define the Lebesgue integral as follow.

Definition. Let f : [a; b] ! [0;C1/ be measurable (and possibly not bounded above). We defineZ[a;b]

f dm D limk!1 Z[a;b]

fk dm D supnZ

[a;b]fk dm : k > 0

o; where fk .x/ D �f .x/ if f .x/ � kk if f .x/ > k

D min. f; k/.x/:(Note as k increases, the integral of fk will increase. So the limit is the same as the supremum, which is in [0;C1]:)

Once the Lebesgue integral of a nonnegative valued measurable function on a closed and bounded interval isdefined, the Lebesgue integral for a real valued measurable function on a measurable set can be defined as follow.

Definitions. (1) For f : R! [0;C1/ measurable, defineZR f dm D limw!1 Z[�w;w]f dm D sup

nZ[�w;w]

f dm : w > 0o:

(As w increase, the integral will increase. So the limit is the same as the supremum, which is in [0;C1]:)(2) For f : A ! [0;C1/ measurable, define

ZA

f dm D ZR fXA dm: (Note fXA is measurable by property (6) of

measurable functions.)

(3) For f : A ! Rmeasurable, define

f C.x/ D max. f .x/; 0/ D �f .x/ if f .x/ � 00 if f .x/ < 0

; f �.x/ D �min. f .x/; 0/ D �� f .x/ if f .x/ � 00 if f .x/ > 0

:In case at least one of

ZA

f C dm; ZA

f � dm is finite, defineZA

f dm D ZA

f C dm � ZA

f � dm:Examples. (1) Let S be an measurable set. On an interval [a; b]; consider the measurable partition Q D f[a; b] \S; [a; b] n Sg: Then L.XS; Q/ D m.S \ [a; b]/ D U.XS ; Q/; which implies

Z[a;b]

XS dm D m.S \ [a; b]/:(2) By definition, example (1) above and the monotone set theorem, we have for c � 0;Z

Ac dm D ZRcXA dm D lim

n!1 Z[�n;n]cXA dm D lim

n!1 cm.A \ [�n; n]/ D cm.A/:CAUTION: If c D 0 and m.A/ D C1; then cm.A/ should be interpreted as lim

n!1 cm.A \ [�n; n]/ D 0:If c < 0; then cC D 0 and c� D �c > 0 and

ZA

c dm D ZA

cC dm � ZA

c� dm D 0� .�cm.A// D cm.A/:(3) Let m.A/ D 0 and f : A ! Rbe any function. Write f D f C � f �: On each [�w;w]; consider the measurablepartition Q A D f[�w;w] \ A; [�w;w] n Ag: Since m.[�w;w] \ A/ D 0 and XA D 0 on [�w;w] n A; we see thatL. f �k XA; Q A/ D 0 D U. f �k XA; Q A/: These imply

Z[�w;w]

f �k XA dm D 0: SoZA

f � dm D ZR f �XA dm D limw!1Z[�w;w]f �XA dm D limw!1� lim

k!1 Z[�w;w]f �k XA dm

� D 0 H) ZA

f dm D 0:123

Theorem (Simple Properties of Lebesgue Integral). Let f; g : A ! R be measurable and their integrals can bedefined.

(1)Z

A. f � g/ dm D Z

Af dm � Z

Ag dm (provided the right side is not of the form1�1),

ZA

cf dm D cZ

Af dm

for every c 2 R(in case the right side is of the form 0 � 1; we will interpret 0 �1 D 0).

(2) Let f � g a.e. on A:ThenZ

Af dm � Z

Agdm;with equality if andonly if f D g a.e.. Also,

����ZA

f dm

���� � ZAj f jdm:

(3) If S; T are disjoint measurable subsets of A; thenZ

S[Tf dm D Z

Sf dm C Z

Tf dm: (In particular, if f is

nonnegative valued and B is a measurable subset of A; thenZ

Bf dm � Z

Bf dm C Z

AnBf dm D Z

Af dm:)

Proof. Using f D f C � f � and g D gC � g�; we can reduce to the case f; g : A ! [0;C1/: Note in this caseZA

f dm D ZR fXA dm D limw!1� limk!1 Z[�w;w]

fkXA dm�: Thus we may further reduce to the case A D [�w;w] and

the functions f; g are nonnegative valued bounded functions. Then the proofs of these properties follow by replacingpartitions of [�w;w] by measurable partitions in the proofs of the simple properties of Riemann integrals. Takinglimits in k and w; we can get the equations and inequalities for the sets. The proof of the equality case of property (2)will be left as an exercise.

Question: How are Lebesgue integration and Riemann integration related?

Theorem (Equality of Integrals). If f is Riemann integrable on [a; b]; then f is a measurable function andZ[a;b]

f dm D Z b

af .x/ dx :

Proof. Let P be the partition fa D x0 < x1 < � � � < xn D bg of [a; b]: Then the collection fE1; E2; : : : ; Eng; whereE1 D [x0; x1/; E2 D [x1; x2/; : : : ; En D [xn�1; xn]; is a measurable partition of [a; b]: For the case f .x/ � 0 on [a; b];.L/ Z b

af .x/ dx D sup

nL. f; P/ : partition P D fa D x0 < x1 < : : : < xn D bg of [a; b]

o� supfL. f; Q/ : measurable partition Q D fE1; E2; : : : ; Ekg of [a; b]g� inffU. f; Q/ : measurable partition Q D fE1; E2; : : : ; Ekg of [a; b]g� infn

U . f; P/ : partition P D fa D x0 < x1 < : : : < xn D bg of [a; b]oD .U / Z b

af .x/ dx :

So if f is Riemann integrable on [a; b]; then .L/ Z b

af .x/ dx D Z b

af .x/ dx D .U / Z b

af .x/ dx and all inequalities

above become equalities. By the Lebesgue integral theorem, f is measurable andZ

[a;b]f dm D Z b

af .x/ dx : The

case f is real valued follows by considering f C; f � � 0 first. As above, they will be measurable and their Riemannand Lebesgue integrals are the same. Then f D f C � f � is measurable and the Riemann and Lebesgue integrals off are the same.

Examples. (4) By property (3), example (3) above and equality of integrals,Z[0;1]nQ x2 dmD Z

[0;1]x2 dm � Z

[0;1]\Qx2 dm| {z }by property .3/ D Z

[0;1]x2 dm| {z }

by example .3/ D Z 1

0x2 dx D 1

3:

124

(5) Let f : [a;C1/! [0;C1/ be locally Riemann integrable. ThenZ[a;C1/ f dm D ZR fX[a;C1/ dm D limw!1 Z[�w;w]

fX[a;C1/ dmD limw!1Z[a;w]f dm D limw!1 Z w

af .x/ dx D Z 1

af .x/:

Remarks. A similar result holds for any other unbounded interval in place of [a;C1/: So for a nonnegative valuedfunction f; the improper Riemann integral of f on an unbounded interval equals the Lebesgue integral on the sameinterval.

(6) If f : R! [0;C1/ is measurable,ZRp f dm <1 and

ZR f 2 dm <1; then show thatZR f dm <1:

Solution. Let A D fx 2 R : 0 � f .x/ � 1g D f �1.[0; 1]/ and B D fx 2 R : f .x/ > 1g D f �1..1;C1//: Then Aand B are measurable, A [ B D Rand A \ B D ;: Since f � p f on A and f � f 2 on B; soZR f dmD Z

Af dm C Z

Bf dm| {z }

by property .3/ � ZA

pf dm C Z

Bf 2 dm| {z }

by property .2/ � ZRp f dm C ZR f 2 dm| {z }by property .3/ <1:

Remarks. If f is only Riemann integrable, then the sets A and B may not be unions of intervals. In that case, theabove integrals on A and B will not be defined for Riemann integrals!

There is a more elegant approach in defining the Lebesgue integral for nonnegative valued functions. It takes carethe bounded and unbounded cases at the same time.

The approach goes as follow. Any function � : A ! R of the form � D n6iD1

ciXSi ; where ci 2 R and Si is

a measurable subset of A for i D 1; 2; : : : ; n; is called a simple function on A: (We remark that we may evenassume the sets Si are pairwise disjoint subsets of A because if the range of � consists of distinct values r1; r2; : : : ; rj ;then � D j6

iD1riX��1.fri g/ and the sets ��1.fri g/ are pairwise disjoint subsets of A.) For such a simple function �;Z

A� dm D n6

iD1ci m.Si / by the remarks and properties above. The Lebesgue integral of a nonnegative measurable

function f : A ! [0;C1/ is taken to beZ

Af dm D sup

�ZA� dm : 0 � � � f; � simple

�(and for real valued f;

we consider f D f C � f � as before). The following theorem will show this approach agrees with our definition.

Theorem. For a measurable function f : A ! [0;C1/; ZA

f dm D sup

�ZA� dm : 0 � � � f; � simple

� :Proof. Let S D �Z

A� dm : 0 � � � f; � simple

� : Since � � f on A impliesZ

A� dm � Z

Af dm; we have

sup S � ZA

f dm: For the reverse inequality, sinceZA

f dm D ZR f XA dm D supnZ

[�w;w]fkXA dm : w; k > 0

o;for every n 2 N; there are wn; kn > 0 such that

ZA

f dm � 1n< Z

[�wn;wn]fknXA dm � Z

Af dm: Next, since an inte-

gral is the supremum of lower sums, there exists a lower sumL. fknXA; Qn/ D j6iD1

mi m.Ei /;where Qn D fE1; : : : ; E jgis a measurable partition of [�wn; wn]; such thatZ

[�wn;wn]fknXA dm � 1

n< L. fknXA; Qn/ � Z

[�wn;wn]fknXA dm:

125

Let �n D j6iD1

miXEi ; then 0 � �n � fknXA � f; �n is simple andZ

A�n dm D L. fknXA; Qn/: ThenZ

Af dm � 2

n� Z

A�n dm � Z

Af dm:

By sandwich theorem,Z

Af dm D lim

n!1 ZA�n dm � sup S:

Step 5: Convergence Theorems.

When expanding a function into a series of functions, the series often does not converge uniformly on the domain.So for functions difficult to integrate, the following theorems are more useful than the integration theorem for uniformconvergence. This again is another important advantage of Lebesgue integration over Riemann integration!

Monotone Convergence Theorem(MCT). Let f1; f2; f3; : : : : A ! [0;C1/| {z }fi .x/�0

be measurable functions. If

(1) limn!1 fn .x/ D f .x/ a.e. on A and

(2) for almost every x 2 A; f1.x/ � f2.x/ � f3.x/ � � � � ;then

ZA

f dm D ZA

limn!1 fn dm D lim

n!1 ZAfn dm:

(Corollary to) Monotone Convergence Theorem. Let g1; g2; g3; : : : : A ! [0;C1/| {z }gk.x/�0

be measurable functions. If16kD1

gk.x/ converges for almost every x 2 A; thenZ

A

16kD1

gk dm D 16kD1

ZA

gk dm:Examples. (1) Find lim

n!1 Z 2

1

x

1C e�nxdx :

Solution. On [1; 2]; as n % 1; we have �nx & �1; 1 C e�nx & 1 and fn .x/ D x

1C e�nx% x : Since fn is

continuous (hence Riemann integrable) on [1; 2]; by equality of integrals and MCT,

limn!1 Z 2

1

x

1C e�nxdx D lim

n!1 Z[1;2]

x

1C e�nxdmD Z

[1;2]lim

n!1 x

1C e�nxdmD Z

[1;2]x dmD Z 2

1x dx D 3

2:

(2)

126

Z 1

0eex

dx D Z[0;1]

eexdm .eex

continuous. Equality of Integrals/D Z[0;1]

16kD0

.ex /kk!| {z }�0

dm .Used ew D 16kD0

wk

k!for every w 2 R/D 16

kD0

Z[0;1]

ekx

k!dm .Terms continuous. Corollary to MCT/D 16

kD0

Z 1

0

ekx

k!dx .Equality of Integrals/D 1C 16

kD1

ek � 1

k!k:

(3)Z.0;1/ x1=3

1� xln

1x

dm D Z.0;1/ 16kD0

x .1=3/Ck ln1x| {z }>0 on .0;1/ dm .Used

a

1� rD 16

kD0ark for r 2 .0; 1//D 16

kD0

Z.0;1/�x .1=3/Ck ln x dm (Terms continuous. Corollary to MCT)D 16kD0

Z 1

0�x .1=3/Ck ln x dx .8� > 0; lim

x!0C x� ln x D 0 ) Riemann integrable on [0; 1]/D 16kD0

�� x .4=3/Ck.4=3/ C kln x

���10C� Z 1

0

x .1=3/Ck.4=3/C kdx

�(Integration by parts)D 16

kD0

9.3k C 4/2:(4) Let f : .0; 1] ! [0;C1/ be improper Riemann integrable. Observe that if A � B; thenXA.x/ D ( 1 if x 2 A

0 if x 2 B n A0 if x 62 B

� XB.x/ D (1 if x 2 A1 if x 2 B n A0 if x 62 B

:Since [ 1

n ; 1] � [ 1nC1 ; 1] H) X[ 1

n ;1] � X[ 1nC1 ;1]; so as n !1; f X[ 1

n ;1] increases to f on .0; 1]: By MCT and equalityof integrals, Z.0;1]

f dm D limn!1 Z.0;1]

fX[ 1n ;1] dm D lim

n!1 Z[ 1n ;1]

f dm D limn!1 Z 1

1=nf dx D Z 1

0f dx :

Remarks. A similar result holds for any other bounded interval in place of .0; 1]: So for a nonnegative valued functionf; the improper Riemann integral of f on a bounded interval equals the Lebesgue integral on the same interval.

Definition. A measurable function f is Lebesgue integrable on A iffZ

Aj f j dm D Z

Af C dm C Z

Af � dm <1:

Lebesgue Dominated Convergence Theorem(LDCT). Let f1; f2; : : : : A ! Rbe measurable functions. If

(1) limn!1 fn .x/ D f .x/ a.e. on A and

(2) there exists a Lebesgue integrable function g : A ! R such that for all n; j fn.x/j � g.x/ a.e. on A;127

thenZ

Af dm D Z

Alim

n!1 fn dm D limn!1 ZA

fn dm:Remarks. The special case when m.A/ <1 and g is a constant function is called the Lebesgue Bounded ConvergenceTheorem (LBCT) in many books.

(Corollary to) Lebesgue Dominated Convergence Theorem. Let g1; g2; : : : : A ! Rbe measurable functions. If16kD1

ZAjgkj dm <1; then

ZA

16kD1

gk dm D 16kD1

ZA

gk dm:The corollary was proved by Beppo Levi. So in many books, this corollary is also called the Beppo Levi theorem.

Examples. (1) Show that Z 1

0sin x ln x dx D 16

kD0

.�1/kC1.2k C 1/!.2k C 2/2 :(Note lim

x!0C sin x ln x D limx!0C sin x

x.x ln x/ D 1 � 0 D 0: So sin x ln x can be extended to a continuous (hence Riemann

integrable) function on [0; 1]:)Solution. Recall sin x D 16

kD0

.�1/k x2kC1.2k C 1/! for all x 2 R: So on .0; 1]; sin x ln x D 16kD0

.�1/k x2kC1 ln x.2k C 1/! : Next, let

gk.x/ D .�1/k x2kC1 ln x.2k C 1/! : Since limx!0C x2kC1 ln x D 0; gk can be extended to a continuous (hence Riemann integrable)

function on [0; 1]; so by equality of integrals,Z[0;1]

jgkj dm D Z 1

0� x2kC1 ln x.2k C 1/! dx D �1.2k C 1/! � x2kC2

2k C 2ln x

���10C� Z 1

0

x2kC1

2k C 2dx

� D 1.2k C 1/!.2k C 2/2and so

16kD0

Z[0;1]

jgkj dm D 16kD0

1.2k C 1/!.2k C 2/2 < 16kD1

1

k2<1 (by the comparison and p-tests). By the corollary

to LDCT,Z 1

0sin x ln x dx D Z

[0;1]

16kD0

gk dm D 16kD0

Z[0;1]

gk dm D 16kD0

Z[0;1]

.�1/kC1jgkj dm D 16kD0

.�1/kC1.2k C 1/!.2k C 2/2 :(2) Show that

Z[0;1/ e�x cos

px dx D 16

kD0

.�1/k k!.2k/! :Solution. Recall cosw D 16

kD0

.�1/kw2k.2k/! for every w 2 R: So on [0;1/; e�x cosp

x D 16kD0

.�1/k xke�x.2k/! : Next, let

gk.x/ D .�1/k xke�x.2k/! : By example (5) of step 4,Z[0;1/ jgkj dm D limw!1 Z[0;w]

jgk j dm D limw!1Z w0

xke�x.2k/! dx D Z 10

xke�x.2k/! dx D k!.2k/! :So

16kD0

Z[0;1/ jgkj dm D 16

kD0

k!.2k/! < 16kD0

1k!D e <1 (by the comparison test). By the corollary to LDCT,Z

[0;1/ e�x cosp

x dx D Z[0;1/ 16

kD0gk dm D 16

kD0

Z[0;1/ gk dm D 16

kD0

Z[0;1/.�1/k jgkj dm D 16

kD0

.�1/kk!.2k/! :128

(3) Find limn!1 Z 1

0

nx ln x

1C n2x2dx :

Solution. Let fn.x/ D �nx ln x1Cn2x2 if 0 < x � 10 if x D 0

; then by l’Hopital’s rule, limn!1 fn.x/ D 0 on [0; 1]:Using the inequality

or by calculus,t

1C t 2� 1

2; we have j fn.x/j � g.x/ D � j ln xj

2 if 0 < x � 10 if x D 0

: Since g D jgj � 0 on [0; 1] and the

improper Riemann integralZ 1

0g.x/ dx D �1

2.x ln x � x/���1

0CD 1

2; so

Z[0;1]

jgj dm D 1

2and g is Lebesgue integrable

on [0; 1]: By LDCT,

limn!1 Z 1

0

nx ln x

1C n2x2dx D lim

n!1 Z[0;1]fn dm D Z

[0;1]lim

n!1 fn dm D Z[0;1]

0 dm D 0:Remarks. The sequence fn does not converge uniformly on [0; 1] to 0 because k fn � 0k[0;1] � j fn.1

n/j D ln n

2!1:

So integration theorem for uniform convergence cannot be applied directly.

The following remarks compare Riemann integrable functions with Lebesgue integrable functions.

Remarks. (1) For proper Riemann integrals, we have the following.

Theorem. If f : [a; b] ! R is Riemann integrable, then (j f j is Riemann integrable on [a; b]: By the equality ofintegrals, the Lebesgue integral of j f j on [a; b] is finite and so) f is Lebesgue integrable on [a; b]:

However, the converse is false. The function f .x/ D �1 if x 2 Q\ [a; b]0 if x 62 Q\ [a; b]

is measurable on [a; b] andZ[a;b]

f dm D 0� 1C .b � a/� 0 D 0: Since j f j D f; f is Lebesgue integrable on [a; b]: However, f is not

Riemann integrable on [a; b]:(2) For improper Riemann integrals, we have the following.

Theorem. If j f j is improper Riemann integrable on an interval I; then f is (improper Riemann integrable by thecomparison test and) Lebesgue integrable on I and the two integrals give the same value. In particular, this holds truefor nonnegative valued functions.

The proof in the unbounded interval case follows from the ideas in example (5) of step 4 and in the boundedinterval case from example (4) of MCT. It is a consequence of equality of integrals, MCT and LDCT (using g D j f j).

If the function is real valued, then it is possible that the function be improper Riemann integrable, but not Lebesgueintegrable as the following examples show.

For the case of a bounded interval, say [0; 1]; we can partition [0; 1] with the points 0; : : : ; 116 ; 1

8 ; 14 ; 1

2 ; 1:Let g : [0; 1] ! R be defined by g.0/ D 0 and g.x/ D .�1/k�12k

kwhen

12k< x � 1

2k�1for k 2 N; i.e.

g.x/ D 16kD1

.�1/k�12k

kX. 1

2k ; 12k�1 ].x/: Then g is improper Riemann integrable on [0; 1] becauseZ 1

0g.x/ dx D limw!0C Z 1w g.x/ dx D 16

kD1

.�1/k�12k

k

� 12k�1

� 12k

� D 16kD1

.�1/k�1

k< C1

by the alternating series test. However, g is not Lebesgue integrable on [0; 1] becauseZ

[0;1]jgj dm D 16

kD1

1

kD C1:

So some improper Riemann integrable functions are not Lebesgue integrable. Observe that gC and g� has infiniteareas under the curves, so the Lebesgue integral for g cannot be defined.

129

For the case of an unbounded interval, say [0;C1/; take g.x/ D 16kD1

.�1/k�1

kX[k�1;k/.x/ D .�1/[x]

[x]C 1; where [x]

denotes the greatest integer less than or equal to x : AgainZ 1

0g.x/ dx D 16

kD1

.�1/k�1

k< C1 and

Z[0;C1/ jgj dm D16

kD1

1kD C1: (Exercise: Another example is

sin x

xonR:)

Appendix 1 to Step 5 : Proofs of Theorems and Corollaries

The two convergence theorems can be proved using a lemma due to Pierre Fatou in 1906.

Fatou’s Lemma. If f1; f2; : : : : A ! [0;C1/ are measurable functions, thenZ

Aliminfn!1 fn dm � liminf

n!1 ZA

fn dm:Since liminf

n!1 fn.x/ may be C1 for some x 2 A; we need to clarify the meaning ofZ

Aliminfn!1 fn dm: Let

S D fx 2 A : liminfn!1 fn .x/ D C1g and T D fx 2 A : liminf

n!1 fn.x/ 2 Rg: Since fn.x/ � 0 on A; the function mk.x/ Dinff fk .x/; fkC1.x/; : : :g exists as real numbers everywhere on A and hence measurable. Then

T D .liminfn!1 fn/�1.R/ D .sup mk/�1�.�1;C1/� D 1[

nD1

.sup mk /�1�.�1; n]� D 1[

nD1

1\kD1

m�1k

�.�1; n]�

is measurable and so is S D A n T : Now we can use the last theorem of step 4 to defineZA

liminfn!1 fn dm D sup

nZA� dm : 0 � � � liminf

n!1 fn ; � simpleo:

So in case m.S/ > 0; the integral will equal C1:Proof of Fatou’s Lemma. Let f D liminf

n!1 fn on A: Consider a nonnegative simple function � D r6iD1

ciXSi � f on A

with the Si pairwise disjoint. Then each ci � f D liminfn!1 fn on Si : For each t 2 .0; 1/; w 2 Si; tci < liminf

n!1 fn.w/: This

implies tci < inff fk .w/; fkC1.w/; : : :g for some k: Then w 2 Bik D 1\jDk

fx 2 Si : tci < fj .x/g: Now Bi1 � Bi2 � � � �and their union contains everyw 2 Si :Hence their union is Si :By the monotone set theorem, lim

n!1m.Bin / D m. 1[nD1

Bin/D m.Si /: Define n D r6iD1

tciXBin :Now forw 2 Si ; n.w/ D �tci if w 2 Bin

0 if w 2 Si n Bin� � fn.w/ ifw 2 Bin

fn.w/ ifw 2 Si n Bin: So n � fn on every Si (hence also on A). We haveZ

Afn dm � Z

A n dm D r6

iD1tcim.Bin /:

Taking the limit infima of both sides, we get

liminfn!1 Z

Afn dm � liminf

n!1 r6iD1

tci m.Bin/ D tr6

iD1ci m.Si / D t

ZA� dm:

Now let t go to 1, then we get liminfn!1 Z

Afn dm � Z

A� dm: Taking supremum over all nonnegative simple � � f; we

get

liminfn!1 Z

Afn dm � sup

�ZA� dm : 0 � � � f; � simple

� D ZA

f dm D ZA

liminfn!1 fn dm:

130

Proof of MCT. Since 0 � fn � f a.e. on A for n D 1; 2; 3; : : : ; so we have 0 � ZA

fn dm � ZA

f dm: Hence

limsupn!1 Z

Afn dm � Z

Af dm: By Fatou’s lemma,

ZA

f dm D ZA

liminfn!1 fn dm � liminf

n!1 ZA

fn dm: Since liminf �limsup; the two inequalities above combine to give lim

n!1 ZAfn dm D Z

Af dm:

Proof of Corollary to MCT. The function f n D n6kD1

gk is measurable. Since limn!1 fn D 16

kD1gk a.e. and fn � fnCgnC1 D

fnC1; by MCT,ZA

16kD1

gk dm D ZA

limn!1 fn dm D lim

n!1 ZAfn dm D lim

n!1 n6kD1

ZA

gk dm D 16kD1

ZA

gk dm:Proof of LDCT. Since j fn j � g a.e. on A implies g C fn ; g � fn � 0 a.e. on A; by Fatou’s lemma,Z

Ag dmCZ

Af dm D Z

A.gC f / dm D Z

Aliminfn!1 .gC fn/ dm � liminf

n!1 ZA.gC fn/ dm D Z

Ag dmCliminf

n!1 ZA

fn dm:ZA

g dm�ZA

f dm D ZA.g� f / dm D Z

Aliminfn!1 .g� fn/ dm � liminf

n!1 ZA.g� fn/ dm D Z

Ag dm�limsup

n!1 ZA

fn dm:Cancelling

ZA

g dm <1; we getZ

Af dm � liminf

n!1 ZA

fn dm � limsupn!1 Z

Afn dm � Z

Af dm: The result follows.

Proof of Corollary to LDCT. The function fn D n6kD1

gk is measurable. Now16

kD1

ZAjgkj dm <1 implies

16kD1

jgk.x/jconverges a.e. on A: (This is because if S D fx 2 A :

16kD1

jgk.x/j D C1g has nonzero Lebesgue measure, then we get

the contradiction that16kD1

ZAjgkj dm D liminf

n!1 ZA

n6kD1

jgkj dm � ZA

liminfn!1 n6

kD1jgk j dm D Z

A

16kD1

jgkj dm � ZS

16kD1

jgkj dm D C1:/So lim

n!1 fn D 16kD1

gk exists a.e. and j fn j � g D 16nD1

jgkj a.e. on A: By MCT,Z

Ajgj dm D 16

kD1

ZAjgkj dm <1; so g is

Lebesgue integrable on A: By LDCT,ZA

16kD1

gk dm D ZA

limn!1 fn dm D lim

n!1 ZAfn dm D lim

n!1 n6kD1

ZA

gk dm D 16kD1

ZA

gk dm:Appendix 2 to Step 5 : Higher Dimensional Lebesgue Measure and Integral

ForRn; there are also Lebesgue measure and integral. Define an n-dimensional open interval (or open rectangularcell) to be any set of the form

I D .a1; b1/ � .a2; b2/� � � � � .an; bn/ D f.x1; x2; : : : ; xn/ : ai < xi < bi for i D 1; 2; : : : ; ng:Similarly, an n-dimensional semi-open interval is any set of the form

J D [a1; b1/ � [a2; b2/ � � � � � [an; bn/ D f.x1; x2; : : : ; xn/ : ai � xi < bi for i D 1; 2; : : : ; ng:As usual, the n-dimensional volume of these intervals is �.I / D �.J / D .b1 � a1/.b2 � a2/ � � � .bn � an/: In Rn;open sets are sets that are unions of n-dimensional open intervals, closed sets are the complements of open sets andcompact sets are closed and bounded sets. (It is a good exercise for the reader to define the volume of an open set Sby first proving a structure theorem that S is a countable disjoint union of semi-open intervals, then taking the sum

131

of the volumes of the semi-open intervals to be the volume of S: The volume of compact and closed sets can bedefined as before.) However, to develop the theory so it can be used in more general settings, it is common to take theCaratheodory approach. For a set B inRn; we define the outer measure of the set B by

m�n.B/ D inf

n 16kD1

�.Ik / : B � 1[kD1

Ik; Ik open intervalo:

Caratheodory’s theorem suggests that for a set S inRn;we can define S to be (Lebesgue) measurable if and only if forevery set X inRn; m�

n.X / D m�n.X \ S/Cm�

n.X n S/: For a measurable set S;we define the Lebesgue measure mn ofS to be its outer measure, i.e. mn.S/ D m�

n.S/:Next we define a function f : A ! R to be (Lebesgue) measurable if and only if A is a measurable set in Rn

and f �1..a; b// is a measurable set for every a; b 2 R: For a nonnegative measurable functions f : A ! [0;C1/;we define the Lebesgue integral on A as in the last theorem in step 4 via simple functions. All the properties of outermeasure, measurable sets, measurable functions, Lebesgue integral and convergence theorems are true by essentiallythe same proofs.

There are additional facts about these Lebesgue measures and integrals.

Theorem. If X is a measurable set in Rn and Y is a measurable set in Rk; then X � Y is a measurable set inRnCk :If f : X � Y ! R is a measurable function, then for each x 2 X; fx : Y ! R defined by fx .y/ D f .x; y/ is ameasurable function and similarly, for each y 2 Y; f y : X ! Rdefined by f y.x/ D f .x; y/ is also a measurablefunction.

Furthermore, we have the following theorems for computing high dimensional integrals by multiple integralsas in multivariable calculus. The first one is for nonnegative valued functions and the second one is for real valuedfunctions.

Fubini-Tonelli’s Theorem. If f : X � Y ! [0;C1/ is measurable, then the functions � : X ! [0;C1/ defined by�.x/ D ZY

fx dmk and : Y ! [0;C1/ defined by .y/ D ZX

f y dmn are measurable andZX

�ZY

fx dmk

�dmn D Z

X�Yf dmnCk D Z

Y

�ZX

f y dmn

�dmk :

Fubini’s Theorem. If f : X � Y ! R is Lesbesgue integrable (i.e.Z

X�Yj f j dmnCk <1), then similar conclusions

as in the Fubini-Tonelli’s theorem are true.

132

Chapter 13. Vector-valued Functions

In this chapter we will be working with the n-dimensional vector space Rn and studying Rm-valued functionsF : Rn ! Rm :Conventions. An element of Rn may be viewed as a point .x1; : : : ; xn/ or as a vector from the origin to the point or

as a column matrix

0@ x1:::xn

1A depending on the context. We will let O D .0; : : : ; 0/ denote the origin ofRn :Definitions. (1) Define the length (or norm) of v D .x1; : : : ; xn/ 2 Rn to be kvk D qx2

1 C � � � C x2n :

(2) For v D .x1; : : : ; xn/; w D .y1; : : : ; yn/ 2 Rn; define the distance between v andw to be

d.v;w/ D kv �wk D p.x1 � y1/2 C � � � C .xn � yn/2:(3) For v D .x1; : : : ; xn/ 2 Rn and r > 0;

S.v; r/ D fw 2 Rn : d.v;w/ D rg is the sphere centered at v of radius r,

B.v; r/ D fw 2 Rn : d.v;w/ < rg is the open ball centered at v of radius r;B.v; r/ D fw 2 Rn : d.v;w/ � rg is the closed ball centered at v of radius r:

(The open ball B.v; r/ is also called the r-neighborhood of v and is also denoted by Nr .v/ in some books.)

Theorem (Triangle Inequality). For u; v; w 2 Rn; we have ku � wk � ku � vk C kv � wk: (Exercise.)

A vector-valued function is a function F : Rn ! Rm; which assigns to every .x1; : : : ; xn/ 2 Rn a uniquevalue F.x1; : : : ; xn/ D .y1; : : : ; ym/ 2 Rm : Each coordinate yi depends on x1; : : : ; xn: Hence each yi is a functionfi : Rn ! R, i.e. yi D fi .x1; : : : ; xn/: The function fi is called the i-th coordinate function of F for i D 1; : : : ;m:Conventions. We will use capital letters for vector-valued functions and the corresponding small letters for theircoordinate functions.

Example. Let F : R2 ! R3 be given by F.x; y/ D .x cos y; x sin y; 0/: Then the coordinate functions of F aref1; f2; f3 : R2 ! Rgiven by f1.x; y/ D x cos y; f2.x; y/ D x sin y; f3.x; y/ D 0: So

F.x; y/ D . f1.x; y/; f2.x; y/; f3.x; y//:For convenience, this is sometimes written as F D . f1; f2; f3/:

Below we will use the abbreviation “iff” for “if and only if”.

Definitions. For p 2 Rn and F : Rn ! Rm; we define limq!p

F.q/ D L iff limkq�pk!0kF.q/� Lk D 0; i.e. for every" > 0; there exists a � > 0 such that 0 < kq � pk < � implies kF.q/ � Lk < ": We say F is continuous at p iff

limq!p

F.q/ D F.p/: We say F is continuous on a set S iff F is continuous at every p 2 S:Theorem (Useful Inequality). For i D 1; : : : ;m; we have jyi j � q

y21 C � � � C y2

m � jy1j C � � � C jym j: (Just squareall three expressions and compare.)

Limit Theorem. Let L D .l1; : : : ; lm/ 2 Rm and p 2 Rn : For a function F : Rn ! Rm; we have

limq!p

F.q/ D L if and only if limq!p

fi .q/ D li for all i D 1; : : : ;m:133

In the case L D F.p/; that means F is continuous at p if and only if all m coordinate functions fi : Rn ! R arecontinuous at p:Proof. Let p 2 Rn be fixed. For q 2 Rn; let x D kq � pk and F.q/ � L D .y1; : : : ; ym/: (Note both x and the yi’sare functions of q:) Taking the i-th coordinate of both sides, we get fi .q/ � li D yi for i D 1; : : : ;m:

If the limit of F is L ; then by definition,0 D limkq�pk!0kF.q/ � Lk D lim

x!0

qy2

1 C � � � C y2m :By the useful inequality

and sandwich theorem, we get 0 D limx!0

jyi j D limkq�pk!0j fi .q/ � li j for i D 1; : : : ;m: So lim

q!pfi .q/ D li : Therefore,

the limit of fi is li for i D 1; : : : ;m:Conversely, if the limit of fi is li for all i D 1; : : : ;m; then 0 D limkq�pk!0

j fi .q/ � li j D limx!0

jyi j for i D 1; : : : ;m:Adding these m limits, we get 0 D lim

x!0.jy1j C � � � C jymj/: By the useful inequality and sandwich theorem, we get

0 D limx!0

qy2

1 C � � � C y2m D limkq�pk!0

kF.q/� Lk: Therefore, the limit of F is L :Example. In the previous example, the coordinate functions f1.x; y/ D x cos y; f2.x; y/ D x sin y; f3.x; y/ D 0 arecontinuous at every .x; y/ 2 R2; so F D . f1; f2; f3/ is continuous onR2 by the theorem above.

Question. What does it mean to say F : Rn ! Rm is differentiable? If it is differentiable, what is the derivative of Fat a point p 2 Rn?

As in the theorem above, it will turn out that F : Rn ! Rm is differentiable at a point p 2 Rn if and only if everycoordinate function fi : Rn ! R is differentiable at p: For every fi ; we know the meaning of the partial derivatives@ fi@x1

; : : : ; @ fi@xn: How are these “partial” derivatives related to the derivative of fi ? Geometrically, for n D 1; the

derivative is the slope of the tangent line. So for n D 2; we may think the derivative should measure the “slopes ofthe tangent plane.” Since two lines are needed to determine a plane, so the derivative in the case n D 2 should involvetwo numbers. In fact, it will turn out that the derivative of a vector-valued function is a linear transformation and thepartial derivatives are the matrix entries of the linear transformation. First we will do a review.

Review of Some Linear Algebra and Calculus Facts

(1) Recall a linear transformation T : Rn ! Rm is a function satisfying

T .v Cw/ D T .v/ C T .w/ and T .cv/ D cT .v/ for all c 2 R; v; w 2 Rn :With respect to the usual basis

e1 D .1; 0; 0; 0; : : : ; 0/; e2 D .0; 1; 0; 0; : : : ; 0/; e3 D .0; 0; 1; 0; : : : ; 0/; : : :ofRn and Rm; the linear transformation T has a matrix representation

T

0@ x1:::xn

1A D 0@ c11 � � � c1n::: : : : :::cm1 � � � cmn

1A0@ x1:::xn

1A for every

0@ x1:::xn

1A 2 Rn :To get the j -th column of the matrix, we apply T to ej : We have

T .ej / D 0@ c1 j:::cm j

1A :Taking the i-th coordinate of both sides, we get ti.ej / D ci j for all i D 1; : : : ;m and j D 1; : : : ; n:

134

(2) The equation of a nonvertical line passing through .x0; y0/ 2 R2 is y D y0Cm.x � x0/: Observe that the functionT : R! Rdefined by T .v/ D mv is a linear transformation. In terms of T; the equation of the line can be putin the form y D y0 C T .x � x0/:Similarly, using a normal vector, we get the equation of a nonvertical plane passing through .x0; y0; z0/ 2 R3 isz D z0 C c1.x � x0/ C c2.y � y0/: In terms of the linear transformation T : R2 ! Rdefined by T .v1; v2/ Dc1v1 C c2v2 the equation of the plane can be put in the form z D z0 C T .x � x0; y � y0/:In Rn; there are translates of .n � 1/-dimensional linear subspaces passing through any point analogous to linesin R2; planes in R3; etc. These are called “hyperplanes.” Using a normal vector, the equation of a nonverticalhyperplane passing through .x0; y0; : : : ; z0; w0/ 2 Rn can be put in the formw D w0C T .x� x0; y� y0; : : : ; z�z0/; where T : Rn�1 ! R is a linear transformation of the form T .v1; : : : ; vn�1/ D c1v1 C � � � C cn�1vn�1:

(3) For F : R! Rto be differentiable at a; the definition requires the existence of limx!a

F.x/ � F.a/x � a

D F 0.a/: This

is equivalent to the existence of a nonvertical tangent line to the graph of F at .a; F.a//: The equation of this lineis y D F.a/ C F 0.a/.x � a/ (or y D F.a/ C T .x � a/ if we let T .v/ D F 0.a/v:) Now observe that

limx!a

F.x/ � F.a/x � a

D F 0.a/ () limx!a

�F.x/ � F.a/

x � a� F 0.a/� D 0() lim

x!a

���� F.x/ � F.a/ � F 0.a/.x � a/x � a

���� D 0() limx!a

jF.x/ � F.a/ � T .x � a/jjx � aj D 0:So F is differentiable at a if and only if there is a linear transformation T : R! Rsuch that

limx!a

jF.x/ � F.a/� T .x � a/jjx � aj D 0:In fact, the matrix of T is . F 0.a/ / ; which contains the slope of the tangent line to the graph of F at .a; F.a//:Similarly, let F : R2 ! Rbe a function. We expect F differentiable at a point to be equivalent to the existenceof a tangent plane at the point. From multivariable calculus, we learn that if the graph of F has a nonverticaltangent plane at .a1; a2; F.a1; a2//; then the equation of this plane is z D F.a1; a2/ C c1.x � a1/ C c2.y �a2/; where c1 D @F@x

.a1; a2/; c2 D @F@y.a1; a2/: (Note c1 is the slope of the plane in the x-direction through.a1; a2; F.a1; a2// and c2 is the slope of the plane in the y-direction through the point.) So we expect F to be

differentiable at .a1; a2/ if and only if there exists a linear transformation T : R2 ! R such that

lim.x;y/!.a1;a2/ F.x; y/ � F.a1; a2/ � T

�x � a1

y � a2

� k.x � a1; y � a2/k D 0

and we also expect the matrix of T to be . @F@x .a1; a2/ @F@y .a1; a2/ / ; which contains all the slope informations ofthe tangent plane. In general, this is how differentiability of vector-valued functions is defined.

Definitions. A vector-valued function F : Rn ! Rm is differentiable at a D .a1; : : : ; an/ 2 Rn iff there is a lineartransformation T : Rn ! Rm such that

limx!a

kF.x/ � F.a/� T .x � a/kkx � ak D 0

�or lim

t!O

kF.a C t/� F.a/ � T .t/kktk D 0 by substituting t D x � a

� :If such a T exists, we call it the derivative of F at a and denoted it by DF.a/:Question. How do we come up with the linear transformation to check differentiability at a point? (This is answeredby the next theorem.)

135

Differentiation Theorem. If F : Rn ! Rm is differentiable at a D .a1; : : : ; an/ 2 Rn and f1; : : : ; fm : Rn ! Rare

the coordinate functions of F; then all the partial derivatives@ fi@x j

.a/ exist for i D 1; : : : ;m and j D 1; : : : ; n: In fact,

the matrix of DF.a/ with respect to the usual bases is0B@ @ f1@x1.a/ : : : @ f1@xn

.a/::: : : : :::@ fm@x1.a/ : : : @ fm@xn

.a/1CA :(This matrix is called the Jacobian matrix of F at a and is denoted by F 0.a/:)Proof. Since F is differentiable at a; there exists a linear transformation T D DF.a/ satisfying the condition in thedefinition of differentiable function at a point. Let x D a C hej D .a1; : : : ; aj C h; : : : ; an/; then

0 D limx!a

kF.x/ � F.a/� T .x � a/kkx � ak D limh!0

F.a C hej / � F.a/ � hT .ej /h

D limh!0

F.a C hej /� F.a/h

� T .ej / ) limh!0

F.a C hej / � F.a/h

D T .ej /:Taking the i-th coordinate of both sides, we get@ fi@x j

.a/ D limh!0

fi .a C hej / � fi .a/h

D ti .ej /:Since ti.ej / is the .i; j /-entry of the matrix of T; @ fi@x j

.a/ exists and its value is the .i; j /-entry of T D DF.a/ for

i D 1; : : : ;m and j D 1; : : : ; n:Example. We will show that the function F : R2 ! Rdefined by F.x; y/ D x2C y2 is differentiable at .0; 0/:We have@ f@x

D 2x and@ f@y

D 2y: By the differentiation theorem, if F is differentiable at .0; 0/; then T D DF.0; 0/ has matrix. @F@x .0; 0/ @F@y .0; 0/ / D . 0 0 / : Now lim.x;y/!.0;0/ kF.x; y/ � F.0; 0/ � T .x � 0; x � 0/kk.x; y/ � .0; 0/k D lim.x;y/!.0;0/px2 C y2 D 0:So, F is differentiable at .0; 0/:Remarks. (1) The contrapositive of the differentiation theorem asserts that if some partial derivative does not exist ata; then the function is not differentiable at a:(2) The converse of the differentiation theorem is false. It may happen that all partial derivatives exist at a point, butthe function is not differentiable there. For example, take

F.x; y/ D n0 if x D 0 or y D 0,1 otherwise.

Since F.x; 0/ D 0 and F.0; y/ D 0; @F@x.0; 0/ D d

dxF.x; 0/���

xD0D 0 and

@F@y.0; 0/ D d

dyF.0; y/���

yD0D 0: By the

differentiation theorem, if F is differentiable, then T D DF.0; 0/ has matrix . @F@x .0; 0/ @F@y .0; 0/ / D . 0 0 / :Since along .x; x/ ! .0; 0/; kF.x; x/ � F.0; 0/� T .x � 0; x � 0/kk.x; x/ � .0; 0/k D 1p

2x2does not have a limit. So, F is not

differentiable at .0; 0/:Question. If F : Rn ! Rm is differentiable at a 2 Rn; is it continuous at a? (Yes, we need a lemma before we canprove it.) In remark (2), it is easier to see F is not differentiable at .0; 0/ by noticing that F is not continuous at .0; 0/than checking the definition of differentiability.

136

Definitions. Let M be the m � n matrix of a linear transformation T : Rn ! Rm with entries ci j .i D 1; : : : ;mI j D1; : : : ; n/ under the usual bases of Rn and Rm: Writing all the entries ci j in a row, we may consider M 2 Rmn and

define the norm of T and the norm of M to be kTk D kMk D sm6

iD1

n6jD1

c2i j :

Continuity Lemma. For every linear transformation T : Rn ! Rm; we have kT .v/k � kTkkvk for everyv D .x1; : : : ; xn/ 2 Rn : Consequently, limv!0T .v/ D 0 by the sandwich theorem.

Proof. Let e1; : : : ; en be the usual basis of Rn and ci j D ti.ej / be the entries of the matrix of T : By the triangleinequality and the Cauchy-Schwarz inequality,kT .v/k D T

� n6iD1

xi ei� D n6

iD1xi T .ei / � n6

iD1jxi jkT .ei /k � r n6

iD1x2

i

rn6

iD1kT .ei /k2 D kvks m6

iD1

n6jD1

c2i j D kTkkvk:

Theorem. If F : Rn ! Rm is differentiable at a 2 Rn; then F is continuous at a:Proof. Since F is differentiable at a, by definition, lim

x!a

kF.x/ � F.a/ � T .x � a/kkx � ak D 0: So

limx!a

kF.x/ � F.a/ � T .x � a/k D limx!a

kF.x/ � F.a/� T .x � a/kkx � ak kx � ak D 0:By the lemma, lim

x!aT .x � a/ D 0:By the triangle inequality,kF.x/�F.a/k � kF.x/�F.a/�T .x�a/kCkT .x�a/k:

By the sandwich theorem, limx!a

kF.x/ � F.a/k D 0: Therefore, limx!a

F.x/ D F.a/:Question. Are there easier methods to see if a function F : Rn ! Rm is differentiable at a point than to check thedefinition? (The C1 theorem below provides an easier way to check differentiabilty.)

Theorem. F : Rn ! Rm is differentiable at a 2 Rn if and only if all coordinate functions fi : Rn ! R of F aredifferentiable at a: Also, DF.a/ D .D f1.a/; : : : ; D fm.a// in the case the functions are differentiable.

Proof. Since the i-th coordinate ofF.x/ � F.a/ � T .x � a/kx � ak is

fi .x/ � fi .a/ � ti.x � a/kx � ak ; by the limit theorem,

limx!a

F.x/ � F.a/ � T .x � a/kx � ak D O if and only if limx!a

fi .x/ � fi .a/ � ti .x � a/kx � ak D O for all i D 1; : : : ;m:So F is differentiable at a with DF.a/ D T if and only if all fi ’s are differentiable at a with D fi .a/ D ti ; i.e.

DF.a/ D T D .t1; : : : ; tm/ D .D f1.a/; : : : ; D fm .a//:Definition. F : Rn ! Rm is continuously differentiable (or C1) at a 2 Rn iff all partial derivatives

@ fi@x jexist in some

neighborhood of a (i.e. in some open ball B.a; r/) and are continuous at a: Furthermore, we say F is C1 near a iff Fis C1 at every point in some open ball B.a; r0/: (In more advanced courses, we also say F is Cn near a iff all partialderivatives of order n or below are continuous in a neighborhood of a:)C1 Theorem. If F : Rn ! Rm is C1 at a D .a1; : : : ; an/; then F is differentiable at a:Proof. By the previous theorem, it suffices to show each coordinate function f : Rn ! R is differentiable at a: Forthe derivative, the differentiation theorem suggests the linear transformation T : Rn ! Rdefined by

T .h1; : : : ; hn/ D . @ f@x1.a/ : : : @ f@xn

.a/ /0@ h1:::hn

1A D @ f@x1.a/h1 C � � � C @ f@xn

.a/hn :137

Since f is C1 at a; all@ f@x j

exist in some neighborhood B.a; r/: For x D .x1; : : : ; xn/ 2 B.a; r/;f .x/ � f .a/ D f .x1; : : : ; xn/ � f .a1; : : : ; an/D �

f .x1; x2; : : : ; xn/ � f .a1; x2; : : : ; xn/� C � f .a1; x2; : : : ; xn/ � f .a1; a2; : : : ; xn/�C � � � C � f .a1; a2; : : : ; an�1; xn/� f .a1; a2; : : : ; an�1; an/�:Applying the mean value theorem, we have

f .x/ � f .a/ D @ f@x1.c1/.x1 � a1/ C � � � C @ f@xn

.cn/.xn � an/;where cj is some point on the segment joining .a1; : : : ; x j; : : : ; xn/ and .a1; : : : ; aj; : : : ; xn/: Thenj f .x/ � f .a/ � T .x � a/jkx � ak D ���� n6

jD1

� @ f@x j.cj /� @ f@x j

.a/�.x j � aj /����kx � ak � n6jD1

���� @ f@x j.cj / � @ f@x j

.a/���� jx j � aj jkx � ak| {z }�1

:As x ! a; we have cj ! a: Since

@ f@x jis continuous at a for j D 1; : : : ; n; the right side of the inequality will go to

0 as x ! a: By the sandwich theorem, the left side will also go to 0 as x ! a: Therefore, f is differentiable at a:Examples. (1) Let F : R2 ! R2 be defined by F.x; y/ D .x2 C y2; sin x y/: Is F differentiable at .2; �/? If so, whatare F 0.2; �/ and DF.2; �/?Solution. The coordinate functions of F are f1.x; y/ D x2 C y2 and f2.x; y/ D sin x y: Now@ f1@x

D 2x; @ f1@yD 2y; @ f2@x

D y cos x y; @ f2@yD x cos x y

are continuous onR2 (in particular at .2; �/). So F is C1 at .2; �/: By the C1 theorem, F is differentiable at .2; �/:By the differentiation theorem,

F 0.2; �/ D � @ f1@x .2; �/ @ f1@y .2; �/@ f2@x .2; �/ @ f2@y .2; �/� D �4 2�� 2

� ;DF.2; �/� x

y

� D �4 2�� 2

��xy

� D �4x C 2�y�x C 2y

� :(2) Let T : Rn ! Rm be a linear transformation. Is T differentiable on Rn? If so, what is the derivative of T at anarbitrary a 2 Rn?

Solution. (If T is differentiable, then the derivative of T is also a linear transformation. What would you guess thederivative is?) We will show T is differentiable at every a 2 Rn and DT .a/ D T : Since T .x � a/ D T .x/ � T .a/; so

limx!a

kT .x/ � T .a/ � T .x � a/kkx � ak D 0; i.e. DT .a/ D T :Remarks. The addition function P : R2 ! Rdefined by P.x; y/ D x C y is a linear transformation. The subtractionfunction Q : R2 ! R defined by Q.x; y/ D x � y is also a linear transformation. So both are differentiable andD P.a/ D P; DQ.a/ D Q for every a 2 Rn :

138

(3) Show the multiplication function R : R2 ! Rdefined by R.x; y/ D x y is differentiable and find DR.a/:Solution. (Note R is not a linear transformation!) We will check that R is C1 at every a 2 R2: Since

@R@xD y and@R@y

D x are continuous onR2; by the C1 theorem, R is differentiable everywhere. By the differentiation theorem,

R0.a1; a2/ D . @R@x .a1; a2/ @R@y .a1; a2/ / D . a2 a1 / :So DR.a/� x

y

� D . a2 a1 /� xy

� D a2x C a1y:(4) Show the division function S.x; y/ D x=y is differentiable for y 6D 0 and find DS.a/ for a D .a1; a2/ 2 R2 witha2 6D 0: (Although S : R� .Rn f0g/! R; the definition of differentiation is the same!)

Solution. For a D .a1; a2/ 2 R2 with a2 6D 0; we will check S is C1 at a: This is true because the partial derivatives@S@xD 1

yand

@S@yD � x

y2exist in some neighborhood of a and are continuous at a: Now

S0.a/ D . @S@x .a/ @S@y .a/ / D � 1a2

� a1

a22

� :So DS.a1; a2/� x

y

� D � 1a2

� a1

a22

� � xy

� D a2x � a1y

a22

:Theorem (Chain Rule). If F : Rn ! Rm is differentiable at a 2 Rn and G : Rm ! Rk is differentiable atb D F.a/ 2 Rm; then G � F : Rn ! Rk is differentiable at a and

D.G � F/.a/ D DG�F.a/� � DF.a/| {z }

composition of linear transformations

; .G � F/0.a/ D G0�F.a/�F 0.a/| {z }matrix multiplication

:Special Case: If F : R2 ! R2; F.s; t/ D �

x.s; t/; y.s; t/�; and G : R2 ! R are differentiable, then H D G � F :R2 ! R is differentiable and. @H@s@H@t / D H 0.s; t/ D G0�F.s; t/�F 0.s; t/ D . @G@x

@G@y /� @x@s@x@t@y@s@y@t

� :So

@H@sD @G@x

@x@sC @G@y

@y@sand

@H@tD @G@x

@x@tC @G@y

@y@t:

Proof of Chain Rule. Let b D F.a/; h D F.x/ � F.a/ D F.x/ � b; T D DG�F.a/� and S D DF.a/: Since F is

differentiable at a; so I D h � S.x � a/kx � ak D F.x/ � F.a/ � S.x � a/kx � ak ! O as x ! a:Also, since G is differentiable

at b; J D G.b C h/ � G.b/ � T .h/khk ! O as h ! O: For h 6D O; let

K D G � F.x/ � G � F.a/ � DG�F.a/� � DF.a/.x � a/D G.b C h/ � G.b/ � T

�S.x � a/�D G.b C h/ � G.b/ � T .h/| {z }DJkhk C T

�h � S.x � a/�:| {z }Dkx�akT .I/

By the triangle inequality, we havekKkkx � ak � (kJk khkkx � ak C kT .I /k if h 6D 0kT .I /k if h D 0.

Nowkhkkx � ak D kF.x/ � F.a/kkx � ak � kF.x/ � F.a/� S.x � a/kkx � ak C kS.x � a/kkx � ak � kIk C kSk:139

As x ! a; we get h D F.x/ � F.a/! O; J ! O; I ! O; T .I /! O andkhkkx � ak bounded, so

kKkkx � ak ! 0

by the sandwich theorem. Therefore, G � F is differentiable at a:Theorem (Differentiation Formulas). If f; g : Rn ! Rare differentiable at a 2 Rn; then f C g; f � g; f g and f=g(provided g.a/ 6D 0) are differentiable at a and

D. f � g/.a/ D D f .a/ � Dg.a/;D. f g/.a/ D g.a/Df .a/C f .a/Dg.a/;D� f

g

�.a/ D g.a/Df .a/� f .a/Dg.a/g.a/2 :

Equality of Second Derivatives

We learned in multivariable calculus that@2 f@x@y

D @2 f@y@xfrequently. Here is a theorem about when this happens.

Second Derivative Theorem. Let f : R2 ! Rbe such that@ f@x; @ f@y

; @2 f@y@xexist at every point of some ball B.p; r/

and@2 f@y@x

is continuous at p: Then@2 f@x@y

.p/ exists and@2 f@x@y

.p/ D @2 f@y@x.p/:

Proof. Let p D .a; b/ andw D @2 f@y@x.p/:By checking the definition of limit,we will prove

@2 f@x@y.p/ D @@x

�@ f@y

�.a; b/D limh!0

.@ f=@y/.a C h; b/� .@ f=@y/.a; b/h

D w:For every " > 0; since lim

p0!p

@2 f@y@x.p0/ D w by continuity at p; there is �0 > 0 such that kp0 � pk < �0 implies��� @2 f@y@x

.p0/ �w��� < "2: Let � D min.�0; r/: Consider any point q D .aCh; bC k/ in B.p; �/:Applying the mean value

theorem to u.x/ D f .x; b C k/ � f .x; b/ and v.y/ D @ f@x.a0; y/ below, we have1 D f .a C h; bC k/ � f .a C h; b/� f .a; b C k/ C f .a; b/D u.a C h/ � u.a/ D u0.a0/h D �@ f@x

.a0; bC k/ � @ f@x.a0; b/�hD �v.b C k/ � v.b/�h D v0.b0/kh D @2 f@y@x

.a0; b0/kh

for some p0 D .a0; b0/ in B.p; �/ with a0 betwee a and a C h and b0 between b and bC k: Then��� 1kh� w��� < "

2: Note

limk!0

1kD @ f@y

.a C h; b/� @ f@y.a; b/: So 0 < jhj < � implies��� .@ f=@y/.a C h; b/ � .@ f=@y/.a; b/

h� w��� D lim

k!0

��� 1kh�w��� � "

2< ":

Remarks. If@2 f@y@x

is not continuous at p; then the conclusion may become false, i.e. the second derivatives above may

not be equal even if they exist. Here is an example. Let p D .0; 0/: Define f .0; 0/ D 0 and f .x; y/ D x y.x2 � y2/x2 C y2

for .x; y/ 6D .0; 0/:140

If .x; y/ 6D .0; 0/; then@ f@x.x; y/ D y.x4 C 4x2 y2 � y4/.x2 C y2/2 and

@ f@y.x; y/ D x.x4 � 4x2 y2 � y4/.x2 C y2/2 : At .0; 0/;@ f@x

.0; 0/ D limh!0

f .h; 0/� f .0; 0/h

D 0 and@ f@y.0; 0/ D lim

k!0

f .0; k/ � f .0; 0/k

D 0:So

@ f@x; @ f@y

exist everywhere. Next, if .x; y/ 6D .0; 0/; then@2 f@y@x.x; y/ D x6 C 9x4 y2 � 9x2y4 � y6.x2 C y2/3 D @2 f@x@y

.x; y/;which has limit 1 as .x; 0/ ! .0; 0/ and has limit �1 as .0; y/ ! .0; 0/: So these second derivatives cannot

be continuoust at .0; 0/: Finally, we have@2 f@x@y

.0; 0/ D limh!0

.@ f=@y/.h; 0/ � .@ f=@y/.0; 0/h

D limh!0

h � 0

hD 1 and@2 f@y@x

.0; 0/ D limk!0

.@ f=@x/.0; k/ � .@ f=@x/.0; 0/k

D limk!0

�k � 0

kD �1:

Inverse Functions

In one variable calculus, if y D f .x/ is differentiable and has an inverse x D g.y/; which is also differentiable,then

1 D d

dxx D d

dx.g � f /.x/ D g0. f .x// f 0.x/ D dx

dy

dy

dx) dx

dyD 1

.dy

dx:

For example, if we restrict y D f .x/ D sin x to the interval .��2; �

2/; then it has an inverse x D g.y/ D arcsin y: So

we getd

dyarcsin y D dx

dyD 1

dy

dx

D 1cos x

D � 1p1� y2

:However, if we restrict y D f .x/ D sin x to an open interval containing

�2; then there cannot be any inverse function,

since about x D �2; the function is not injective.

In multivariable calculus, we are familiar with changing coordinates, such as F.r; �/ D .x; y/; where x D r cos �and y D r sin �: We can invert by G.x; y/ D .r; �/; where r D p

x2 C y2 and � D arctan.y=x/; provided x 6D 0:Question. For a function F : Rn ! Rn and q 2 Rn; how can we know if the function has an inverse function near q?

If so, what is the analogous formula fordx

dyD 1

.dy

dx?

As the tangent hyperplane at q is close to the graph near q; the injectivity of F near q should be related to theinjectivity of DF.q/; which is equivalent to det F 0.q/ 6D 0: If F�1 exists, then F�1

�F.x/� D x : Applying the chain

rule, we can get the formula�F�1

�0�F.x/�F 0.x/ D I; which implies

�F�1

�0�F.x/� D �

F 0.x/��1:Inverse Function Theorem. If F : Rn ! Rn is C1 near q 2 Rn and det F 0.q/ 6D 0; then there exists an open ballB.q; r/ such that F restricted to B.q; r/ has a C1 inverse function F�1 and

�F�1

�0�F.x/� D �

F 0.x/��1for every

x 2 B.q; r/:We will give a sketch of the proof of the inverse function theorem in the appendix.

Example. InR2; define .u; v/-coordinates by u D eyCx; v D ex� y;where .x; y/ is the usual rectangular coordinates.Is it possible to express .x; y/ as a differentiable function of .u; v/ near the origin .x; y/ D .0; 0/? If so, what are@x@u; @x@v ; @y@u

; @y@v when .x; y/ D .0; 0/; i.e. at .u; v/ D .1; 1/?141

Solution. Consider F : R2 ! R2 given by F.x; y/ D .u; v/: Now@u@x

D 1; @u@yD ey; @v@x

D ex ; @v@yD �1 are

continuous near q D .0; 0/: So F is C1 near q D .0; 0/: Also, det F 0.0; 0/ D ���� 1 e0

e0 �1

���� D �2 6D 0: By the inverse

function theorem, near q D .0; 0/; F has a C1 inverse function F�1.u; v/ D .x; y/: Now applying the formula in thetheorem at q D .0; 0/; we have� @x@u .1; 1/ @x@v .1; 1/@y@u .1; 1/ @y@v .1; 1/� D .F�1/0.F.q/| {z }D.1;1// D �

F 0.q/��1 D � @u@x .0; 0/ @u@y .0; 0/@v@x .0; 0/ @v@y .0; 0/��1 :Then

� @x@u .1; 1/ @x@v .1; 1/@y@u .1; 1/ @y@v .1; 1/� D �1 e0

e0 �1

��1 D 1�2

� �1 �e0�e0 1

�: So@x@u.1; 1/ D 1

2; @x@v .1; 1/ D 1

2; @y@u

.1; 1/ D 12

and@y@v .1; 1/ D �1

2:

Remark. Note in the above example,@x@u

6D 1.@u@x

:Implicit Differentiation

In one variable calculus, given an equation F.x; y/ D 0:We can finddy

dxby implicit differentiation,which assumes

the equation can be solved for y as a differentiable function of x : Below we will discuss implicit differentiation forvector-valued functions. Recall O D .0; : : : ; 0/ is the origin ofRm :Consider a function F : Rn ! Rm with coordinatefunctions f1; : : : ; fm : Rn ! R: The vector equation F.x/ D O is equivalent to the following system of m equationsin n variables:

f1.x1; : : : ; xn/ D 0;::: :::fm .x1; : : : ; xn/ D 0:

In the case when n > m; i.e. more variables than equations, it is a common experience that we may be able to solvefor m of the variables in terms of the other n � m variables. For example, in the linear case when fi .x1; : : : ; xn/ Dai1x1 C � � � C ainxn (the ai j ’s are constants), the system becomes

a11x1 C � � � C a1nxn D 0::: :::am1x1 C � � � C amnxn D 0:

Moving the terms involving xmC1; : : : ; xn to the right sides, we have

a11x1 C � � � C a1mxm D �.a1mC1xmC1 C � � � C a1nxn/;::: :::am1x1 C � � � C ammxm D �.ammC1xmC1 C � � � C amnxn/:

We can put this in matrix form as0@ a11 � � � a1m::: : : : :::am1 � � � amm

1A| {z }call this matrix M

0@ x1:::xm

1A D 0@ �.a1mC1xmC1 C � � � C a1nxn/:::�.ammC1xmC1 C � � � C amnxn/1A :142

We may solve for

0@ x1:::xm

1A by multiplying by M�1 if M is invertible (or equivalently if det M 6D 0). In that case, each

of the variables x1; : : : ; xm will be a function of the remaining n �m variables xmC1; : : : ; xn :To deal with the case when the functions are not linear, we should first look at the condition det M 6D 0 in terms

of the fi ’s. Since ai j D @ fi@x j; so

det M 6D 0 () det

0B@ @ f1@x1� � � @ f1@xm::: : : : :::@ fm@x1� � � @ fm@xm

1CA 6D 0:The matrix on the right above is important in the sequel. So we will introduce the following notation.

Notation. If f1; : : : ; fm are functions of the variables x1; : : : ; xn with n � m and p 2 Rn; then we denote@. f1; : : : ; fm/@.x1; : : : ; xm/ .p/ D det

0B@ @ f1@x1.p/ � � � @ f1@xm

.p/::: : : : :::@ fm@x1.p/ � � � @ fm@xm

.p/1CAand call it the Jacobian of f1; : : : ; fm with respect to x1; : : : ; xm at p:Question. In the general case, when a function F : Rn ! Rm (n > m) has a root p 2 Rn; i.e. F.p/ D O; and F isC1 near p (recall that means F is C1 in some open ball B.p; r/), can we solve the vector equation F.x/ D O for someof the variables near p? For example, we can solve the equation x2 C y2 � 25 D 0 near .3; 4/ to get y D p

25� x2:Near .3;�4/, we can solve it to get y D �p25� x2: However, near .5; 0/; y cannot be expressed as a differentiablefunction of x :

The following theorem tells us a sufficient condition when we can solve a (vector) equation for some of itsvariables as a differentiable function of the other variables.

Implicit Function Theorem. Let

(1) F : Rn ! Rm .n > m/ be C1 near p D .p1; : : : ; pn/ 2 Rn;(2) F.p/ D O and

(3)@. f1; : : : ; fm /@.x1; : : : ; xm/ .p/ 6D 0:Then there exist an open ball B.p; r/ and a C1 function G : B� ! Rm such that for every x D .x1; : : : ; xn/ 2

B.p; r/;F.x1; x2; : : : ; xn/ D O if and only if .x1; : : : ; xm/ D G.xmC1; : : : ; xn/:

(Here B� D f.xmC1; : : : ; xn/ 2 Rn�m : .x1; : : : ; xn/ 2 B.p; r/g is the projection of B.p; r/ onto the last n � mcoordinate space.)

In this case, we say x1; : : : ; xm are implicit functions of xmC1; : : : ; xn near p:Proof. Let k D n � m: Define H : RmCk ! RmCk by H .x; y/ D .F.x; y/; y/ for x D .x1; : : : ; xm/ 2 Rm; y D.xmC1; : : : ; xn/ 2 Rk : Since F is C1 near p; H is also C1 near p: From row operations, we can see that

det H 0.p/ D det�

F 0.p/O I

� D @. f1; : : : ; fm/@.x1; : : : ; xm/ .p/ 6D 0:By the inverse function theorem, there exists an open ball B.p; r/ such that H restricted to B.p; r/ has a C1 inversefunction H�1: In order to be the inverse of H near p; we must have H�1.x; y/ D .K .x; y/; y/ for some C1 functionK defined near H .p/ D .O; pmC1; : : : ; pn/:

143

Let � : RmCk ! Rm be the function�.x; y/ D x for x 2 Rm; y 2 Rk; then � �H D F:Define G.y/ D K .O; y/:Since K is C1 near H .p/ D .O; pmC1; : : : ; pn/; it follows G will be C1 near .pmC1; : : : ; pn/: Now if F.x; y/ D Ofor .x; y/ 2 B.p; r/; then H .x; y/ D .O; y/: So.x; y/ D H�1.O; y/ D .K .O; y/; y/ D .G.y/; y/ H) x D G.y/:Conversely, if x D G.y/; then

F.x; y/ D F.G.y/; y/ D F.K .O; y/; y/ D F.H�1.O; y// D .� � H /.H�1.O; y// D �.O; y/ D O:Examples. (1) Given .1; 2/ is a solution of the equation x6 C x2 y2 C y6 D 69: Find

dy

dxat .1; 2/:

Solution. (Step 2) We perform implicit differentiation to get

6x5 C 2x y2 C 2x2ydy

dxC 6y5 dy

dxD 0:

Solving fordy

dx; we get

dy

dxD �6x5 C 2x y2

2x2 y C 6y5: At .1; 2/; we get

dy

dx.1; 2/ D � 1

14:

Comments: To do implicit differentiation means to apply chain rule to the terms x2 y2 and y6 assuming y is adifferentiable function of x near .1; 2/: To see y is a function of x near .1; 2/, you may say “we can sketch the graphof the equation and see that a small piece of the graph near .1; 2/ is the graph of some function (because every verticalline cuts the piece in exactly one point.)” This is not too rigorous. Further, why is it differentiable? Also, it is hard todraw graphs when there are more than three variables!

(Step 1) To justify y is a differentiable function of x near .1; 2/; we can use the implicit function theorem as follow.Consider F : R2 ! R (here n D 2;m D 1) given by F.x; y/ D x6 C x2 y2 C y6 � 69 and p D .1; 2/: ThenF.1; 2/ D O: Near p; @F@x

D 6x5 C 2x y2 and@F@y

D 2x2 y C 6y5

are continuous. So F is C1 near p D .1; 2/: Now@F@y

.1; 2/ D 196 6D 0: By the implicit function theorem, near .1; 2/;there is a C1 (in particular, differentiable) function G such that

x6 C x2 y2 C y6 � 69 D 0 if and only if y D G.x/:(2) For the system wx yz D 0; w4 C x4 C y4 C z4 D 18; is it possible to express .x; y/ as a differentiable function of.w; z/ near the solution p D .w; x; y; z/ D .�1; 0; 1; 2/? If so, what is

@x@w.p/ and@y@w.p/?

Solution. (Step 1) Consider F : R4 ! R2 given by

F.w; x; y; z/ D . f1; f2/; where f1.w; x; y; z/ D wx yz and f2.w; x; y; z/ D w4 C x4 C y4 C z4 � 18:Since near p D .�1; 0; 1; 2/;@ f1@w D x yz; @ f1@x

D wyz; @ f1@yD wxz; @ f1@z

D wx y; @ f2@w D 4w3; @ f2@xD 4x3; @ f2@y

D 4y3; @ f2@zD 4z3

are continuous, so F is C1 near p: Now@. f1; f2/@.x; y/ .p/ D ���� @ f1@x .p/ @ f1@y .p/@ f2@x .p/ @ f2@y .p/ ���� D �����2 0

0 4

���� D �8 6D 0: By the implicit

function theorem, near p; there is a C1 (hence differentiable) function G such that F.w; x; y; z/ D 0 iff .x; y/ DG.w; z/:

144

(Step 2) Since near p; .x; y/ is a differentiable function of .w; z/; holding z constant in differentating the equationswith respect to w; we have

0 D @@w.0/ D @@w.wx yz/ D x yz C w @x@w yz Cwx@y@w z;

0 D @@w.18/ D @@w .w4 C x4 C y4 C z4/ D 4w3 C 4x3 @x@w C 4y3 @y@w :At p D .w; x; y; z/ D .�1; 0; 1; 2/; the equations simplifies to 0 D �2

@x@w.p/ and 0 D �4C 4@y@w.p/: Therefore,@x@w.p/ D 0 and

@y@w.p/ D 1:Appendix : Proof of the Inverse Function Theorem

Contractive Mapping Theorem. Let F : B.0; r/ ! B.0; r/ be such that kF.x/ � F.y/k � 12kx � yk for every

x; y 2 B.0; r/: Then F has a unique fixed point w 2 B.0; r/; i.e. F.w/ D w: (Here1

2may be replaced by any

positive constant less than 1.)

Proof. Since F.x/ D x and F.y/ D y imply kx � yk D kF.x/ � F.y/k � 1

2kx � yk; which implies kx � yk D 0;

i.e. x D y: So there is at most one fixed point.

To get a fixed point, let x1 D 0 and xnC1 D F.xn/ 2 B.0; r/ for n D 1; 2; 3; : : : : Now kxn � xnC1k DkF.xn�1/ � F.xn/k � 1

2kxn�1 � xnk: Repeating this n � 1 times, we easily get kxn � xnC1k � 1

2n�1kx1 � x2k: So

for n < m;kxn � xmk � kxn � xnC1k C � � � C kxm�1 � xmk � � 1

2n�1C � � � C 1

2m�2

�kx1 � x2k � 1

2n�2kx1 � x2k;

which implies the sequence x1; x2; x3; : : : is a Cauchy sequence. Hence the sequence has a limit w: Now kxnk � rimplies kwk � r; i.e. w 2 B.0; r/:

Next from 0 � limx!a

kF.x/ � F.a/k � limx!a

12kx � ak D 0; we see that F is continuous. So xn ! w implies

xnC1 D F.xn/! F.w/; but xnC1 ! w: Therefore, F.w/ D w:Theorem (Mean Value Inequality). If F : Rn ! Rm is differentiable at every point of B.p; r/ and kF 0.a/k � k forevery a 2 B.p; r/; then kF.x/ � F.y/k � kkx � yk for every x; y 2 B.p; r/:Proof. For x; y 2 B.p; r/ and t 2 [0; 1]; let .t/ D .1� t/y C t x and g.t/ D F

� .t/�: By exercise 13(b),kF.x/ � F.y/k D kg.1/� g.0/k � kg0.c/k.1 � 0/ D kF 0� .c/�.x � y/k � kF 0� .c/�kkx � yk � kkx � yk:Proof of Inverse Function Theorem. First we consider the special case q D O; F.O/ D O and F 0.O/ D I:(Step 1) Since F is C1 near O; all the partial derivatives are continuous near O: Considering the n2 partial derivatives

as coordinate functions, we see that the function F 0 : Rn ! Rn2is continuous near x D O: So for " D 1

2> 0;

there is a � > 0 such that kx � Ok < � implies kF 0.x/ � F 0.O/k D kF 0.x/ � Ik < 1

2: Let G : Rn ! Rn be

145

given by G.x/ D F.x/ � x : On B.O; �/; kG0.x/k D kF 0.x/ � Ik < 1

2: By the mean value inequality, we getkG.x/ � G.y/k � 1

2kx � yk for every x; y 2 B.O; �/: By triangle inequality, for every x; y 2 B.O; �/;kF.x/ � F.y/k D kx � y��G.y/ � G.x/�k � kx � yk � kG.y/ � G.x/k � 1

2kx � yk: .�/

(Step 2) Let 0 < r < �=2: For each t 2 B.O; r/; we will find a s 2 B.O; �/ such that F.s/ D t : This is thesame as finding a fixed point in B.O; �/ for the function E : Rn ! Rn defined by E.s/ D s C �

t � F.s/�: NowkE 0.s/k D kI � F 0.s/k � 1

2for every s 2 B.O; �/: By the mean value inequality, kE.x/ � E.y/k � 1

2kx � yk

for every x; y 2 B.O; �/: Since 2r < �; so for every s 2 B.O; 2r/; we have kE.s/ � E.0/k � 1

2ksk: ThenkE.s/k � 1

2jsj C kE.0/k � r C jt j � 2r: So E takes B.O; 2r/ to B.O; 2r/: By the contractive mapping theorem,

there is a unique s 2 B.O; 2r/ � B.O; �/ such that E.s/ D s; then F.s/ D t : Note there is exactly one suchs 2 B.O; �/ satisfying F.s/ D t by (*).

(Step 3) Now for every t 2 B.O; r/; define F�1.t/ D s;where s 2 B.O; �/ is as in step 2. In particular, F�1.O/ D O:By (*), we have (**) kF�1.t/ � F�1.t0/k � 2kt � t0k; which implies F�1 is continuous. Since F is differentiable atq D O; F.O/ D O and DF.O/ D I; so

r.x/ D F.x/ � F.O/ � DF.O/.x � O/kxk D F.x/ � xkxk ! O as x ! O:By (**), as x ! O;kF�1.x/ � F�1.O/ � I .x � O/kkxk D kF�1

�F.x/ � r.x/kxk� � O � xkkxkD kF�1

�F.x/ � r.x/kxk� � F�1

�F.x/�kkxk � 2k.F.x/ � r.x/kxk/ � F.x/kkxk D 2kr.x/k ! 0:

So F�1 is differentiable at O:(General Case) Define �; : Rn ! Rn by �.x/ D x � q and .y/ D DF.q/�1

�y � F.q/�: Note that � and are

C1 functions and have C1 inverses given by ��1.x/ D x C q and �1.y/ D DF.q/.y/ C F.q/: Let H : Rn ! Rn bedefined by H .x/ D � F � ��1.x/: Then H .O/ D .F.��1.O/// D .F.q// D O and

H 0.O/ D 0�F.q/�F 0.q/.��1/0.0/ D .F 0.q//�1 F 0.q/I D I:By the special case above, H has an inverse near O; which is differentiable at O: Then F has the inverse near q givenby F�1 D ��1 � H�1 � and it is differentiable at F.q/: If in the above argument,we replace q by another point q 0near q where F is C1; then similarly we will get F�1 is also differentiable at F.q 0/: Finally, F�1 is C1 near F.q/because the chain rule gives .F�1/0�F.x/� D �

F 0.x/��1., .F�1/0.y/ D �F 0.F�1.y//��1

if we set y D F.x/). Sothe partial derivatives of F�1 are polynomials of compositions of the partial derivatives of F with the function F�1;which are continuous near q:Remarks. If F is only C1 at q 2 Rn; then from the proof above, we will see that F has an inverse function near q;which is differentiable at F.q/: However, the inverse function may not be C1 at F.q/ as the partial derivatives of theinverse function are not known to exist near F.q/:

146

Practice Exercises for Math 301

1. Prove the triangle inequality: for u; v; w 2 Rn; we have ku � wk � ku � vk C kv � wk: (Hint: Write outcoordinates for u � v and v � w: Recall the Cauchy-Schwarz inequality.)

2. Let L D .l1; l2; : : : ; ln/ 2 Rn : For i D 1; 2; 3; : : : ; let vi D .xi;1; xi;2; : : : ; xi;n/ 2 Rn : For sequence v1; v2; v3; : : :inRn; define lim

i!1 vi D L if and only if limi!1 jjvi � L jj D 0: Use the useful inequality and the sandwich theorem

to prove thatlim

i!1 vi D L if and only if limi!1 xi; j D lj for all j D 1; 2; : : : ; n:

3. Let F : Rn ! Rm be differentiable at p 2 Rn : Explain why there cannot be two different linear transformationsboth satisfy the definition of derivative of F at p:

4. Show that F : R3 ! R2 given by F.x; y; z/ D .x2 C ey; z cos x/ is differentiable at every .a; b; c/ 2 R3: FindF 0.a; b; c/ and DF.a; b; c/ for .a; b; c/ 2 R3:

5. (1997 Midterm) Let F : R3 ! R2 be defined by F.x; y; z/ D .x y C cos.y C z/;�zex/:(i) Show that F is differentiable at every .a; b; c/ 2 R3:(ii) Find F 0.a; b; c/ and DF.a; b; c/:

6. Let F : R2 ! Rbe defined by F.x; y/ D �x2 C y2 if x 6D 0 and y 6D 00 if x D 0 or y D 0

: At which .a; b/ 2 R2 is F differen-

tiable? Explain your answers.

7. (2000 Midterm) Show that f .x; y/ D 8<: x y2

x2 C y2if .x; y/ 6D .0; 0/

0 if .x; y/ D .0; 0/ is not differentiable at .0; 0/ by checking

the definition.

8. (2000 Midterm) Find all .x; y/ 2 R2 where the function f .x; y/ D � 12 x2 C y2 if x > 0x2 C 1

2 y2 if x � 0is differentiable. Be

sure to show work.

9. (2000 Final) Determine if f .x; y/ D 8<: x2 � y2

x � yif x 6D y

x2 C 2x otherwiseis differentiable at .0; 0/ or not by checking the

definition.

10. (1999 Midterm) (a) Show that F.x; y/ D jx yj is not C1 at .0; 0/:(b) Show that F.x; y/ D jx yj is differentiable at .0; 0/:

11. If f; g : Rn ! Rare differentiable at a 2 Rn; then show that f C g; f � g; f g and f=g (provided g.a/ 6D 0) aredifferentiable at a: Show that

D. f g/.a/ D g.a/D f .a/ C f .a/Dg.a/:(Hint: Let F : Rn ! R2 be defined by F.x/ D . f .x/; g.x//: For f C g; consider P � F;where P is the functionP.x; y/ D x C y; then use the chain rule.)

12. For a; b 2 Rm; the line segment joining a to b consists of all points .t/ D .1 � t/a C tb 2 Rm for t 2 [0; 1]:If F : Rm ! R is continuous on every point on the line segment joining a to b and F is also differentiableon the line segment (except perhaps at the endpoints), show that there is a c on the line segment such thatF.b/ � F.a/ D F 0.c/.b � a/:

147

13. (a) For F : R! R2 defined by F.t/ D .cos t; sin t/; show that there is no c on [0; 2�] such that F.2�/� F.0/ D.2� � 0/F 0.c/:(b) Show that if F : R! Rn is continuous on [a; b] and differentiable on .a; b/; then there is a c 2 .a; b/ suchthat jjF.b/ � F.a/jj � jjF 0.c/jj.b � a/: (Hint: Consider �.t/ D .F.b/ � F.a// � F.t/ for t 2 [a; b]: Here thedot is the dot product inRn :)

14. (2000 Midterm) Consider the system of equations

u D x2ey C cos x y C x and v D xexy C y2:Show that near .x; y; u; v/ D .0; 1; 1; 1/; .x; y/ can be expressed as a differentiable function of .u; v/ and showthat @u@y

.0; 1/ 6D 1@y@u.1; 1/ :

15. Consider the system of equations x D u C v and y D u2 C v: Using the inverse function theorem, show that near.u; v/ D .1;�1/; u and v can be expressed as differentiable functions of x and y: Find@u@x; @2u@x2

in terms of u:16. (1999 Midterm) Consider the system of equations

x D v C 2wC evw and y D e�vw Cw C 2v:Show that near .v;w; x; y/ D .0; 0; 1; 1/; .v;w/ can be expressed as a differentiable function of .x; y/ and@v@x

.1; 1/C @v@y.1; 1/ D @w@x

.1; 1/ C @w@y.1; 1/:

17. (1999 Midterm) Consider the system of equations

x D w cos v C v cosw C ew and y D w sin v C v sinw C ev :Show that near .v;w; x; y/ D .0; 0; 1; 1/; .v;w/ can be expressed as a differentiable function of .x; y/ and findthe value of @v@x

.1; 1/C @v@y.1; 1/C @w@x

.1; 1/C @w@y.1; 1/:

18. (1999 Midterm) Consider the system of equations

x D cos u C sin v and y D sin u C cos v:Show that near .u; v; x; y/ D .0; 0; 1; 1/; .u; v/ can be expressed as a differentiable function of .x; y/ and�@u@x

�2 C �@u@y

�2 D �@v@x

�2 C �@v@y

�2:19. (1997 Midterm) Given .1; 1/ is a solution of the equation x3 C x y5 C y2 D 3: Show that near .1; 1/; the equation

can be expressed as y D f .x/ for some differentiable function f: What isdy

dxat .1; 1/?

148

20. Given that p D .u; v; w/ D .1; 0;�1/ is a solution of the system of equations euv C w D 0 and u sin vw D v:Using the implicit function theorem, show that v and w can be expressed as differentiable functions of u near p:Find

dvdu.p/ and

dwdu.p/:

21. Consider the system of equations ex C ye � u � v D 1 and eu � ve � x C y D 1:(i) Show that near .x; y; u; v/ D .0; 1; 0; 1/; .x; y/ can be expressed as a function of .u; v/:(ii) Find

@x@u.0; 1/; @y@u

.0; 1/:22. Suppose f1; f2 : R4 ! Rare differentiable functions. Consider the system of equations

f1.w; x; y; z/ D 0 and f2.w; x; y; z/ D 0:If near a solution .w0; x0; y0; z0/; .x; y/ can be expressed as a differentiable function of .w; z/; then show that at.w0; x0; y0; z0/; @x@w D �@. f1; f2/@.w; y/ .@. f1; f2/@.x; y/ and

@y@w D �@. f1; f2/@.x; w/ .@. f1; f2/@.x; y/ :(Hint: Just use chain rule and Cramer’s rule.)

Remarks. For n differentiable functions fi : RnCm ! Rwith variables x1; : : : ; xn; w1; : : : ; wm; the formula is@xi@wjD � @. f1; : : : ; fn/@.x1; : : : ; xi�1; wj; xiC1; : : : ; xn/.@. f1; : : : ; fn/@.x1 : : : ; xn/ :

23. (2000 Midterm) Consider the system of equations

exCyCzCw D cos.x C y/ C sin.z C w/ and e�x�y�z�w D sin.x C y/ C cos.z C w/:Show that near .x; y; z; w/ D .1;�1; 1;�1/; .x; z/ can be expressed as a differentiable function of .y; w/: At.x; y; z; w/ D .1;�1; 1;�1/; find

@x@w and@z@w :

24. (1999 Final) For the system x3 C y3 C z3 C 3y D xw3 and x5 C y5z3 D x yz C x3w5; show that .x; y/ can be

expressed as a differentiable function of .w; z/ near the solution p D .w; x; y; z/ D .1; 1; 0; 0/: Find@x@w.p/ and@y@w.p/:

25. (1998 Midterm) (a) Given .w; x; y; z/ D .0; 0; 0; 0/ is a solution of the system of equations

cosw C sin x C tan y D z C 1 and ew C x D y C e�z:Near .0; 0; 0; 0/; show that .x; y/ can be expressed as a differentiable function G of .w; z/:(b) Find

@x@w.0; 0/ and@y@w.0; 0/:

26. Given F : R3 ! R is C1 near .0; 0; 0/;F.0; 0; 0/ D 0; @F@x

.0; 0; 0/ 6D 0; @F@y.0; 0; 0/ 6D 0; @F@z

.0; 0; 0/ 6D 0:149

Show that near .0; 0; 0/; the equation F.x; y; z/ D 0 can be expressed in each of the form x D f .y; z/;y D g.z; x/; z D h.x; y/: Show that near .0; 0; 0/;@x@y

@y@z

@z@xD �1:

(Here@x@y

D @ f@y; @y@z

D @g@z; @z@x

D @h@x:)

27. Prove that if limsupn!1 xn D x D liminf

n!1 xn (i.e. L D fxg), then limn!1 xn D x :

Remarks. Think carefully. This was not explained in the notes!

28. Find the limit superior and the limit inferior of each of the following sequences.

(a) xn D n2.1� cos e�n/I(b) xn D f .en cos.n�=2/�1/; where f .x/ D x ln x :

29. (1997 Final) Find the limit superior and limit inferior of each of the following sequences:

(i) xn D n.1� cos 1n / sin n for n D 1; 2; 3 : : : I

(ii) xn D .cos n�/ cos 2n�5 for n D 1; 2; 3; : : : :

30. (1997 Final) Find the limit superior and limit inferior of each of the following sequences:

(i) xn D [n=2] cosn�n

for n D 1; 2; 3 : : : ; where [x] is the greatest integer less than or equal to xI(ii) x1 D 1 and xnC1 D xn

�1C 1�2n�1

�for n D 1; 2; 3; : : : :

31. (1998 Final) Find the limit superior and limit inferior of each of the following sequences:

(i) xn D n sin.e�n/ for n D 1; 2; 3 : : : I(ii) xn D 2n cos.n�=7/ for n D 1; 2; 3; : : : :

32. Find limsupn!1 ����anC1

an

���� ; liminfn!1 ����anC1

an

���� ; limsupn!1 n

pjanj and liminfn!1 n

pjanj; where

ak D �k=2k if k is oddk2=2k if k is even

:(Remarks. This is an example to show lim

n!1 npjanj may exist, but lim

n!1 ����anC1

an

���� may not.)

33. (1999 Midterm) (a) Let fang and fbng be sequences of real numbers. Show that

limsupn!1 .an C bn/ � limsup

n!1 an C limsupn!1 bn;

provided the left or right side is not of the form1�1:(b) Give an example of sequences fang and fbng of real numbers so that

limsupn!1 .an C bn/ 6D limsup

n!1 an C limsupn!1 bn

and the right side is not of the form1�1: Give reasons to support your example.

150

34. (a) Let fang and fbng be sequences of real numbers. Show that

liminfn!1 an C liminf

n!1 bn � liminfn!1 .an C bn/ � liminf

n!1 an C limsupn!1 bn � limsup

n!1 .an C bn/ � limsupn!1 an C limsup

n!1 bn;provided none of the expressions is of the form1�1:

(b) If liminfn!1 an � 1

2; liminf

n!1 bn � 1

2and lim

n!1.an C bn/ D 1; then prove that limn!1 an D 1

2D lim

n!1 bn :35. Let fang and fbng be sequences of nonnegative real numbers. Show that

limsupn!1 .anbn/ � .limsup

n!1 an/.limsupn!1 bn/;

provided the right side is not of the form 0 � 1:36. Give an example of sequences fang and fbng of nonnegative real numbers so that

limsupn!1 .anbn/ 6D .limsup

n!1 an/.limsupn!1 bn/

and the right side is not of the form 0 � 1:37. For a sequence a1; a2; a3; : : : of real numbers, let�n D a1 C a2 C � � � C an

n:

Show thatliminfn!1 an � liminf

n!1 �n � limsupn!1 �n � limsup

n!1 an :38. Find the domain of each of the following series of functions, i.e. the largest set where the series converges

pointwise. For those that are power series, state the radii of convergence.

(a)16

kD1

.x � 1/k.2k/! (b)16

kD2

xk

ln k

(c)16

kD1

.2x � 3/k4k

(d)16

kD1

ekx

k C 1

39. Find the first three nonzero terms of the Taylor series of tan x with center at c D 0: Next, find the first threenonzero terms of the Taylor series of sin x at c D 0 divided by the Taylor series of cos x at c D 0 by long division.

Remarks. Although the Taylor series of sin x and cos x at c D 0 have infinite radii of convergence, the Taylorseries of their quotient tan x at c D 0 does not have infinite radius of convergence since tan.�=2/ D1:

40. Show by an example that two power series with the same center may have radii of convergence equal 1, but theirsum may have infinite radius of convergence.

41. Find the Taylor series of cos2 x with center c D 0 and show that the Taylor series converges pointwise on.�1;C1/ to cos2 x :42. Let f : R! Rbe defined by f .x/ D �

e�1=x2if x 6D 0

0 if x D 0: Show that f .n/.0/ D 0 for n D 1; 2; 3 : : : : (Hint: First

show limt!1 t ne�t2 D 0 for n D 0; 1; 2; : : : by induction. Then, for x 6D 0; show f .n/.x/ D Pn

�1x

�e�1=x2

for some

polynomial Pn.x/ by induction. Use the definition of derivative to compute f .n/.0/:)151

43. Show that each of the following sequences of functions converges uniformly on the intervals indicated.

(a) fn.x/ D xn

1C xnon [0; 0:99] (b) gn.x/ D nxe�nx2

on [0:1;1/(c) hn.x/ D cos xn on [0; r]; where 0 < r < 1

44. Does the sequence of functions Sn.x/ D xn converge uniformly on [0; 1/? Give explanations to support youranswer.

45. Show that each of the following series of functions converges uniformly on the intervals indicated.

(a)16

kD1

sin kx

ekon .�1;C1/ (b)

16kD2

xk

k.ln k/2 on [�1; 1]

(c)16

kD1ke�kx on [1;C1/ (d)

16kD1.x ln x/k on .0; 1]

(e)16

kD1k2xk on [�r; r]; where 0 < r < 1 (f)

16kD1

xk

kon [0; r/; where 0 < r < 1

(g)16

kD1

kx

ekxon [r;C1/; where r > 0

46. Show that each of the following sequences or series of functions converges pointwise on the intervals indicated,but fails to converges uniformly there.

(a) Sn.x/ D sinn x on [0; �] (b)16

kD0xk.1� x/ on .0; 1]

(c) Sn.x/ D nxe�nx2on [0; 1]

47. Give an example of a sequence of functions Sn : [0; 1] ! R that does not converge uniformly on [0; 1]; but thesequence of their squares S2

n.x/ converges uniformly on [0; 1]:Give reasons to support your example.

48. Construct sequences fn and gn which converges uniformly on some set E , but such that fn gn does not convergeuniformly on E : (Hint: Try fn.x/ D x and gn.x/ D 1

n onR:)49. If fn converges uniformly on E to f; gn converges uniformly on E to g and both f and g are bounded on E; then

show that fn gn converges uniformly on E to f g:50. Prove that if fn converges uniformly on a set E to f and each fn is a bounded function on E (i.e. there is Mn 2 R

such that j fn.x/j � Mn for every x 2 E/; then the sequence fn is uniformly bounded on E in the sense that thereis M 2 R such that j fn.x/j � M for every n and every x 2 E :

51. If fn and gn converges uniformly on a set E to f and g; respectively, then prove that fn C gn converges uniformlyon E to f C g: If, in addition, fn and gn are bounded functions on E; then prove that fn gn converges uniformlyon E to f g: (Hint: For the second statement, use the previous exercise.)

52. (2000 Midterm) (a) If16

kD1gk.x/ converges uniformly on [0; 1]; then show that

16kD1

cos gk.x/ cannot converge

uniformly on [0; 1]: Can16

kD1cos gk.x/ converge pointwise on [0; 1]?

(b) If the sequence of functions Sn.x/ converges uniformly on [0; 1] to S.x/; then show that cos Sn.x/ mustconverge uniformly on [0; 1] to cos S.x/:

53. Let F0.x/ be a bounded Riemann integrable function on [0; 1]: For n D 1; 2; 3; : : : ; define Fn.x/ on [0; 1]

by Fn.x/ D Z x

0Fn�1.t/ dt : Prove that the sequence Fn.x/ converges uniformly on [0; 1]: Prove that the series16

kD0Fk.x/ also converges uniformly on [0; 1]:

152

54. If I .x/ D n 0 if x � 01 if x > 0

; if xn is a sequence of points in interval .a; b/ and if16

nD1cn converges absolutely, prove

that the series f .x/ D 16nD1

cn I .x � xn/ converges uniformly on .a; b/ and f is continuous at every x 6D xn :(Remarks: If .a; b/ D .�1;C1/ and Q D fx1; x2; x3; : : :g; then f is a function that is continuous at everyirrational number, but discontinuous at every rational number.)

55. Prove that16

iD1

� 16jD1

ai j

� D 16jD1

� 16iD1

ai j

�if ai j � 0 for all i and j (the case C1 D C1 may occur).

56. (1998 Midterm) Let [x] be the greatest integer less than or equal to x; (for example, [p

2] D 1; [3 12 ] D 3 and

[5] D 5:) Define f .x/ D x�[x]: Consider the graph of f: Show that g.x/ D 16kD1

12k

f .kx/ is (Riemann) integrable

on [0; 1]:57. Prove that the series

16kD1.�1/k x2 C k

k2converges uniformly on every bounded interval, but does not converge

absolutely for any value of x :58. For n D 1; 2; 3; : : : ; x real, put fn.x/ D x

1C nx2: Show that fn converges uniformly on R to a function f and

that the equation f 0.x/ D limn!1 f 0n .x/ is correct if x 6D 0; but false if x D 0:

59. Find limx!0

1� cos.x2/x4

and show thatZ 1

0

1� cos.x2/x4

dx D 16kD1

.�1/kC1.2k/!.4k � 3/ :60. (1999 Midterm) Find lim

x!0

ex2 � 1

xand show that

Z 1

0

ex2 � 1

xdx D 16

kD1

1

k!.2k/ :61. (2000 Midterm) (a) Find the maximum value of f .x/ D jx ln.2x/j on the interval .0; 1

2 ]: Be sure to show workto support your answer.

(b) Show thatZ 1

2

0

x ln.2x/1C x

dx D 16kD0

.�1/kC1

2kC2.k C 2/2 :62. (a) Use

d

dx

� 1

1� x

� D 1.1 � x/2 to find a power series for1.1 � x/2 on the open interval .�1; 1/:

(b) Show thatZ 1

2

0

xp

2.1� x/2 dx D 16kD0

k C 1.p2C k C 1/2p2CkC1:

63. (a) Show that limx!0C sin x ln x exists by considering sin x ln x D sin x

x.x ln x/:

(b) Show that16

nD0

.�1/n x2nC1 ln x.2n C 1/! converges uniformly on [0; 1]: (Here at x D 0; we interpret x2nC1 ln x as 0.)

(c) Use the Taylor series of sin x to show thatZ 1

0sin x ln x dx D 16

nD0

.�1/nC1.2n C 1/!.2n C 2/2 :64. (1998 Midterm) (a) Prove that the Taylor series of f .x/ D sin2 x D 1

2 .1 � cos 2x/ about c D 0 equals f .x/ forevery x 2 R:(b) Prove that

Z 1

0

sin2 x

x2dx D 16

kD1

.�1/kC122k�1.2k/!.2k � 1/ : (At 0, the integrand is set to 1.) Be sure to give reasons to

support the steps.

153

65. (1999 Midterm) Given that ln.1� x/ D � 16kD1

xk

kfor �1 < x < 1:

(a) Show thatZ 2

3

13

ln.1� x2/x2

dx D � 16kD1

22k�1 � 1

k.2k � 1/32k�1:

(b) Show that16

kD1

1k.k C 2/2kC2

converges and find its sum by considering f .x/ D 16kD1

xkC2

k.k C 2/ :66. (2000 Midterm) (a) Find the value of

Z 1

0u2 sin u du: Show work.

(b) Find16

kD0

.�1/k.2k C 1/!.k C 2/ : Be sure to explain the steps. (Hint: Consider the integral in part (a).)

67. Find limx!0

tan x � x

x.1 � cos x/ and limx!0

x � sin x

tan x � x(by power series method).

68. Suppose f .x/ D 16kD0

ak.x � c/k for every x 2 .a; b/; where c 2 .a; b/: Show by induction that ak D f .k/.c/k!

for

k D 1; 2; 3; : : : : (Hint: f .nC1/ D . f 0/.n/:)Remarks. This means if a function can be written as a power series on an open interval (with the center on theinterval), then it is the Taylor series of the function on the interval.

69. Show that16

kD1

1kek

converges and find its sum by considering f .x/ D 16kD1

xk

k:

70. (a) Find the domain of the power series f .x/ D 16nD1

.�1/nC1x2nC1.2n � 1/.2n C 1/ :(b) Given arctan x D x � x3

3C x5

5� x7

7C � � � D 16

nD1

.�1/nC1x2n�1

2n � 1for every x 2 .�1; 1/: Find f 0.x/ for x 2.�1; 1/ and justify.

(c) Show f .1/ D 16nD1

.�1/nC1.2n � 1/.2n C 1/ D �4� 1

2:

71. Given that all the solutions of the differential equation f .4/.x/� f .x/ D 0 are of the form f .x/ D aex C be�x Cc cos x C d sin x; where a; b; c; d 2 R: Show

16kD0

1.4k/! converges and find its sum by considering16

kD0

x4k.4k/! :72. On E D [�1; 1]; define polynomials P1.x/ D 0 and for n > 1;

Pn.x/ D Pn�1.x/ C x2 � P2n�1.x/

2:

Show that Pn.x/ converges uniformly on E to jxj: (Hint: Show 0 � Pn.x/ � PnC1.x/ � jxj on E by induction.Use jxj � Pn.x/ D �jxj � Pn�1.x/��1� 1

2.jxj C Pn�1.x//� to show jxj � Pn.x/ � jxj.1� 12 jxj/n�1 < 2

n on E :)Remarks. This gives an example of a sequence of differentiable functions converging uniformly on an interval toa nondifferentiable function. In this example, the limit function jxj is not differentiable at x D 0 only. Applyingthe Weierstrass approximation theorem to a continuous, but nowhere differentiable function, it is possible that thelimit function can be nowhere differentiable!

73. If f is continuous on [0; 1] and ifZ 1

0f .x/xn dx D 0 for n D 0; 1; 2; : : : ; prove that f .x/ D 0 on [0; 1]: (Hint:

The integral of f with any polynomial is zero. Use the Weierstrass approximation theorem to show that theintegral of f 2 is 0.)

154

74. Show that inR; if S 6D ; and S 6D R; then S cannot be both open and closed.

75. For each part below, determine if the set is open, closed, compact or none of these. (You may use graphs.)

(i) A D f �1..2; 9]/; where f : R! R; f .x/ D x2:(ii) B D g�1.[�1; 2/ [ .3; 4//; where g : .0;C1/! R; g.x/ D ex :

(iii) C D h�1.K /; where h : [0; 1] ! R; h.x/ D px and K is the Cantor set.

(iv) D D i� 1[

nD1

.1n; 2C 1

n2/�; where i : .0;1/! R; i.x/ D ln x :

(v) E D j�Rn [

n2Z.2n�; .2n C 1/�/�; where j : R! R; j .x/ D sin x :(vi) F D �1

n: n 2Zand n 6D 0

:(vii) G D f0g [ �[

n2N[1

n2 C 2; 1

n2 C 1]�:

76. Find the length of each of the following sets.

(i) H D 1[nD1

�1

n� 1

20; 1

nC 1

20

�: (Be sure to explain where overlaps occur.)

(ii) I D [0; 1] n 1[kD1

. 12k C 1

; 12k/:

77. Let A and B be sets inR:(i) Show that Int.A \ B/ D Int.A/ \ Int.B/:

(ii) Must Int.A [ B/ D Int.A/ [ Int.B/?78. In the construction of the Cantor set, at the n-th stage, let us remove an open interval of length

1

3n2in the middle

of each remaining intervals instead, show that this new set is compact. Show that its length is positive, but it doesnot contain any interval of positive length (i.e. it has no interior point).

Remarks. The existence of such a positive length Cantor-like set will be used in Math 501. From this, we can geta sequence of continuous functions converging pointwise on [0; 1] to a function not Riemann integrable.

79. For each part below, show that the set is measurable and find its Lebesgue measure.

(a) A D fx 2 [��;p2/ : x is irrationalg:(b) B D fx� C y

p2 : x; y 2 Qg:

(c) C is the set of all numbers x such that x has a distance less than1

5nfrom some rational number

m

n; where

m; n 2 N have highest common factor equal 1.

(d) D is the set of all numbers in the Cantor set that are irrational.

80. Show that the set

A D fx 2 [0; 1] : x has a decimal representation not containing the digit 5gis measurable. What is m.A/?

155

81. (1997 Final) (i) Show that if C is the Cantor set, then the set S D fx C n : x 2 C; n D 1; 2; 3; : : :g is measurable.

(ii) Show that m.S/ D 0; where S is the set in part (ii).

82. (1999 Final) Let G be a nonempty open set. Prove that not every number in G is equal to a number of the formr C k; where r is a rational number and k is a number in the Cantor set.

83. Use countable additivity to prove the first part of the monotone set theorem, namely for measurable sets A1 ; A2; : : : ;if A1 � A2 � : : : ; then m

� 1[kD1

Ak� D lim

k!1m.Ak /:84. Define the symmetric difference of two sets A and B to be A4B D .An B/[ .B n A/: Show that if A is measurable

and m.A4B/ D 0; then B is measurable.

85. (1998 Final) Write the rational numbersQD fq1; q2; q3; : : :g: Define G D 1[nD1

�qn � 1

n2; qn C 1

n2

� :(i) Show that G is measurable and m.G/ <1:(ii) Show that if F is a closed set andQ� F; then F D R:(iii) For every closed set F; show that either m.G n F/ > 0 or m.F n G/ > 0:

86. (2000 Final) For X; Y � R; define X C Y D fx C y : x 2 X and y 2 Y g:(a) Give the definition of an interior point of a set S inR:(b) Let A be a nonmeasurable set and B be a nonempty open set. Show that A C B is a measurable set.

(c) Let A be a nonmeasurable set and C be a nonempty closed set. Give two examples, one showing AC C maybe a measurable set and one showing AC C may be a nonmeasurable set. Explain your examples.

(d) If X; Y and X C Y are nonempty measurable sets, must m.X C Y / D m.X / C m.Y /? Give explanations.

87. This is a structure theorem for measurable sets. Let A be a bounded measurable set, show that there is a sequence

of open set S1; S2; S3; : : : and a set C of measure 0 such that A D � 1\kD1

Sk� n C: (Hint: Take open set Sn such that

A � Sn and m.Sn/ � m�.A/ C 1n :)

Remarks. A set which is a countable intersection of open sets is called a G� set. Thus, the exercise says everybounded measurable set is a G� set minus a set of measure 0. Similarly, there is a description of bounded measurablesets using closed sets. It asserts that every bounded measurable set is a union of closed sets W1;W2;W3; : : :and a set of measure 0. A set which is a countable union of closed set is called a F� set. Thus, every boundedmeasurable set is the union of a F� set and a set of measure 0.

88. (1998 Final) Give an example of two functions f; g : R! Rsuch that 0 < g.x/ � f .x/ for every x 2 Rand fis not measurable, but g is measurable.

89. Let A be a measurable set. Show that the following are equivalent:

(a) f : A ! R is a measurable function.

(b) for every open set S; f �1.S/ is measurable.

(c) for every closed set W; f �1.W / is measurable.

(d) for every compact set K ; f �1.K / is measurable.

Remarks. There are measurable functions f; g : A ! R such that for some measurable set B; f �1.B/ isnonmeasurable and g.B/ is nonmeasurable. See fact (2) of Appendix 2 to Step 3 of Chapter 12.

156

90. (1997 Final) Let f : R! Rbe a function and f 2.x/ D . f .x//2 .

(i) Show that even if the function f 2 is measurable, f may still be not measurable.

(ii) Show that if the function f 2 is measurable and the set S D fx 2 R : f .x/ � 0g is measurable, then f ismeasurable. (Hint: XS is measurable.)

91. For a; b; c 2 R; define mid.a; b; c/ to be the middle value when a; b; c are arranged in increasing order. Iff1; f2; f3 : R! Rare measurable functions, show that g : R! Rdefined by g.x/ D mid. f1.x/; f2.x/; f3.x//is a measurable function.

92. (2000 Final) Let p; q; r; s : R! Rbe measurable. Show that the function t : R! Rdefined by

t .x/ D �r.x/ if p.x/ < q.x/s.x/ otherwise

is a measurable function.

93. Show that if f : R! R is differentiable, then f 0 is a measurable function.

94. Show that if f : R! R is a monotone function, then f is a measurable function.

95. (a) Show that if f; g : R! Rare measurable functions, then S D fx 2 R : f .x/ D g.x/g is a measurable set.

(b) Let f1; f2; f3; : : : : R ! [0; 1] be a sequence of measurable functions. Show that S D fx 2 R :limn!1 fn.x/ existsg is measurable.

96. (2000 Final) Let f; g : R! Rbe continuous. Show that A D fx 2 R : f .x/ < g.x/g is an open set. Give anexample to show that the set A may also be a closed set.

97. (1999 Final) Let f; g : R! Rbe measurable functions. Show that the set

S D fx 2 R : f .x/ D n C g.x/ for some n 2Zg is measurable.

98. (1998 Final) Let [x] be the greatest integer less than or equal to x : Let f : R! Rbe a measurable function.

(i) Show that g.x/ D [ f .x/] is a measurable function.

(ii) Show that h.x/ D f .[x]/ is a measurable function.

99. (1999 Final) Let K be the Cantor set. Let f : R! Rbe defined by f .x/ D n1 if x 62 K0 if x 2 K

and let g : R! Rbe continuous.

(a) Prove that the composition function g � f is measurable.

(b) Prove that the composition function f � g is measurable.

100. Let C be the Cantor set. Prove that C CC D fx C y : x; y 2 Cg is a measurable set. (Hint: Show it is an interval.Need to use base 3 representation.)

101. (1999 Final) Find limn!1 Z[ �2 ; 3�

2 ]

sin x

1C e�nxdm: Be sure to give reasons, not just answer.

102. (1997 Final) (i) For x > 0; show that limn!1�1C x

n

�n D ex : (Hint: Let t D 1n and take log.)

(ii) For x > 0 and positive integer n; show that�

1C x

n

�n � ex :157

(iii) Find limn!1 Z 1

0

�1C x

n

�ne�2x dx :

103. (1998 Final) (i) Find the maximum of y.s/ D s1=s for s 2 [1;C1/:(ii) Find lim

n!1 Z 11

e�x.nx/ 1nx dx :

104. (1997 Final) Show thatZ 1

0

x

ex � 1dx D 16

nD0

1.n C 1/2 : (Hint: Note e�x < 1 on .0;1/:)105. (1998 Final) Show that

Z.0;1/.ln x/ ln.1 � x/ dm D 16nD1

1n.n C 1/2 :

106. (2000 Final) Show thatZ e

1ex ln x dx D 16

nD0

nenC1 C 1n!.n C 1/2 :

107. Use the Monotone Convergence Theorem to prove the fact that if A1; A2; A3; : : : are measurable sets, A1 � A2 �A3 � � � � ; then m

� 1[nD1

An� D lim

n!1 m.An /:108. Use the Lebesgue Dominated Convergence Theorem to prove the fact that if A1; A2; A3; : : : are measurable sets,

A1 � A2 � A3 � � � � and m.A1/ <1; then m� 1\

nD1

An� D lim

n!1m.An /:109. Let g.x/ D � 0 if x 2 [0; 1=2]

1 if x 2 .1=2; 1]: Let f2k .x/ D g.x/ and f2kC1.x/ D g.1 � x/ for x 2 [0; 1]: Show that

liminfn!1 fn.x/ D 0 on [0; 1]; but

Z 1

0fn.x/ dx D 1

2for every n: So the inequality in Fatou’s lemma can be strict.

110. Show that f .x/ D n sin xx if x 6D 0

1 if x D 0is a measurable, but not Lebesgue integrable function onR:

111. Show by an example that Fatou’s lemma may become false if the functions are real-valued instead of nonnegativevalued.

158