Real Analysis Rudin Solution Manual(1-8)(10-13)

Real Analysis - Math 630Homework Set #1

by Bobby Rohde9-7-00

Problem 1

� If A and B are two sets in � with A � B, then mA � mB.

� ProofSince � is a �-algebra, we know that C = B � A

� � �, with C � A = � and C � A = B. Hence mB

= m(C � A) = mC + mA � mA. � mB � mA, QED.

Problem 2

� Let <En> be any sequence of sets in �. Then m(� En) � � mEn.

� ProofBy Proposition 1.2, � a sequence <An> of sets in � with � En = � An and An � Am = �, � n m.Thus m(� En) = m(� An) = � mAn. If we construct An from En according to the algorithim usedin the proof of Proposition 1.2 then we have An En � n, and from Problem 1 therefore mAn �mEn. Hence we have � mAn � � mEn � m(� En) � � mEn. QED.

Problem 3

� If there is a set A in � such that mA < �, then m� = 0.

� ProofBy Way of Contradiction (BWOC). Suppose m� = 0, and let mA = � < �. A � � = �, henceA and � are disjoint. Thus by Property 3, m(A � �) = mA + m� = � + . But A � � = A, so m(A� �) = mA = �. Thus � = � + > �. Which is a contradiction, hence m� = 0. QED.

Problem 4

� Let nE be � for an infinite set E and equal to the number of elements in E for a finite set. Show that n is a countably additive set function that is translation invariant and defined for all sets of real numbers.

� Proof

� Countably AdditiveWe need to show that � disjoint sequences <Ai> of sets in �, n(� Ai) = � nAi.

If Ai is infinite for any i then n(� Ai) = � nAi = �, so we may as well assume that Ai is finite � i.However since Ai is disjoint that implies that the number of elements in Ai � A j = nAi + nA j, � i j. Thus by repeated applications it is clear that n(� Ai) must equal � nAi. QED.

� Translation InvariantWe need to show that n(A + y) = nA � y.

Clearly the definition of A + y = {x + y : x � A}, establishes a 1-1 correspondance betweenelements of A and elements of A + y, thus A and A + y must have the same number of elements andhence n(A + y) = nA � y. QED

� Defined on All Sets of �Clearly it is in the nature of sets that they must have either an infinite number of elements orexactly one well-defined finite number of elements so it is impossible to construct a set on � whichdoes not have a unique value under operation by n. Thus all sets on � are measurable by n. QED.

MATH630-1.nb 2

Problem 5

� Let A be the set of rational numbers between 0 and 1, and let {In} be a finite collection of open intervals covering A. Show that � l(In) � 1.

� ProofLet (an, bn) denote the interval In with end points an < bn.

BWOC Suppose 0 < an, � n, then since rationals are dense, � a rational number � between 0 andmin{an}, not covered by In, which is a contradiction. So we know � n such that an � 0. Similarlywe must have an m such that bm � 1.

BWOC Suppose � In with bn < 1 such that � m n, am � bn < bm is false � � an open interval C =(bn, min({an: an > bn} � {1})), which is not in � In. But since the rational numbers are dense, � arational number in C, which contradicts the fact that {In} covers A. Thus � In with bn < 1, � msuch that am � bn < bm.

Consider T = (� In) � {bn} � {an}. Clearly T covers A since � In covers A. Furthermore sinceam � bn < bm is true under the conditions just stated, this implies that including {an} and {bn} willclose any gaps between intervals. Since there are a finite number of open intervals, we are addingonly a finite number of points to T, and hence m*T = � l(In). But T is now a continuous intervalwith (0, 1) � T, using the fact that � n, m such that an � 0 and bm � 1. Thus m*T � 1 - 0 = 1. � � l(In) � 1. QED.

Problem 6

� Prove that: Given any set A and any � > 0, � an open set O such that A � O and m*O � m*A + �. There is a G G such that A � G and m*A = m*G.

� ProofSince inf

A� In

�� l�In� = m*A we know by definition of infimum that � a collection of open intervals

{In} where � l�In� � m*A + � � � > 0. Thus we have that O = � In is an open set with m*O =� l�In� � m*A + �. Which gives us the first part of the Proposition

For the second part we merely have to consider G = �n�1� O 1��n

, where O� is the O associated with aspecified � in the first part. Since A O� � � > 0, we know that A � G. Also since G is an

MATH630-1.nb 3

intersection of a countable collection of open intervals, G � G�. Finally since m*A � m*O � m*A +�, with � � 0 in our construction of G, we can therefore conclude that m*A = m*G. QED.

Problem 7

� Prove that m* is translation invariant.

� ProofLet A be a set in �, then we need to show that m*(A + y) = m*A � y.

Let {In} be a countable collection of open intervals that cover A, then m*A = infA� In

�� l�In�.

Clearly {In+ y} will cover A + y, with m*(A + y) = infA� In

�� l��In � y��, but l(�An � y�) = l(An), so

infA� In

�� l�In� = infA� In

�� l��In � y�� m*(A + y) = m*A. QED.

Problem 8

� Prove that if m*A = 0, then m*(A � B) = m*B.

� ProofIf A B then A � B = B which is trivially true. So we may assume A � B. Let {In} denote opencovers of A, and {Jn} denote open covers of B. Thus {In} � {Jn} will cover A � B. Furthermorem*(A � B) � inf

A� B� �In � Jn�� l�In� � l�Jn�� = inf

A� B� �In � Jn�� l�In�� l�Jn�� =

infA� In

�� l�In� + infB� Jn

�� l�Jn� = m*A + m*B = m*B.

But also B A � B so m*B � m*(A � B). Thus we must have that m*(A � B) = m*B. QED.

MATH630-1.nb 4

Real Analysis - Math 630Homework Set #2 - Chapter 3


Problem 9

� Show that if E is a measurable set, then each translate E + y of E is also measurable.

� ProofE is measurable so � A, we have m*A = m*(A � E) + m*(A � E

�). We also know from Problem 7

that outer measure is translation invariant.Thus m*(A + y) = m*A = m*(A � E) + m*(A � E

�) = m*((A � E) + y) + m*((A � E

�) + y), � y.

m*((A � E) + y) + m*((A � E�) + y) = m*((A + y) � (E + y)) + m*((A + y) � (E

� + y)), from the

nature of intersection. But any set B may be written as A + y, thus � B, m*B = m*(B � (E + y)) +m*(B � (E

� + y)) = m*(B � (E + y)) + m*(B � ~(E + y)), since ~(E + y) must equal (E

� + y) from

the fact that both ~ and + are 1-1 operations on sets.The last equality thus shows that E + y is measurable, QED.

Problem 10

� Show that if E1 and E2 are measurable, then m(E1 � E2) + m(E1 � E2) = mE1 + mE2

� ProofSince each is measurable we know that mA = m(A � E1) + m(A � E1

�), mA = m(A � E2) + m(A �

E2�

), � A so mE2 = m(E2 � E1) + m(E2 � E1�

) and mE1 = m(E1 � E2) + m(E1 � E2�

)Thus mE1 + mE2 = 2 * m(E2 � E1) + m(E2 � E1

�) + m(E1 � E2

�).

However A � B = (A � B�) � (A

� � B) � (A � B).

On inspection we see that the three groupings on the right hand side of the last expression arepairwise disjoint and hence m(A � B) = m(A � B

�) + m(A

� � B) + m(A � B). Thus 2 * m(E2 � E1)

+ m(E2 � E1�

) + m(E1 � E2�

) = m(E2 � E1) + m(E2 � E1). � m(E1 � E2) + m(E1 � E2) = mE1 +mE2, QED.

Problem 11

� Show that the condition mE1 < � is neccesary in Proposition 14 by giving a decreasing sequence <En> of measurable sets with � = � En and mEn = � � n.

� ExampleLet En = (n, �), � n � �. Thus mEn = � � n � �, En�1 � En, En is measurable and � En = �.

Problem 14

� a) Show that the Cantor ternary set has measure 0.

� ProofLet C0 = [0, 1] and let Cn denote the Cn1 with open intervals of length 3n removed from thecenter of each of the 2n1 continuous interval in Cn1. Thus Cn is the nth stage in the constructionof the Cantor set � and � � Cn, � n. Since Cn1 � Cn and mC0 = 1 < �, by Proposition 14, wehave that m(�n�0

� Cn) = limn�� m C�n. However � � �n�0� Cn since � � Cn, � n � 0 � m(�) �

m(�n�0� Cn) = limn�� m C�n = limn�� 2��3 �

n = 0. Thus m(�) = 0. QED

MATH630-2.nb 2

� b) Let F be a subset of [0, 1] constructed in the same manner as the Cantor ternary set except that each of the intervals removed at the nth step has length �3�n with 0 < � < 1. Then F is a closed set, F

� dense in [0, 1] and mF

= 1 - �.

� ProofLet Fn denote the construction at the nth step.

� ClosedConsider Fn

� which must be open � n. Since F

� = � Fn

�, we know that F

� must be open since infinite

unions of open sets are open. Thus F��

= F must be closed since its the complement of an open set.QED

� F�

is dense in [0, 1]Suppose not, then � a, b � F

� � [0, 1], with a < b, such that (a, b) � F

� = �. Thus (a, b) � F, with a,

b � F. But this contradicts the contruction of F since every continuous interval must have itsmiddle removed for some Fn and (a, b) never did since the entire interval is in F. Thus � a, b � F

�

� [0, 1], with a < b, � c � (a, b) � F�. Hense F

� is dense in [0, 1]. QED

� mF = 1 - �At each phase 2n1 intervals are removed of size �3n. Hence mF = m([0, 1]) -

�m�Intervals Removed� = 1 - �n�1� 2n1��3n = 1 - ��2 �n�1

� � 2��3 �n = 1 - ��2 *

2��3��1 2��3

= 1 - �. � mF = 1 - �.

QED

Problem 15

� Show that if E is measurable and E � P, then mE = 0.

� ProofSuppose BWOC that mE > 0. Then we know that mE = m(E �

� y), � y. Let <Ei> be a sequence of

disjoint sets defined by Ei = E �� ri, with � ri �n�1

� being an enumeration of the rational numbers in[0, 1). Clearly � Ei � [0, 1] so m(� Ei) � m[0, 1] = 1, but since Ei are pairwise disjoint, m(� Ei) =� mEi = �. Which gives a contradiction. Hence mE = 0. QED

MATH630-2.nb 3

Problem 16

� Show that, if A is any set with m*A > 0 then there is a non-measurable set E � A.

� ProofLet An denote A � [n, n + 1), � n � �Since An are disjoint we know that m*A = m*(� An) = � m*An � � n such that m*An > 0. Let Pdenote the standard non-measurable set, <ri> be an enumeration of the rationals and Pi = P �

� ri.

Let B = (An - n). Let Bi = B � Pi. By problem 15, if Bi is measurable then it must must havemeasure 0. Since {Bi} are pairwise disjoint and � Bi = B we must have that 0 < m*B � �i�1

� m*Bi =0. Which is a contradiction thus � a set � B which is not measurable. Finally by translationinvariance of m* we know that � a set � An � A which must be not measurable.

MATH630-2.nb 4



Problem 18

� Show that (v) does not imply (iv) in Propostion 18 by constructing a function f such that {x : f(x) > 0} = E, a given non-measurable set, and such that f assumes each value at most once.

� Counter-ExampleLet E = the standard non-measurable set on [0, 1]

Let f(x) = �x, x � E�x, x � E defined on domain [0, �)

Thus {x : f(x) > 0} = E, and f(x) = � has at most one solution for any �, namely � or -�.Hence {x : f(x) = �} is measurable � � but {x : f(x) > 0} is not measurable.

Problem 19

� Let D be a dense set of real numbers. Let f be an extended real-valued function on � such that {x : f(x) > �} is measurable � � � D. Show that f is measurable.

� ProofWe wish to show that {x : f(x) > �} is measurable � � � D is equivalent to the condition {x : f(x) >�} is measurable � �.

Take � � D then define An = {x : f(x) > �n , with �n � (�- 1n , �) � D}.

We know that (�- 1n , �) � D is non-empty � n since, D is dense.

Consider �n1� An � �n1

� �x : f �x� � � � 1n � = {x : f(x) �}. But {x : f(x) �} � �n1� An

since � n, �n < �. Thus �n1� An = {x : f(x) �} which means that {x : f(x) �} is measurable and

hence by Proposition 18, {x : f(x) > �} is measurable, so {x : f(x) > �} is measurable � �. QED

Problem 20

� Show that the sum and product of two simple functions are simple.

� Proof

� �A�B � �A � �B

�A� B = �1, if x � A� B0, if x � A� B

�A � �B = �1, if x � A and x � B0, if x � A or x � B

But x � A and x � B � x � A � B and, x�A or x� B � x � A � B, thus �A� B �A � �B,QED.

MATH630-3.nb 2

� �A�B � �A � �B �A � �B

�A� B = �1, if x � A� B0, if x � A� B

�A � �B � �A � �B = �A � �B � �A� B = �1, if x � A or x � B0, if x � A and x � B

But x � A or x � B � x � A � B and, x�A and x�B � x � A � B, thus�A� B �A � �B � �A � �B . QED.

� �A

� 1 �A

�A� = �

0, if x � A1, if x � A

1� �A = �1 � 1 0, if x � A1� 0 1, if x � A

Thus �A� 1� �A. QED

Problem 21

� a) Let D and E be measurable sets and f a function with domain D � E. Show that f is measurable if and only if its restrictions to D and E are measurable.

� Prooff is measurable � �x : f �x� � �, x � D � E� is measurable � �.

A � �x : f �x� � �, x � D � E� = B � �x : f �x� � �, x � D� � C � �x : f �x� � �, x � E�

Clearly if B and C are measurable then A is measurable as the union of measurable sets.

Since D is measurable, we have that A ~ D is measurable if A is measurable, but A ~ D = C, so Cwould be measurable, and similarly A ~ E = B gives B measurable. Thus if A is measurable B andC are measurable.

Since B and C are equivalent to f restricted to D and E respectively, we must have that f ismeasurable if and only if its restrictions to D and E are measurable. QED.

MATH630-3.nb 3

� b) Let f be a function with measurable domain D. Show that f is measurable iff the function g defined by g(x) = f(x) for x � D and g(x) = 0 for x � D is measurable.

� Prooff is measurable � �x : f �x� � �, x � D� is measurable � �.

� � 0, {x : g(x) > �} = �x : f �x� � �, x � D�, thus {x : g(x) > �} is measurable iff f ismeasurable.

� � < 0, �x : g�x� � �� = �x : f �x� � �, x � D� � �x : x � D�.

However the first term on the right is clearly measurable iff f is measurable, and the second is just x� D�

, but D�

is measurable since D is measurable. Thus �x : g�x� � �� is measurable for � < 0 iff f ismeasurable.

� �x : g�x� � �� is measurable � � iff f is measurable, so g is measurable iff f is measurable. QED

Problem 23

� Prove Proposition 23 by establishing the following lemmas:

� a) Given a measurable function f on [a, b] that takes the values ± � only on a set of measure 0, and given > 0, � M such that |f| � M except on a set of measure less than / 3.

� ProofWe know that A = �x : f �x� �� and B = �x : f �x� � �� are measurable for all �.

Thus C� = A � B is measurable = �x : f �x� � ��.

We also know that C��

� C��1�

, since � < � + 1.

Furthermore we know that since f is infinite on only a set of measure 0, then � �0 such that mC�0

� <

�. Thus by Proposition 14, we have that m�n1� C�0�n

� = limn�� m�C

��0�n = 0. Hence by

definition of limit, � D = C�0�n�

, for some n, with mD < � / 3. QED.

MATH630-3.nb 4

� b) Let f be a measurable function on [a, b]. Given > 0 and M, � a simple function � such that |f(x) - �(x)| < except where |f(x)| � M. If m � f � M, then we may take � so that m � � � M.

� ProofLet � be a simple function with values of n*(� / 2) where n � � � (-(M+1)*2/�, (M+1)*2/�). Thussimply choose the value for n that satisfies |f(x) - �(x)| < � when |f(x)| < M � -� - f(x) < - �(x) < -f(x)+ � � � + f(x) > �(x) > f(x) - � � � + f(x) > n*(� / 2) > f(x) - � � 1 + f(x) / � > n / 2 > f(x) / � - 1 � 2 +2 * f(x) / � > n > 2 * f(x) / � - 2.

Such an n being guaranteed to exist by the fact that � at least one integer in every range 1 + a > x >a - 1. Thus we have defined � such that it has the neccesary property. This construction willclearly yield results of the form m � � � M, when m � f � M, except possibly within � / 2 of thebounds where we may choose the bound itself to satisfy our criterion.

� c) Given a simple function � on [a, b], � a step function g on [a, b] such that g(x) = �(x) except on a set of measure less than / 3. If m � � � M then we can take g such that m � g � M.

� ProofLet � = b�a2n , define [am, bm] = [a + (m-1)*�, a + m*� ], with m � � � [0, 2n]. Thus [am, bm]partition [a, b] into n intervals of width �.

We know that �(x) = i1N �i*�Ai, with Ai = {x : �(x) = �i}.

Define g�x� ��m�, � x � �am, bm�, with �m = �i such that m(Ai � [am, bm)) is the max over all i.

Clearly g(x) is a step function and in the limit n � � we have that g(x) = �(x). Also we have thatm({x : g(x) = �(x)}) = m1

2nm(Ai � [am, bm)), for the Ai associated with �m.

Hence Kn = m({x : g(x) � �(x)}) = m12n

m(Ai�

� [am, bm)).

However Kn forms a strictly decreasing series since for successive n, each pair of new partitionpieces must cover at least as well as the original. And we know that limit Kn � 0 so, � n such thatKn < � / 3 � � > 0. Thus � some step function g conforming to the same bounds as � that will haveg(x) = �(x) except on a set of measure less than � / 3.

MATH630-3.nb 5

� d) Given a step function g on [a, b], � a continuous function h on [a, b] such that h(x) = g(x) except on a set of measure less than / 3. If m � g � M then we can take h such that m � h � M.

� ProofClearly for any interval [c, d], we may construct a continuous curve joining g(c) and g(d).Since g is a step function, � a partition x0 < x1 < ... < xN such that for each interval (xn, xn�1), g(x)assumes only one value. Let �n�x� be any continuous function joining g(xn �

�6��N�1� ) and g(xn �

�6��N�1� ), which does not exceed its boundary points, for n � � � [1, N-1].

Define h(x) = �g�x�, if � x� xn � � �6��N�1� , � n

vn�x� , if � x� xn � � �6��N�1�

Thus h(x) is neccesarily a continuous function and h(x) = g(x) except for N intervals of length2� �6��N�1� , which has total measure ��N3��N�1� < �3 . Thus by explicit construction we have found aacontinuous h satisfying the conditions. QED.

Problem 29

� Give an example to show that we must require mE < � in Proposition 23.

� ExampleLet E = (-�, �). Let fn�x� xn . Thus fn�x� � 0.

Take � > 0. Then fn�x�� 0 fn�x� � � n some N implies that | x | < N*�.

Hence B = (-N*�, N*�) is the collection of all x such that fn�x� � � n N. Thus B�

� A = (-�,-N*�] � [N*�, �) is the smallest set such that � x � A

� and � n N, fn�x� �. This is true

because if we were to remove any element of A, we would add an element to A�

with fn�x� �.Hence the measure of any set, K, with the neccesary property of Proposition 23 will be larger thanmA = �. Thus � � > 0 we have mK = � > �. Hence showing a counter-example.

MATH630-3.nb 6

Problem 30

� Prove Ergoroff's Theorem: If < fn> is a sequence of measurable functions that converge to a real-valued function f a.e. on a measurable set E of finite measure, then given � > 0, � a subset A � E with mA < � such that fn converges to f uniformly on E ~ A.

� ProofDefine Ai � E for i � � with mAi < �i = 2�i�! and a smallest Ni such that � x � Ai, and all n Ni, |fn�x� � f �x�| < �i = 1i . We know that Ai exists by Proposition 24.

Let A = � Ai. Thus mA � mAi < 2�i�! = !.

Since A� = ~ � Am = � Am

� � � x � E ~ A that fn converges to f since | fn�x� � f �x�| < �m � 0.

Thus we have fn converging with some Nm depending only on �m, when fn(x) restricted to A. So itdemonstrates uniform convergence. QED

MATH630-3.nb 7



Problem 1

� a) Show that if

f �x� � � 0, x is irrational1, x is rational

Then

R��

a

bf �x�� x � b � a and R� �

��a

b f �x�� x � 0

� ProofR��

a

bf �x�� x � inf �a

b��x�� x, � step functions �(x) � f(x), � x.

But each step function is composed of a finite collection of open intervals, and in the domain eachopen interal � an rational number since the rationals are dense. Thus �(x) � f(x), � x implies thatthe value over any open interval of the step function �(x) is � 1. Thus the infimum of the integral

of all such step functions must be when �(x) = 1, � x.

R��

a


b��x�� x � �b a��1 � b a.

R� ��a

b f �x�� x � sup �ab��x�� x, � step functions �(x) f(x), � x. Since irrationals are dense, each

interval in the composition of � has an irrational in its domain. Thus �(x) f(x), � x � that overany open interval in �, �(x) 0. Thus the supremum of the integral of all such step functions must

be when �(x) = 0. Hence R� ��a

b f �x�� x � sup �ab��x�� x � 0. QED

� b) Construct a sequence < fn> of nonnegative, Riemann integrable functions such that fn increases monotonically to f. What does this imply about changing the order of integration and the limiting process.

� ConstructionLet fn = x

1��n over (0,1], so limit n � � we have f = 1. �01

fn�� x = n�1��n . So

limn�� 01

fn�� x � 1 � �01

f �� x � �01limn�� fn�� x.

Thus we have that the limit of the integral is the integral of the limit.

MATH630-4.nb 2

Problem 2

� a) Let f be a bounded function on [a, b], and let h be the upper envelope of f. Then R��

��

a

bf � �a

bh.

� ProofR��

a


b��x�� x, � step functions �(x) � f(x), � x.

We know from Problem 2.51(a) that h(x) � f(x), and from 2.51(b) we have that h(x) is upper semi-continuous so that h(y) � limx�y

��h�x� = inf

��0sup

0��xy��h�x�.

From 2.51(a), h�x� � f �x� if and only if f is upper semi-continuous at x, so �(x) � h(x), exceptpossibly where f(x) is not upper semi-continuous � f(y) < limx�y

��f �x�. However

h�y� � inf��0

sup�xy��

f �x� so h(y) = limx�y��

f �x�, when f(y) < limx�y��

f �x�. Let A = {x : �(x) < h(x)}

Suppose A � �. Let � � A. So h(�) > �(�), and h(�) = limx��

f �x�. Let � = h(�)-�(�). Bydefinition of lim sup we have that � � > 0 such that | sup f(x) - h(�) | < � / 2 and 0 < | x - � | < �, � �in that range such that �� f �� 2� h�� 2 � �� f �� 2� � h�� 2 � f �� h�� f �� h�� h�� f �� . Which means that � and �must be in differant intervals of the partition of �, however this works � � > 0 which implies that �must be a boundary point between intervals of the partition. However since � is a step function it iscomposed of a finite number of distinct intervals and thus has a finite number of boundary points.So A has at most a finite number of items in it.

Thus � � h except on a set of measure 0, so R��

a

bf �x�� x = inf �a

b��x�� x � �a

bh�x�� x.

Define <�n�x�> to be a sequence of step function such that �n is over a equipartition of [a, b] into 2n

pieces, with the value of �n over each piece to be sup h(x) over that piece.

BWOC Suppose limn�� n�x� � h(x). That implies that � � > 0 such that � n, | �n�x� - h(x) | � � �� partition intervals including x, � � in that interval with h(�) � h(x) + �. However h(x) �lim��x��

h��, which gurantees that for some small distance � > 0, h(x) + � > h(�) � � � (x-�, x+�).Thus we have a contradiction since h(�) must be < h(x) + � for � within � of x.

Thus limn�� n�x� = h(x) and clearly �n is bounded since h and f are bounded. Thus by

MATH630-4.nb 3

Proposition 6 we have that �abh�� x = limn��a

b�n�� x. Furthermore �n are each step functions � f

so that limn��ab�n�� x � inf�a

b�� x = R��

��

a

bf �x�� x, where � is any step function � f.

Hence �abh�� x � R��

��

a

bf �x�� x. So using the previous bound we have R��

��

a

bf � �a

bh. QED.

� b) Use Part (a) to prove that: A bounded function f on [a, b] is Riemann integrable if and only if the set of points at which f is discontinuous has measure 0.

� ProofA bounded function f is Riemann integrable if and only if R��

��

a

bf = R� ��a

b f .

Since f is bounded we know R��

a

bf � �a

bh, where h is the upper envelope from (a). By symmetry

of defintion, interchanging the supremums and infimums we know that R� ��a

b f � �abg, where g is

the lower envelope.

So f is Riemann integrable if and only if �abh = �a

bg. However 2.51(c) gives us that g(x) = h(x) if

and only f is continuous at x. So by Proposition 5(i), we know that �abh = �a

bg � �a

b�h g� � 0.

But since h�x� � g�x�, � x we also know that �ab�h g� � 0 if and only if, the set of points were

h�x� � g�x� has measure 0, since otherwise � some � > 0 such that m(�x : h�x� g�x� � ��) > 0,

and hence �ab�h g� � �*(measure of that set) > 0. But the set of points h�x� � g�x� is identical to

the set of the discontinuities in f. f is Riemann integrable if and only if the set of points at which fis discontinuous has measure 0. QED.

MATH630-4.nb 4

Prove parts (iii) - (v) of Proposition 5

� (iii) If f � g a.e., then

�E f � �Eg

Hence

��E f � � �E�� f ��

� Proof�E f �Eg � �E f �Eg 0 � �E� f g� 0. However f g a.e. � f g 0 a.e.

Thus for simple functions � f g 0 a.e., sup �E � 0. Since � 0 except on a set of measure0. However we know that the integral must exist since the integral of f and g exist. since it existsit must equal sup �E �, so �E� f g� 0 � �E f �Eg. QED.

It follows immediately that �E�� f �� E f , since | f | � f. But also | f | = | -f | � -f, so �E�� f �� E f� �E�� f �� E f , � �E�� f �� E f �E�� f �� E f �E�� f ��. QED

� (iv) If A � f(x) � B, then

A �m�E � �E f � B �mE

� ProofSince A f(x) B, we know from (iii) that �EA �E f �EB. However the integral of a constantis just that constant times the measure of the set integrated over, so �EA � A�mE and�EB � B�mE. Hence A�m�E �E f B�mE. QED.

� (v) If A and B are disjoint measurable sets of finte measure, then

�A�B f � �A f � �B f

� Proof�AB f � � f � AB � � f � � A � B�, since AB = A � B.However from (i) we have that � f � � A � B� = � f � A � � f � B � �A f � �B f

MATH630-4.nb 5



Problem 3

� Let f be a nonnegative measurable function. Show that � f � 0 implies that f � 0 a.e.

� ProofBWOC, suppose f � 0 a.e. Then � some set E with mE > 0 such that f �x� � 0, � x � E.

Since �x � f �x� � 0� � �n�� x � f �x� � 1��n � , we know that the countable sum of the measures of thesets on the righthand side is greater than or equal to the measure of the left hand side which is non-zero. So � N such that if A = �x � f �x� � 1��N �, then mA > 0.

Define h�x� � min� f �x�, 1��N �, � x. Thus h�x� f �x�, � x. Let F � A, such that 0 < mF < �.

We know � f � supg f

� g, and � f � �F f , the latter coming from the fact that since f is

nonnegative and �F f � � f �F with f �F f , � x. This implies

� f � �F f � supg f

�F g � �F h � 1��N mF � 0. Thus � f � 0, which contradicts the

assumption that � f � 0, which implies that f � 0 a.e. QED

Problem 4

� Let f be a nonnegative measurable function.

� a) Show that � an increasing sequence � �n � of nonnegative simple functions each of which vanishes outside a set of finite measure such that f � lim�n.

� Construction

Define �n = �m�1��2n , � x � �n, n such that m�1��2n f �x� � m��2n and � m � � with m 22n

0, otherwiseThus �n assumes exactly 22n-1 non-zero values over a set of measure 2n. Thus �n is anonnegative simple function and vanishes except on a set of finite measure.

Furthermore we can show that �n�1 � �n � x and n as follows. Suppose �n�x� = 0 then clearly�n�1�x� � 0, thus suppose �n�x� = m�1��2n for some integer m with 0 < m 22n. This implies thatm�1��2n f �x� � m��2n � 2�m�1��2n�1 � m�1��2n f �x� � m��2n � 2m��2n�1 , but if m� 1 22n, then clearly2�m� 1� 2 22n � 22�n�1�, so thus we have that �n�1�x� � 2�m�1��2n�1 � m�1��2n � �n�x�, so its anincreasing sequence.

Now we need only show that f = lim �n.

Let [x] denote the least natural number � x, and log2 x = to the logarithim base 2 of x, with log2 0 =-�.

Given � > 0 and some x0, we may choose N = max{[�x0 �], �log2 �, [log2 f �x0�]}.We wish to show that � n > N, ��n�x0� � f �x0� � < �.

Since n > �x0 � � �x0 �, we have that x0 � �n, n so x0 meets the first condition for assigning anon-zero value to �n�x0�.

Since n > [log2 f �x0�] > log2 f �x0�, we know that 2n � 2log2 f �x0� � f �x0�. Sof �x0� � m��2n 22n

��2n � 2n, for some m � �, m 22n. Thus x0 meets the second condition forassigning a non-zero value to �n�x0�.

Finally since n > �log2�] and n meets the requirements to assign x0 a particular non-zero value,we know that � �n�x� � f �x� � to the spacing of the values of �n = 1��2n < 1��2�log2 � 1��1 = �. Thus� �n�x� � f �x� � < �. � �n converges pointwise to f � x. So f = lim �n.

MATH630-5.nb 2

Thus we have constructed a �n satisfying all required properties, QED.

� b) Show that � f � sup � � over all simple functions � � f.

� ProofBy definition � f � sup

h f� h, where h is a bounded measurable function with m�x : h�x� � 0� is

finite.

We may of course limit ourselves to taking the sup over all h'�x� such that h'�x� � max�h�x�, 0�.However all such h'�x� are nonnegative measurable functions. The measurability of h'�x� derivingfrom the fact that �x : h�x� � 0� and the set �x : h�x� � 0� must both be measurable.

Hence we may apply part a) and conclude that � h'�x�, � an increasing sequence � �n � such that�n� h'�x�. Thus by Proposition 10 we have that � h' � lim � �n which equals the sup � �n since itis an increasing sequence. Thus the sup over some subset of all such simple functions � must be =

suph f

� h = � f . So � f sup � �.

However � f, and since we are considering a sup we may assume that � � 0. Thus by Proposition8 we have that � � � f . So � f � sup � �.

� � f � sup � �, over all simple functions � f. QED

MATH630-5.nb 3

Problem 5

� Let f be a nonnegative integrable function. Show that the function F defined by

F�x� � ��x

f

is continuous by using Theorem 10.

� ProofIn order to show continuity we need that F�a� = limx�a F�x�, � x. � We need ��

af = limx�a ��

xf .

Define fn � � f �x�, x � a� 1��n

0, otherwise then fn is an increasing sequence with fn� f as n�� x � (-�,

a).

� x < a, we have that ��x

f � ��a

fNfor some N such that x � a� 1��N � a � N > 1��a�x .

Thus limx�a� ��x

f limn�� a

fn, and by Theorem 10 we have that limn�� a

fn � ��a

f . Butalso we know that � x such that 0 < a - x < � < 1, we have that ��

xf � ��

afM , � M < 1�� , thus as �

� 0 in the limit x � a�, we get that limx�a� ��x

f � limn�� a

fn.

So limx�a� ��x

f � limn�� a

fn � ��a

f .

Now consider limx�a� ��x

f � ��a

f � limx�a� �ax

f , by Proposition 12. We know that since f isintegrable f < � a.e. Since if f = � on some set, A, of measure > 0, we would have� f � mA � � �, which is false since � f � �, by definition of integrable. Thus without lossof generality we may assume that f < � everywhere since this won't change it's integral. Now sincef is finite over a closed interval we know that f is bounded over that interval. Let N(x) be themaximum of f �y� � y � [a, x]. Hence, �a

xf m[a, x]*N(x), by Proposition 5. However it is clear

that N�x1� N�x2�, if a � x1 x2. Thus limx�a� �ax

f limx�a� ma, x N�x� = 0.

Thus limx�a� ��x

f � ��a

f � limx�a� �ax

f � ��a

f . Hence ��a

f = limx�a ��x

f � F�a� =limx�a F�x�. � F is continuous.

MATH630-5.nb 4

Problem 6

� Let � f n � be a sequence of nonnegative measurable functions that converge to f, and suppose fn � f n. Then

� f � lim � fn

� ProofBy Fatou's Lemma we know that � f lim�� fn. Also since fn f � n, we know that � fn � f ,� n, and thus lim

�� fn � f .

However lim�� fn lim��

� fn by definition of lim sup and lim inf. So by squeezing,lim��

� fn � � f � lim�� fn � lim � fn. QED

Problem 7

� a) Show that we may have strict inequality in Fatou's Lemma.

� ProofConsider the sequence � f n � defined by fn�x� = 1 if n x � n� 1, with fn�x� � 0 otherwise.

� x, � N > x, such that fn�x� � 0, � n > N. Thus fn� f = 0. So � f � 0, however lim�� fn � 1,

since � n, we have � fn � 0 � �nn�1

1 � 1. Thus 0 � 1 � � f � lim�� fn. QED

� b) Show that the Monotone Convergence Theorem need not hold for decreasing sequences of functions.

� ProofLet fn�x� = 0 if x � n and fn�x� = 1 for x � n.

� x, � N > x, such that fn�x� � 0, � n > N. Thus fn � f = 0. Hence � f � 0. However consider

lim � fn. � fn � �nn�1

fn � 1. Hence lim � fn � 1 � 0. Hence lim � fn � � f . QED

MATH630-5.nb 5



Problem 10

� a) Show that if f is integrable over E then so is � f � and

��E f � � �E �� f �

Does the integrability of � f � imply integrability of f?

� Prooff is integrable � f � and f � are each integrable. However � f � = f � + f �, so by Proposition 15ii,we have that � f � is integrable and �E � f � � �E f � � �E f �. Also��E f � � ��E f � � �E f � � ��E f � ��E f � �, but since f � and f � are nonnegative we can drop theasbsolute value signs, so ��E f � � �E f � � �E f � � �E � f � . So we have shown the first part.

� f � � f � � f �, So � f � integrable � ( f � + f �) is integrable. However the set A � �x � f ��x� � 0�is disjoint from the set B � �x � f ��x� � 0�, and A and B must be measurable iff f is a measurablefunction. Thus by Proposition 15, �E� f � � f �� A�B� f � � f �� A f � � �B f �. Thus f � andf � are each integrable iff f is a measurable function.

But this implies that �A f � � �B f � � �A�B� f � � f �� E f , is integrable. Thus � f � impliesintegrability of f iff f is a measurable function.

� b) The improper Riemann integral of a function may exist without the function being integrable. If f is integrable, show that the improper Riemann integral is equal to the Lebesgue integral whenever the former exists.

� ProofLet R��a

bf �� x, be a Riemann integrable function with improper limit point a, and f defined for all x

(a, b]. Note that if any other point in (a, b] is improper, we may into two or more integrals eachof which with one improper limit point so we need only assume one limit is improper. In particularwe know from analysis that improper points must be disconnected in order for a function to beRiemann integrable. WLOG I assume that a < b, but all parts of the proof may be reversed for a > b,and will still hold accordingly.

� Case |a| < �

Then consider the sequence of measurable integrable functions f n �, such that

fn�x� � f �x�, a� 1 n � x � b0, otherwise

. Then lim fn�x� = f �x� a.e. Also we have that � fn � � f . Thus by

Theorem 16 we have that �ab

f � lim �ab

fn. But by construction we must have that fn�x� is

bounded, which allows us to apply Proposition 4 to get that �ab

fn = R��ab

fn, and hence

lim �ab

fn � lim R��ab

fn � R��ablim fn � R��a

bf . So �a

bf � R��a

bf .

� Case a = -�

Then consider the sequence of measurable integrable functions f n �, such that

fn�x� � f �x�, �n � x � b0, otherwise

. Then lim fn�x� = f �x� a.e. Also we have that � fn � � f . Thus by

Theorem 16 we have that �ab

f � lim �ab

fn. But by construction we must have that fn�x� is

bounded, which allows us to apply Proposition 4 to get that �ab

fn = R��ab

fn, and hence

lim �ab

fn � lim R��ab

fn � R��ablim fn � R��a

bf . So �a

bf � R��a

bf . QED

MATH630-6.nb 2

Problem 11

� If � is a simple function, we have two definitions for � �, � � � � �� and � � � �i�1

n �ai�mAi. Show that they are equal.

� ProofConsider � � � � �� , By problem 4b, we know that nonnegative measurable functions f,� f � sup � �, with the sup taken over all simple functions � f, and the right hand side evaluatedaccording to the old rule for integrating step functions. Since it follows immediately from thedefinition of f � and f � that �� and �� are simple, we can thus deduce that� �� sup � � � �i�1

n1 �bi�mBi, � �� sup � � � �i�1n2 �ci�mCi, where �� i�1

n1 bi� �Bi and�� i�1

n2 ci� �Ci . So � � � � �� i�1n1 �bi�mBi � �i�1

n2 �ci�mCi, also we know thatBi � C j � � i, j and each bi corresponds to some a j � 0 and each ci corresponds to somea j 0, so �i�1

n1 �bi�mBi � �i�1n2 �ci�mCi � �i�1

n1 �bi�mBi � �i�1n2 ��ci��mCi � �i�1

n �ai�mAi. Since thesets Ci and Bi, must follow the same correspondence. QED.

Problem 12

� Let g be an integrable function on a set E and suppose that f n is a sequence of measurable functions such that � fn�x� � � g�x� a.e. on E. Then

�E lim�� fn � lim��E fn � lim��

��E fn � �Elim��

� fn

� ProofWe know for free that lim��E fn � lim

��E fn, since this is true for any lim sups and infs.

We only need to show that �E lim�� fn � lim��E fn and lim��

��E fn � �Elim��

� fn.

fn is measurable and bounded by an integrable function, therefore fn is integrable. Also we knowthat lim�� fn � fn n thus By Proposition 15iii, we have that �E lim�� fn � �E fn n ��E lim�� fn � lim��E fn. A similar argument for lim sup fn � fn gives us that lim

��E fn � �Elim

�� fn,

and thus we are done. QED

MATH630-6.nb 3

Problem 15

� Let f be integrable over E. Then given � > 0,

� a) a simple function � such that

�E � f � � � �

� ConstructionSince �E � f � � � � ��E f � � � � ��E f � �E � � � ��E f � � �E f � � �E � �, it suffices to show that thelatter is less than �. If we further specify that A � �x � f ��x� � 0� and B � �x � f ��x� � 0�, thenwe may clearly choose � = 0, x E � �A � B�. So we may further simplify the aboveexpression to ��E f � � �E f � � �E � � ��E f � � �E f � � �A � � �B � � � ��E f � � �A � ��E f � � �B � �,which again all that we need to show is that this new expression is less than �.

By Problem 4, we know that nonnegative measurable functions f we have that � f � sup � �,taken over all simple functions � � f. Since f � and f � are nonnegative measurable functions weknow there must exist simple functions �S and �I respectively such that 0 � � f � � � �S � 2and 0 � � f � � � �I � 2. The second implies that 0 � � f � � � ��I� � 2. Thus if wechoose � over A = �S and � over B = ��I , then we have that��E f � � �A � ��E f � � �B � � � 2 � � 2 � �. Thus we have shown the construction asrequired.

� a) a step function � such that

�E � f � � � �

� ConstructionBy the Question 3.23c, which part of the proof to Proposition 22, we know that � a step function �such that � = a simple function � except on a set of less than � / 3 and if M bounds � then M bounds�, � > 0. Let � > 0 and � be a step function such that �E � f � � � � 2. Let N be a number suchthat ��x� � N, x. This is possible since simple functions are always bounded. Choose � suchthat � = � except on a set of measure ��4�N with the set denoted by A, and � bounded by N. Thus�E � f �� E�A� f � � � � �A � f �� < �E�A � f � � � + �A� f � � � = �E�A � f � � � +�A� f � � � � � � � � �E�A � f � � � + �A �� f � � � � �� E�A � f � � � + �A� f � � � + �A�� / 2 + 2*N *mA = � / 2 + 2*N* ��4�N = �.

Therefore �E � f �� .

MATH630-6.nb 4

� a) a continuous function g such that

�E � f � g � �

� ConstructionBy question 3.23d, we have that � a continuous function g such that g = a step function � except ona set of measure less than � / 3, which agress with the bounds on �. With this we may replace theoccurances of � in the above proof with � and � with g and the proof will precede identically,except that we have to infer the boundedness of � from the boundedness of the preceding �. Thuswe may construct g such that �E � f � g � �.

MATH630-6.nb 5



Problem 1

� Let f be the function defined by f �0� � 0 and f �x� � x � sin� 1��x � for x � 0.

Find D�� f �0�, D�� f �0�, D�� f �0�, and D�� f �0�.

� D�� f �0�

D�� f �0� = limh�0�

�� f �h�� f �0��h = limh�0�

�� h�sin� 1��h ��h = limh�0�

��sin� 1��h � = 1, since for arbitrarily small positive h we

can find a value of 1��h which gives sine its maximal value.

� D�� f �0�

D�� f �0� = lim��h�0�

f �h�� f �0��h = lim��h�0�

h�sin� 1��h ��h = lim��h�0�

sin� 1��h � = -1, since for arbitrarily small positive h we

can find a value of 1��h which gives sine its minimal value.

� D�� f �0�

D�� f �0� = limh�0�

�� f �0�� f ��h��h = limh�0�

�� h�sin� 1��h ��h = limh�0�

��sin� 1��h � = 1, since for arbitrarily small positive h

we can find a value of 1��h which gives sine its maximal value.

� D�� f �0�

D�� f �0� = lim��h�0�

f �0�� f ��h��h = lim��h�0�

�h�sin� 1��h ��h = lim��h�0�

sin� 1��h � = -1, since for arbitrarily small positive h

we can find a value of 1��h which gives sine its minimal value.

Problem 2

� a) Show that D�� f �x�� = �D�� f �x�

� ProofD�� f �x�� = lim

h�0�

�� f �x�h�� f �x��h = limh�0�

�� f �x�h�� f �x��h = inf

h�0�sup

0�h�� f �x�� f �x�� =

infh�0�

�� inf0�h

f �x�� f �x�� = �� sup

h�0�

inf0�h

f �x�� f �x��

� �� = � lim��

h�0�

f �x�h�� f �x��h = �D�� f �x�. QED.

� b) If g�x� = f ��x�, then D��g�x� = �D�� f ��x�

� Proof�D�� f ��x� = � lim��

h�0�

f ��x�� f ��x�h��h =� lim��h�0�

g�x��g�x�h��h = lim��

h�0�� g�x��g�x�h��h = lim

��

h�0��g�x� h� � g�x��h =

D��g�x�. QED

MATH630-7.nb 2

Problem 3

� a) If f is continuous on [a, b] and assummes a local maximum at c (a, b), then

D�� f �c� D�� f �c� 0 D�� f �c� D�� f �c�

� ProofWell D�� f �c� = lim

h�0�

�� f �c�h�� f �c��h � D�� f �c� = lim��h�0�

f �c�h�� f �c��h , since lim sups are neccesarily greater

than lim infs. Similarly D�� f �c� D�� f �c�. So we only really need to establish thatD�� f �c� 0 D�� f �c�. Since f �c� is a local maximum � an open interval I �a, b� such that c� I and f �c� is the maximum of f over I. WLOG we may consider h to be sufficiently small suchthat c� h and c� h � I.

Consider D�� f �c� = limh�0�

�� f �c�h�� f �c��h , since c� h � I, f �c� h� � f �c� � f �c� h�� f �c� � 0.

Furthermore since h > 0, we know that f �c�h�� f �c��h � 0. By definition limh�0�

��g�h� � inf

h�0sup

0��hg��,

but as h � 0 , the supremums form a nonincreasing sequence, and hence so long as we permit ± �as acceptable limits, lim

h�0�

��g�h�must exist.

Thus we have that limh�0�

�� f �c�h�� f �c��h 0, since each f �c�h�� f �c��h < 0, and the limit exists.

Similarly D�� f �c� � lim��h�0�

f �c�� f �c�h��h has f �c�� f �c�h��h > 0 over I, and the limit must exist over the

extended reals, so lim��h�0�

f �c�� f �c�h��h � 0. Hence D�� f �c� 0 D�� f �c�. QED.

� b) What if f has a local maximum at a or b?

� AnswerIf f is maximal at a or b then the two derivates which are still inside the interval must exist and obeythe above inequality, however the other two may not even exist. Thus with a < b and f �a� a localmaximum we have that D�� f �a� D�� f �a� 0, and if f �b� is a local max then0 D�� f �b� D�� f �b�.

MATH630-7.nb 3

Problem 4

� Prove: If f is continuous on �a, b� and one of its derivates (say D�) is everywhere nonnegative on �a, b�, then f is nondecreasing on �a, b�; i.e. f �x� f �y� for x y.

� Proof

� Lemma - For a function g�x� such that D��g�x� � � > 0, the above property holds.Suppose � x � y, such that g�x� � g�y�. Since g is continuous we may apply the normal knowledgeof continuous functions to conclude that either g has a local maximum value or g is alwaysdecreasing. If g were always decreasing then we know that D��g�x� = lim

h�0�

�� g�x�h��g�x��h < 0 since

g�x� h�� g�x� � 0, � x. Which is a contradiction, thus g has a local max.

However by problem 3, D��g�c� 0, where c is the location of the local max, but D��g�x� � 0, � xwhich gives a contradiction. Thus we know that g is nondecreasing.

� Consider f �x� � g�x� � �� x, where g�x� is defined as in the lemma.Therefore D�� f �x� = lim

h�0�

�� f �x�h�� f �x��h = limh�0�

�� g�x�h��x�h��g�x��x��h = limh�0�

�� g�x�h��g�x��h � �h��h � =

limh�0�

�� g�x�h��g�x��h � � = limh�0�

�� g�x�h��g�x��h �� 0.

So if f is an arbitrary function of the type given in the problem then we know that � > 0,f �x� � � x, is a nondecreasing function. Suppose � x � y such that f �x� � f �y�, yet we know thatf �x� � � x f �y� � � y. So this means that f �x�� f �y� � 0 and f �x� � f �y� � �y� x�. Let � =f �x� � f �y�, then � > 0, and � � �y� x�. However we may choose = ��y�x��2 > 0. Which givethat � *(y-x) = ��2 . Which is a contradiction. Hence f �x� must be nondecreasing.

MATH630-7.nb 4

� Now all we need to show is that D�� f �x� � 0 iff the other derivates are � 0.However we know that nondecreasing is equivalent to D�� f �x� � 0, and it follows immediatelythat f �x� h� � f �x� and f �x�� f �x� h� are each � 0 for h > 0. However this means that the termsfor which we are considering the limit of in each derivate must be always � 0. So by definition allthe derivates are � 0 if D�� f �x� � 0.

Now we need only look at the other direction. Clearly for both D�� f �x� � D�� f �x� and D�� f �x� �D�� f �x�, so if respectively D�� f �x� and D�� f �x� are � 0 than the other two will be. So in order tocomplete the other direction we only need that D�� f �x� � 0 implies that the function isnondecreasing. However if we replace D�� f �x� with D�� f �x� and the related definitions in theabove two steps its easy to see that the remainder of the proof will preceed verbatim. HenceD�� f �x�� 0 implies that f �x� is nondecreasing. This completes the proof. QED.

MATH630-7.nb 5



Problem 7

� a) Let f be of bounded variation (BV) on �a, b�. Show that for each c � �a, b� the limit of f �x� exists as x � c� and also as x � c�. Prove that a monotone function (and hence a function of BV) can have only a countable number of discountinuities.

� Proof

� Lemma: For a monotone BV function g on �a, b�, the number of discontinuies such that � � � 0, �g�x � �� g�x � �� 1

n , � n � �, is finite.

BWOC Suppose this number is infinite. Since it is a BV, we know that g is strictly real valued.However we can start at x � a, and proceed towards b such that each discontinuity we choose �succifiently small that � infintely many disconinuties between the current location and b. Thus weexperience arbitrarily many increases of at least 1��n , which since g is monotonic implies thatg�b� � �, which is a contradiction. Hence the number is finite.

� The number of discontinuities of g as above is at most countable.Let An � �x ��g�x� �� g�x� �� 1��n , � > 0}. It is clear that any discontinuity of a monotonefunction must be in some An for n sufficiently large. So the set of all discontinuities is �n�1

� An,which must be countable since it is a countable union of finite sets. Thus the number ofdiscontinuities is countable.

� The limit of f �x� exists as x � c� and x � c�

Consider a fixed c. Suppose � � > 0 such that f over �c� �, c�, contains no discontinuties, then f iscontinuous in that region and limit of f �x� as x c�exists.

Thus we may assume that � > 0 � a discontuinity in f over the region in question. We need toshow that given � > 0, � � > 0, such that x with 0 c� x � � � f �x�� f �c� � �. However weknow that f is BV and thus the sum of two monotone functions g, h. Thus by the first lemma thereexist only finitely many points with �g�x�� g�c� � �1��n . So we may take � sufficiently small such thatall points of differance > ��2 are more than � away. Similarly for h. Since they contribute withopposite sign, � f �x� � f �c� � �, were � is the minimum of the two �'s needed.

Hence limit x c� exists. The proof however clearly proceeds likewise for x c�. QED

� b) Construct a monotone function on �a, b� which is discontinuous at each rational point.

� Construction

Let the sequence ri �i�1� , be an enumeration of the rationals in the interval �0, 1�. Define

fr�x� � � r, x � r0, x r

.

Let f �x� � i�1� fri�x� � 1��2n . Claim that f is such a function as required.

Clearly i�1� fri�x� � 1��2n � i�1

� 1��2n � 1. Thus since the lefthand side is a monotonically increasingbounded series it must converge to some number less than infinity. Furthermore it should be clearthat x, y � �0, 1� with x y, we have that � some q � �x, y� �, since rationals are dense. Sof �y�� f �x� � fq�y�� 1��2N , for some N since all terms for which fri�x� � 0 yields fri�x� � fri�y�, andfq�x� � 0 � fq�y�. Thus f �x� is monotonicly increasing.

Furthermore p � �, and h > 0. f �p� h� � f �p� h� � fp�p� h�� 1��2N , using the above property.However since p � � implies that fp�p� h� � fp�p� � p. Thus h > 0,f �p� h� � f �p� h� � p��2N � 0. Which implies that f �p� h�� f �p� h� � p��2N , so limit h 0,limxp� f �x� � limxp� f �x� � p��2N . Hence f �x� is discontinuous at p since the limits can not beequal.

Thus f satisfies all the conditions desired. QED.

MATH630-8.nb 2

Problem 8

� a) Show that if a c b, then Tab � Ta

c � Tcb and that hence Ta

c Tab.

� Proof

� Tab Ta

c � Tcb

By definition Tab � sup t, where sup is taken over all possible partitions of �a, b�. However

t � i�1k � f �xi� � f �xi�1� �. Then either c � xi, for some i, or c � �xi, xi�1�, for some i.

If c � xi, then clearly we may split the partition over �a, b� into two partitions, one over �a, c� andthe other over �c, b�, such that tab is equal to tac + tcb, for those partitions.

If c � �xi, xi�1� then consider the partition a � x0 x1 ... xi c xi�1 ... xk � b. In thispartition t2 is � to the original t � t1, since � f �xi�1� � f �xi� � � � f �xi�1�� f �c�� f �c�� f �xi� � �� f �xi�1�� f �c� � � �� f �c�� f �xi� �, by triangle inequality. If we break the new partition as above wehave two partitions about �a, c� and �c, b�, respectively with t taken over these partitions summed =t2 � t1. Hence Ta

c � Tcb � Ta

b.

� Tab � Ta

c � Tcb

Every partition p1, p2 over �a, c� and �c, b� respectively can be used to form a partition over �a, b�by joining the two at c. However Ta

b = sup t over all possible partitions, so this includes thepartition formed by the union of p1 and p2 and hence Ta

b � Tac � Tc

b.

Thus Tab � Ta

c � Tcb. Showing that Ta

c � Tab.

� b) Show that Tab� f � g� Ta

b� f �� Tab�g�, and Ta

b�c f � � �c � Tab� f �.

� Proof

� Tab� f � g� Ta

b� f �� Tab�g�

Tab� f � g� � sup t� f � g�, where t� f � g� � i�1

k �� f � g��xi�� f � g��xi�1� � =i�1

k � f �xi�� g�xi� � f �xi�1� � g�xi�1� � � i�1k �� f �xi� � f �xi�1� ��g�xi�� g�xi�1� �� by the triangle

inequality, however this last term is just t� f �� t�g�.

Thus sup t� f � g� � sup �t� f � � t�g��, hence Tab� f � g� � Ta

b� f �� Tab�g�. QED

MATH630-8.nb 3

� Tab�c f � � �c � Ta

b� f �Ta

b�c� f � � sup t�c� f �, where t�c� f � � i�1k �c� f �xi�� c� f �xi�1� � = i�1

k �c � �� f �xi�� f �xi�1� �= �c � �i�1

k �� f �xi�� f �xi�1� � = �c � � t� f �.

Thus sup t�c� f � � sup � c � � t� f � � �c � � sup t� f �, hence Tab�c� f � � �c � �Ta

b� f �. QED

Problem 10

� a) Let f be defined by f �0� � 0 and f �x� � x2�sin� 1x2 �, for x � 0. Is f of BV

on ��1, 1�?

� AnswerNo. Over any interval �a, b�, the diffence from x2�sin� 1��x2 �, will be = a2�sin� 1��a2 � � b2�sin� 1��b2 �. Suppose a is a minimum, thus 1��a2 � n� � � ��2 , for some integer n. Thus � an adjacent maximum atb such that 1��b2 � n� � � ��2 . So a2�sin� 1��a2 � � b2�sin� 1��b2 � = 1��n�� 2

��1�� 1��n�� 2�1=

�2��2�n�1�� 2��2�n�1�� = �8�n��4�n2�1�� . However as the number of partitions gets arbitrarily large we mayadd arbitrarily many terms of this form with n increasing. SoT�1

0 � f � � n�1� � �8�n��4�n2�1�� n�1

� 2� 1��n�� , since the last series is a constant times theharmonic series. Therefore f is not a BV.

MATH630-8.nb 4

� b) Let g be defined by g�0� � 0 and g�x� � x2�sin� 1x �, for x � 0. Is g of BV

on ��1, 1�?

� AnswerYes. Since �g�x� � is symmetric in x we know that it is suffiecent to show that g�x� is BV on ��1, 0�.

Consider a single transition for a minimum to an adjacent maximum. It is clear that it will proceedmonotonically in this region. On inspection it is clear that t over a monotonic region will beindependant of the choice of partition and have the value equal to the differance of the end points.(Either p or n is 0).

From problem 8a) and an obvious induction it is clear that Tab may be broken into finitely many

pieces. We know that minima and maxima occur at sin� 1��x � � � 1 � x = 1��n�� 2. It is obvious that

T�1

�2�� <�, purely by inspection and our knowledge of basic calculus. Proceed by breaking Tab into m

pieces such that the first piece is T�1

�2�� , and the proceeding m-2 pieces follow for decreasing valuesof n, and the last piece represents the tail going to 0.

Now we need to evaluate T over one these pieces that have been set up to run from a local max to alocal min or vice versa. So the points in question have the form � 1��n�� 2

, 1��n�1�� 2 . So

�� 1��n�� 2�2 � sin�n�� 2 � � � 1��n�1�� 2

�2 � sin��n� 1� � � � ��2 � � = �� 1��n�� 2�2 � � 1��n�1�� 2

�2 � =

� 2��n��2�2�� 2 �2��n�� 2 �2��n�1�� 2 �2� < � 2��n2� 1��4 ��n�1�2��n�2�2 � < � 2��n2� 1��4 ��n�2�4 �, note that the inequalities depend on the fact

that n < 0.

Now we want to consider the total T�10 � T�1

�2�� +n��1��m�2� � 2��n2� 1��4 ��n�2�4 �� Ttail

0 , where Ttail0 is the part not

covered in m steps. Taking limit m �, the tail contribution must go to 0. So we only need that

the n��1�� 2��n2� 1��4 ��n�2�4 � < �. However this may be rewritten as n�1

� � 2��n2� 1��4 ��2�n�4 �. However this is

bounded above by 3��n2 for n large, which is a convergent sequence. Thus our sum must converge.Hence T�1

0 �. So T�11 � � x2�sin� 1��x � is BV. QED.

MATH630-8.nb 5

Problem 14

� a) Show that the sum and differance of two AC functions are also AC.

� ProofLet f and g be AC. Let � > 0. Then � �1, �2 such that i�1

n � f �xi�� f �xi� � ��2 and

i�1n � g�xi

�� g�xi� � ��2 finite collection of intervals ��xi, xi��, such that

i�1n �xi

� � xi � � � min��1, �2�.

However i�1n � f �xi

�� g�xi�� f �xi� � g�xi� � � i�1

n � f �xi�� f �xi� � � ��g�xi

�� g�xi� �, by triangleinequlaity. Further i�1

n � f �xi�� f �xi� � � ��g�xi

�� g�xi� � < ��2 � ��2 = �.

Hence we may choose � � min��1, �2� in order to satisfy the condition and show that f � g is AC.

Since � f �xi�� f �xi� � = �� f �xi

�� f �xi�� = �� f �xi�� f �xi�� , we know that f is AC implies

that -f is AC. So f � g must also be AC.

� b) Show that the product of two AC functions is AC

� Proof

� Lemma: f is AC implies that � f �2 is AC.Let f be AC with respect to �a, b�. Then we know that f is BV, and hence that since �a, b� is aclosed interval, that � f � is bounded. Let M > 0 be a bound on � f �. Let � > 0. Since f is AC, � � >0such that i�1

n � f �xi�� f �xi� � ��2�M , finite collection of intervals ��xi, xi

��, such thati�1

n � xi� � xi � �.

Consider i�1n �� f �xi

��2 � � f �xi��2 � = i�1n � f �xi

�� f �xi� � �� f �xi�� f �xi� � �

i�1n � f �xi

�� f �xi� � � �� f �xi�� f �xi� �� i�1

n � f �xi�� f �xi� � �2�M < ��2�M �2�M = �.

Thus � f �2 is AC. Concluding the Lemma.

MATH630-8.nb 6

� If f , g are AC over �a, b� then f g is AC.

f � g � � f �g�2�� f �2��g�2��2 , but that both sums, differances and squares of AC functions are AC, so weonly need that constant multiples of AC functions are AC. But for c� f , where f is AC it followsimmediately that for any � > 0 we may choose f constrained by ��c , and the c will carry through tomake c� f , AC. (Note c = 0 is obviously true.)

Hence f � g can be gotten as a sequence of operations that perserve AC, so it is also AC. QED.

� c) If f is AC on �a, b� and if f is never zero there, then the function g �1f is

also AC on �a, b�.

� ProofSince f is never 0 over a closed interval, we know that � m > 0 such that � f � � m everywhere. Let �> 0. Since f is AC, � � > 0, such that i�1

n � f �xi�� f �xi� � � �m2.

Consider i�1n � 1��f �xi

�� 1��f �xi� � = i�1

n � f �xi�� f �xi��f �xi

�� f �xi� � � i�1n � f �xi�� f �xi

��m2 � < ��m2��m2 = �.

Thus g � 1��f is AC. QED.

MATH630-8.nb 7



Problem 7

� a) For 1 � p � �, we denote l p the space of all sequences � �v �v�1� such

that �v�1� ��v ��

p � �. Prove the Minkowski ineqality for sequences

� � �v � �v � �p � � � �v � �p � � � �v � �p.

Here we have 1 � p � �,

� � � �v � �p�p � �v�1

� ��v ��p and � � �v � �� sup � �p �

� Proof

Case 1 � p � �

� � �v � �p � ��v�1� ��v ��p� 1��p � � � �v � v � �p = ��v�1

� �� v � v ��p�� 1��p .

If either �v or v has norm 0, the proof is trivial so we may assume that each has norm greater than1 and then define real numbers , � > 0 such that � � �v�� p � 1, � � v�� p � 1.

��v�1� �� v � v ��p�� 1��p � ��v�1

� �� v ��v ��p�� 1��p = ��v�1� � � �v�� v�� p��

1��p =

��v�1� � � ��p ��

�v�� v�� p��

1��p� �v�1

� � � ��p �� v�� p

� � �� v�� p��

1��p ,

since p � 1 and the quantities inside the distribution are each less than 1.

However expnading this expression and summing we see that it is in turn = � � �� , with � � 1.So ��v�1

� �� v � v ��p�� 1��p � + � = � � �v � �p+ � � v � �p. QED.

Case p � �

� � v � �v � �� sup � �v � p � � sup ��v �� v �� sup ��v � + sup �v � =� � �v � �� v � ��. QED.

� b) Show that if � �v � l p and � �v � lq with 1��p � 1��q � 1, then �v�1

� ��v ��v � � � � �v � �p � � � �v � �q

� Proof

Case p � 1, q � �

�v�1� � �v �v � � �v�1

� � �v � � sup � v � =�v�1� ��v � � � � v � �� = � � �v � �1 � � � v � ��.

Case 1 � p � �

Define � � vq��p . We may chosoe to assume that �v and v are positive since the norm is dependant

only on their absolute value. By Lemma 3 � p� t � �v � � = p� t � �� p�1 � �v � t � �v�p � vp

Summing both sides. �v�1� p� t � �v � v � �v�1

� �v � t � �v�p � vp =

� � v � t � �v � �pp- � � v � �p

p � � � � v � �p � � � t � �v � �p�p � � � v � �pp

Differentiating with respect to t we have that �v�1� p� �v � v � p� � � �� p � � � � � �p

p�1 =� � �v � �p � � � v � �q. QED.

Problem 10

� Let � f n � be a sequence of functions in L�. Prove that � f n � converges to f in L� if and only if there is a set E of measure 0 such that fn converges uniformly to f on E

.

� Prooffn � f in L� imples � f n � f �� 0. Thus � � � 0 � N such that � n � N, � f n � f �� . �except on a set E of measure 0, � fn � f � � �. Thus fn is uniformly converges to f on E

�.

If fn uniformly converges to f except on a set E of measure 0 then � � > 0, � N such that � x � E�

and n � N, � fn�x� � f �x� � � �. � except on E, � f �x� � < � fn�x� �� ess sup fn�x� � �. Thus f isbounded a.e. and hence f is in L�.

MATH630-10.nb 2

Problem 11

� Prove that L� is complete.

� ProofIn order to show this we need that every Cauchy sequence of functions � f n � in L� converges tosome f in L�.

Given � > 0, � N such that � m, n � N, � f n � fm � � �, by definition of Cauchy. Since this is L�

norm we know that fn � fm is bounded a.e. by �.

Define the pointwise limit f �x� � limn�� fn�x�, if the limit exists�, otherwise

.

It suffices to show that ess sup f � �, since if fn converges then it must converge to f a.e.

Since fn � fm is bounded a.e. by �, except on a set M of measure 0. However if we consider thecollection of all such M � m > n, we have a countable collection of sets of measure zero so thereunion is also a set of measure 0. Thus � fn�x�� f �x� � � � if limn�� fn�x� exists. Since fn is in L� itis bounded by some ess sup . Then where the limit exists, f is bounded by + �.

Thus we only need to show that the limit must exist upto a set of measure 0.

MATH630-10.nb 3

Problem 16

� Let � f n � be a sequence of functions in Lp, 1 � p � �, which converge a.e. to a function f in Lp. Show that � f n � converges to f in Lp if and only if � f n �� f �.

� ProofIf � f n � � � f � � � f n � � � f � � 0 �

�� fn��p�1��p � �� f ��p��

1��p � � �� fn��p � �� f ��p��1��p � � 0, since p � 1. However �� f n�p�� f ��p � �

�� fn � � � f ��p �, for p � 1

� f n � converges to f in Lp � � f n � f �� 0 � �� fn � f ��p�1��p � 0. However, �� fn � f ��p�

�� fn �� f ��p.

??????????????????????????????????

Demonstrate Proposition 8 fails at p = �.

� Show that � f , g L�, such that � step functions � and continuous functions �, � f � � �� 0 and � g � � �� 0.

� Counter-example on step functions.

Define f �x� � sin� 1��x �, x � 00, x � 0

.

Clearly f �x� is bounded by 1 and thus in L�. Consider a step function � on partition� � 0 � �0 � �1 � �2 �... � �N � 1�. Let � denote the size of the smallest interval in the partition.Consider intervals of the form An = � 1��n�� , 1��n�1�� , for n � �. Clearly An � �0, 1�, and � infinitelymany An of length < �. So choose n, m such that An � ��m, �m�1� and length An � �. Clearly byconstruction � f �An� � � �0, 1�, so regardless of the value of � over ��m, �m�1�, � f � � � � 1��2 over An.Hence � f � � �� 1��2 . Thus showing the counter example to Proposition 8.

MATH630-10.nb 4

� Counter-example on continuous functions.

Define g�x� � 0, x � 1��2

1, x � 1��2.

Clearly g is in L�. Consider a continuous function approximating g. By continuity at x � 1��2 , � �> 0 � � > 0 such that � �x� 1��2 � � �, � �x� � � 1��2 � � � �.

Suppose that � 1��2 � � 0, then choose � = 1��2 and � the interval � 1��2 , 1��2 � �� where �x� � 1��2 , yet g = 1over this interval so � g � �� 1��2 .

Suppose that � 1��2 � � 0, then choose � = �1��2 ��2 and � the interval � 1��2 � �, 1��2 � where �x� � � 1��2 ��2 .

So � g � �� 1��2 ��2 . Thus � g � �� > 0. Thus showing the counter example for Proposition

8 at p = �. QED.

MATH630-10.nb 5

Real Analysis - Math 630Homework Set #11 - Chapter 6 & 11


Problem 6.21

� a) Let g be an integrable function on �0, 1�. Show that � a bounded measurable function f such that � f � � 0 and

� f � g � � g �1 � � f ��

� Proof

� Case � g �1 � 0

Let f �x� � sgn�g�x��, where sgn�� , � � 0

0, � � 0. Since � g �1 � 0, that implies that

m��x : g�x� � 0� � 0 � ess sup f(x) = 1.

Hence � f � g = � g � sgn�g� = � �g � = � g �1= � g �1 � � f �. QED

� Case � g �1 � 0Let f �x� � � � 0, x. Then � f � � �. � f � g = �� g � �� g � � 0 finite �. Thus� f � g = � g �1 � � f �= 0. QED.

� b) Let g be a bounded measurable function. Show that � � > 0 � an integrable function f such that

� f � g � � g �� f �1

� ProofBy definition ess sup g(x) = inf �M : m��t : f �t� � M� � 0�. Given � > 0. LetE � �x : g�x� � � g � ��. Thus mE > 0. Let f �x� � �E. Then � f � g = � �E � g �� E � � � g � �� by definition of E. Hence we have this = mE � � � g � �� =� � g � �� f �1. Thus � f � g � � � g � �� f �1. QED.

Problem 6.22

� Find a representation for the bounded linear functionals on l p, 1 � p � �.

� Construction

Problem 6.24

� Show that the element g in Lp given by Theorem 13 is unique.

� Proof

MATH630-11.nb 2

Problem 11.7

� Prove Proposition 4: If �X , �, � is a measure space, then we can find a complete measure space �X , �0, 0� such that

i. � � �0.ii. E � � � E � 0�Eiii. E � �0�E � A �B where B � � and A � C, C � �, C � 0.

� ProofFirst we must show that the �0 defined by (iii) is in fact a �-algebra. Thus we need that E1, E2 � �0, � E1 E2 � �0 and E1

�� 0 and the union of any countable collection of sets in �0

is in �0.

Clearly E1 � �0 � E1 � A1 B1 where B1 � � and A1 � C1, C1 � �, �C1 � 0. And similarly forE2. Thus E1 E2 = �A1 A2� �B1 B2�. And we know that �B1 B2� � �, since � is a �-algebra. Furthermore �A1 A2� � �C1 C2�. �C1 C2� � �, since � is a �-algebra and0 � ��C1 C2� � �C1 � �C2 � 0. So ��C1 C2� � 0.

E1�

= � �A1 B1� = A1� B1

� = �C1� �A1 B1�� B1 C1��. However B1 C1 � �, and

hence ��B1 C1� � �. Also C1� �A1 B1� � C1. Hence E1�

has the appropriate form to be in �0.

Thus we need only show that the union of a countable collection of sets in �0 is in �0. Considern�1

En, En � �0 � �An Bn� = �Bn� � An�, we know that Bn � �, so we only needthat An is the subset of some set of measure 0. However An � Cn, with �Cn � 0, and the unionof countable collection of things of meaurse 0 is still measure 0. Hence it suffices to consider theunion of the Cn and we are done. Thus �0 is a �-algebra.

Consider E � �0 with E � A B as above so that A is a subset of a set of measure 0. We wish toshow that �B is not dependant on the choice of B � �. Given two such representationsA1 B1 � A2 B2. Let A1 � C1, with �C1 � 0 and A2 � C2, with �C2 � 0. Thus,B1 � A1

� �A2 B2� ��B1 � �C2 � �B2 = �B2, however by symmetry of argument, we have�B2 � �C1 � �B1 = �B1, and hence �B1 � �B2.

Thus we may define �0�E � �B, without concern that the value will change depending on ourchoice of B. So that �0�A � 0 if A � C, �C � 0.

MATH630-11.nb 3

It then follows immediately that �0 is a measure from the fact that � is a measure and that we haveonly added sets which have measure 0 and the sets generated by their unions with sets of �, whichwill neccesarily preserve the additivity of the measure.

Thus we have created a complete measure with the neccesary properties.

Problem 11.10

� Prove Proposition 7: Let f be a nonnegative measurable function . Then � a sequence � �n � of simple functions with �n�1 �n such that f � lim�n at each point of X. If f is defined on a �-finite measurable space, then we may choose the fucntions �n so that each vanishes outside a set of finite measure.

� ConstructionDefine En,k � �x : k �2 n � f �x� � �k � 1��2 n� and �n � 2 n��k�0

22�nk � �En,k , with respect to each

pair of integers � n, k �.

I wish to show that �n thus defined has the neccesary properties.

Since f is finite, for a given � n, k �, n � 0, k � 0, mEn,k � . Thus m�k�122�n

En,k � , and so �n =0, except on that set which is finite. So each function vanishes off a set of finite measure.

Furthermore it should be clear that for a given n, the sets En,k are mutually exclusive and that if fora given n and k, f �x� is bracketed by �k0 �2 n, �k0 � 1� �2 n� then for the next n, we will have f �x�bracketed by either �2� k0 �2 �n�1�, �2� k0 � 1� �2 n�1� or��2� k0 � 1� �2 �n�1�, �2� k0 � 2� �2 �n�1��. And the lower bound of the interval is the same as thevalue returned by the sum, it is then clear that �n�1 � �n.

Furthermore, since the width of the interval between k �2 n and �k � 1� �2 n goes to 0 as n � 0,while the range goes to both 0 and as k � 0 and k � 22�n. This allows that for any value of f �x�,it will eventually be in some interval and the value can never exceed f �x�. Thus since the intervalsuccessively approximates f �x� with an � width equal to that of the interval hence the limit must goto f �x�.

MATH630-11.nb 4

Problem 11.17

� Prove Proposition 15: If f and g are integrable functions and E is a measurable set, then

i. �E�c1 � f � c2 � g� � c1 � �E f � c2 � �Egii. If �h � � � f � and h is measurable then h is integrable.iii. If f g a.e., then � f � g.

� Proof

� �E�c1 � f � c2 � g� � c1 � �E f � c2 � �Eg

Simply break f and g into + and - parts and apply Proposition 13.

� If �h � � � f � and h is measurable then h is integrable.If we can first show part (iii), then by applying that result we can conclude that � � � f � � � h � � � f � � , and thus h in integrable. So let us proceed to part (iii).

� If f g a.e., then � f � g.

If we consider � f � inf�� f

�� , then clearly � a sequence of monotone increasing simple functions

� �n � which converge to g which will be less than f a.e. By montone convergence theorem

� g � lim�� n � inf�� f

�� f . Hence � f � � g. QED.

MATH630-11.nb 5



Problem 12.2

� Assume that � Ei � is a sequence of disjoint measurable sets and E = �Ei. Then � A we have that

��A � E� � ��A � Ei�� Proof

If � Ei � has only one non-empty set element then this is trivial. Suppose � Ei � has a finitenumber of non-empty sets, then by the assumption of the induction hypothesis say thatE�� i�1

n�1 Ei , conforms to the property that ��A� E� � � �i�1

n�1 ��A� Ei�.

Consider E � �i�1n Ei, then ��A� E� = ��A� E � En�� A� E � En

�, since En ismeasurable. However E En, and exploitning the fact that each Ei is disjoint, we arrive at��A� E� = ��A� En�� A� E

� � = �i�1n ��A� Ei�. Thus the induction is proved. Hence we

know for any finite sequence this property will hold.

So � N, ��A� ��i�1N Ei�� i�1

N ��A� Ei�. However A� ��i�1N Ei� � A� ��i�1

Ei�, so��A� E� � ��A� ��i�1

Ei�� i�1N ��A� Ei�. The left hand side is now independant of N so

we make take the right hand side to infinity. So ��A� E�� i�1 ��A� Ei�, yet we know form

the properties of outer measure that ��A� E�� i�1 ��A� Ei�. Hence

��A� E� � � ��A� Ei�. QED.

Problem 12.4

� Prove Proposition 9 - Let � be a semialgebra of sets and � a nonnegative set function defined on � with �� 0��if � ��. Then � has a unique extension to a measure on the algebra � generated by � if the following conditions are met:

i) If a set C � is the union of a finite disjoint collection �Ci� of sets in �, then �C � ��Ci

ii) If a set C in � is the union of a countable disjoint collection �Ci� of sets in �, then �C � ��Ci

� a) Condition (i) implies that if A is the union of each of two finite disjoint collections �Ci� and �Di� of sets in �, then ��Ci � ��D j.

� Proof�A � � �Ci by condition (i), but also by condition (i) we have that �A � � �D j, so be thetransitivity of equality, � �Ci � � �D j.

� b) Condition (ii) implies that � is countably additive on �.

� ProofBy Problem 5, we have that all sets in � take the form of countable unions of sets in �. However itis always possible to generate a disjoint sequence of sets from an arbitrary sequence of sets, whichpreserve unions, provide that we may take finite intersections and complements, as we can on asemialgebra. Thus all elements of � may be expressed in terms of the union of a countable dijointcollection of sets.

This implies that condition (ii) is equivalent to � A � �, �A �� Ci, for any disjoint sequence� Ci �, such that �Ci � A. Which is exactly what is required to show that the measure iscountably additive.

In conclusion, property (i) gives us that the notion of measure on � is well defined and unique onfinite sets, and property (ii) gives us that countable additivity is preserved and that the value of �Ais still uniquely defined over countably additive unions.

MATH630-12.nb 2

Problem 12.5

� Let � be a semialgebra of sets and � the smallest algebra of sets containing �.

� a) Show that � is comprised of sets of the form A = �i�1n Ci with Ci �.

� ProofClearly it is neccesary that � be closed under finite unions and thus it will contain finite unions ofelements in �. We need to show that this is sufficient to be closed under finite intersections andcomplements.

Let A, B � �, such that A � �i�1n Ai, and B � �i�1

m Bi, where each set Ai and Bi is in �. ThenA� B ��i�1

n Ai� � ��i�1m Bi� = �i�1

n �Ai �� j�1m B j� = �i�1

n � j�1m �Ai � B j�, but each Ai � B j � �,

since � is closed under finite intersections. Thus A� B is a finite union and hence an element of �of the appropriate form. By induction � is closed under finite intersections.

Consider A

= �i�1n Ci

. However by definition of semialgebra, Ci

� �k�1

ni Di,k, with Di,k � �. ThusA��i�1

n �k�1ni Di,k, but �k�1

ni Di,k � �. Thus since � is closed under finite intersections fromabove, we must have that A

� �.

Hence � is an algebra of sets. That it is the smallest algebra of sets containing � is clear from thefact that any algebra of sets must contain the elements of the form �i�1

n Ci with Ci � �, yet no otherelements have been added besides these.

� b) Show that �� , so that �� and �� may be replaced in theorems by �� and �� , respectively.

� Proof� �, thus �� . However �� contains all elements of the form �i�1

Ci, where Ci � �.Hence for each element in �, A = �i�1

n Ci, A � �i�1 Ci. Thus � � �� = ��. Thus

�� = ��. QED.

MATH630-12.nb 3

Problem 12.6

� Let � be a collection of sets which is closed under finite unions and finite intersections; an algebra of sets, for example.

� a) Show that �� is closed under countable unions and finite intersections.

� ProofWell by definition �� is the set of all sets generated by countable unions of sets in � so all thatremains to be shown is that it is closed under finite intersections. GivenB � �� with B � �, C � �. Then we want to show that B�C � ��. We know that there mustexist a sequence of sets � Bi � , with each set in � such that �Bi � B, since it appears in theclosure of � with respect to unions.

Define � Ei � to be the sequence of sets such that Ei � Bi �C � �, since � is closed under finiteintersections. However the �Ei � ��Bi �C� � B� C. Thus the intersection of any element of�� with one of � is in ��. By induction we may intersect any finite number of elements in �with elements of ��.

Given two elements B, C � ��, B, C � �, then we know that B� C � B� ��Ci�, where Ci arethe elements of a generating sequence for C. This is equivalent to ��B� Ci�, however we knowthat B�Ci � �� from above, and thus taking the countable union will also be in ��. � �� isclosed under finite intersections.

� b) Show that each set in �� is the intersection of a decreasing sequence of sets in ��

� ConstructionClearly if the set in question is a member of �� then the result is trivial, since we just take everyterm in the sequence to be that set.

Given B � ��, B � �, then � a sequence � � Bi ��i�0 of sets in �� such that �Bi � B.

Define Ai � �� j�i B j� � ��k�0

i�1 �Bk�, with A0 � �Bi. Given any element x � B, � x � Bk, � k,hence x � Ai, � i. Given y � B, y � BN for some N and hence y � Ai, � i � N. If N is the leastsuch N that works then y � Ai, � i � N, and thus Ai's form a decreasing sequence and � Ai � B.Thus we have a countable intersection of decreasing sets in A� that yields B � A��.

MATH630-12.nb 4

Problem 12.7

� Let � be a finite measure on an algebra �, and �� the induced outer measure. Show that a set E is measurable if and only if for each � > 0 there is a set A � , A � E, such that ��E�A� � �.

� Proof

MATH630-12.nb 5



Problem 12.19

� Let X � Y be the set of positive integers, � � � � ��X�, and let � � � be the measure defined by setting ��E� equal to the number of points in E if E is finite and � if E is an infinite set. (This measure is called the counting measure.) State the Fubini and Tonelli Theorems explicitly for this case.

� AnswerWhile I am unclear on exactly what suffices as response to this question, I am going to say when wecan apply the Fubini and Tonelli theorems and then simpify the conclusions of the theorems asapplicable to this specific case.

Let f �x, y� be a real-valued function from X �Y .

First considering the Tonelli theorem, we know that the set of positive integers is �-finite, so wemay apply the Tonelli theorem provided that f is nonnegative everywhere.

In the case of Fubini theorem we need that �X�Y f �� . However since X �Y is a set ofcardinality 0 we know that an enumeration of the points in X �Y . Let � ri �i�1

� be such anenumeration. Then we can equate f �x, y� = �i�1

� f �ri� ��ri��x, y�. If we further define�n�x, y� � �i�1

n f �ri� ��ri��x, y�, then the sequence of functions are each simple and clearlyconverge to f. Suppose we consider the positive and negative parts of f. Then f � � �n

�, and byapplication of the monotone convergence theorem, � f � � limn�� n

�. However the right handside is equivalent to limn��i�1

n f ��ri� ��ri�� = �i�1� f ��ri�, since the counting measure over a

set of one element is 1. Thus saying that �X�Y f �� is equivalent to asking that�i�1� f ��ri� � � and �i�1

� f ��ri� � �.

Let us now consider the implication of the theorems which are of course exactly the same. Forsimplicity I will not state the results generate by interchange of x and y.

� (i) for almost all x the function fx defined by fx�y� � f �x, y� is a measurable function on Y.Well fx�y� ��i�1

� f �ri� ��ri��x, y�, where x is fixed. Which is equal to �i�1� f �x, i� ��i��y�.

Consider �y : fx�y� � �� = A�. We need only show that �A� is well defined for all �. However A�is some subset of Y, and thus A� � ��Y�, so A� is measurable, since it is in the collection ofmeasurable sets, and hence fx�y� is a measurable function by Proposition 11.5.

� (ii) �Y f �x, y�� y� is a measurable function of X.Let us explicitly compute �Y f �x, y�� y� = �Y fx�y�� = �Y�i�1

� f �x, i� ��i��y�� . Using thesame caveat as above we may consider this as positive and negative pieces of f. Since integration isa linear operation, we will only consider nonnegative functions f. In the nonnegative case we mayexploit the fact that the sum is the limit of a sequence of increasing simple function and apply themonotone convergence theorem to say that

�Y fx�y�� = �i�1� f �x, i� ��i�� = �i�1

� f �x, i�, since ��i�� 1, � i . That the function �i�1

� f �x, i� is measurable follows immediately form the fact that x � X , and � =�(X). The argument is the same as the 2nd paragraph in (i).

� (iii) �X�Y f �x, y�� X�Y f �x, y�� Working from the conclusion of (ii), that �Y fx�y�� i�1

� f �x, i�, we will then integrate withrespect to �, to find that �X �Y f �x, y�� = �i�1

� � j�i� f �i, j��i��. (Note that technically we

msut first consider this as the limit of finite sums, but the functions in the finite sums are increasingnonnegative functions whose limit is the � sum, and the integrals are thus the same by Monotoneconvergence theorem.) However ��i�� 1 since � is the counting measure, and thus�X�Y f �x, y�� = �i�1

� � j�i� f �i, j�. However we also know that �X�Y f �x, y�� =

�X�Y�i�1� f �ri� ��ri��x, y�� = �i�1

� f �ri� ��ri�� = �i�1� f �ri�, since

��ri�� 1, � i . However since the ri's range over ever pair � j, k� of integers exactly once,this is identical to the above result.

Hence �X �Y f �x, y�� X�Y f �x, y�� = �i�1� � j�i

� f �i, j�.

MATH630-13.nb 2

Problem 12.22

� Let h and g be integrable functions on X and Y, and define f �x, y� � h�x��g�y�. Then f is integrable on X �Y and

�X�Y f �� Xh�� Y g��

� ProofWithout loss of generality we may suppose that f , g and h are nonnegative functions. If not wemay rewrite them as the differance of their negative and positive parts and by the linearity ofintegration we only need to show the proof for nonnegative functions.

Consider �X�Y f �� . If the integral exists then by definition of integration �X�Y f �� =

inf�� f

��X�Y �� i�1n ci��Ei � �X �Y ��, where ��x, y� � �i�1

n ci� �Ei�x, y� is a

nonnegative simple function and each Ei � X �Y .

However g, h are integrable functions, so they each admit a representation

�Xh�� inf�h�h

��X �h�� i�1m1 di��Hi � X � and �Y g�� inf

�g�g��Y �g�� i�1

m2 fi��Gi � Y�, with

�h�x� � �i�1m1 di� �Hi�x� and �g�y� � �i�1

m2 fi� �Gi�y�.

Since f �x, y� � h�x� g�y� � that every �g �h is a possible function for � ��i�1

n ci� �Ei�x, y� � �i�1m1 di� �Hi�x� �i�1

m2 fi� �Gi�y� = �i�1m1 � j�1

m2 di� f j� �Hi�x�� G j�y�. However�Hi�x�� G j�y� = �Hi�G j�x, y�.

This implies that f admits a representation as a sum of n � m1 m2 elements, with eachEi � H j�i��Gk�i� and ci � d j�i� fk�i�, where the j, k depend on i and each pair of j, k are usedexactly once over the m1 m2 terms.

This that �i�1n ci��Ei � �X �Y�� = �i�1

m1 m2 d j�i� fk�i��H j�i��Gk�i�� X �Y ��. We knowthat each G j and Hk is in X and Y respectively since these are the basis sets, so this devolves to�i�1

m1 m2 d j�i� fk�i��H j�i��Gk�i��. However H j�Gk is a rectangle in X �Y , so��H j�i��Gk�i�� = ��H j�i�� Gk�i��

� �i�1m1 m2 d j�i� fk�i��H j�i��Gk�i�� = �i�1

m1 m2 d j�i� fk�i��H j�i�� Gk�i��, but by definition ofthe j�i� and k�i� this is just equal to �i�1

m1 di��Hi� �i�1m2 fi��Gi� = �X �h�� Y �g��.

MATH630-13.nb 3

Thus inf�� f

��X�Y �� inf�h�h

��X �h�� inf�g�g

��Y�g��. Evaluating the infs we arrive at

�X�Y f �� X h�� Y g�� , assuming the integral over f exists. However there is nothingspecific in this argument to the fact we choose inf, except the last inequality, hence by the sup oversimple functions less than the given function approach we can arrive at�X�Y f �� X h�� Y g�� , if the integral over f exists.

However what we have in fact shown is that

�Xh�� Y g�� sup�� f

��X�Y �� inf�� f

��X�Y �� Xh�� Y g��, the middle

inequality is always true of the sups and infs used to define integrals. �

�Xh�� Y g�� sup�� f

��X�Y �� inf�� f

��X�Y �� , which implies that �X�Y f ��

exists and its value is �X h�� Y g�� . QED.

Author's note: I am all but sure that there is a better way to do this, but I don't know what it is so Idecided to do it based on first principles.

MATH630-13.nb 4

Problem 12.24

� The following example shows that we cannot remove the hypothesis that f be nonnegative from the Tonelli Theorem or that f be integrable from the Fubini Theorem. Let X � Y be the positive integers and � � � be the counting measure. Let

f �x, y� � �2 2x, if x � y

2 � 2x, if x � y � 10, otherwise

� ContradictionFirst it is neccesary to notice that f �x, y� � �2� 2�x�� A�x, y�� 2� 2�x�� B�x, y�, whereA � ��x, y� : x � y� and B � ��x, y� : x � y� 1�.

In order to disprove the Fubini and Tonelli thereoms it suffices to show that property (iii) in eachfails to hold. In other words I wish to show that �X�Y f �x, y�� Y�X f �x, y�� .

Let us first consider �X�Y f �x, y�� . So computing �Y f �x, y�� at a particular x, we have that� �2� 2�x��Ax�� 2� 2�x��Bx�. Furthermore we know that for a fixed x, Ax has only onepoint determined by x � y, and similarly Bx has only one point. Hence the counting measureevaluated on these sets is just 1. Thus �Y f �x, y�� = �2� 2�x�� 2� 2�x� � 0. Hence�X�Y f �x, y�� 0.

Now let us consider �Y�X f �x, y�� that we first evaluate �X f �x, y�� for a fixed y. Thusgives the integral as �2� 2�y��Ay�� 2� 2�y�1��By�, note the substitution in the exponent interms of the fixed coordinate. Again each of these has exactly one item in its set, with theexception of By at y � 1, which is empty because x � 0 is not in the domain of positive integers. Sofor y � 2, �X f �x, y�� 2� 2�y� � ��2� 2�y�1� = 2�y�� 1��2 � 1� � �2�y�1. Also at y � 1,�X f �x, y�� = �2� 2�1� � 3��2 .

Now in order to compute the integral over Y, it is not hard to see that by restricing the integral tothe first N integers we are in fact integrating over a simple function. Thus �Y�X f �x, y�� =��1,2,...�,N��

3��2 � ��1� ��2�y�1 ��1�� = 3��2 ��1��i�2N �2�i�1 ��i�� Since we are using the

counting measure, each evaluation is in fact one. Simplifying the geometric series we have

MATH630-13.nb 5

�3��2 �1��2��

1� 1��2 = 1��2 . Thus �Y�X f �x, y�� 1��2 � 0 � �X �Y f �x, y�� . Hence neither the

Fubini nor the Tonelli theoreoms will hold for this particular fucntion and choice of measure. QED.

Problem 12.25

� The following example shows that we cannot remove the hypothesis that f be integrable from the Fubini Theorem or that � and � are �-finite from the Tonelli Theorem: Let X � Y be the interval �0, 1�, with � � � the class of Borel sets. Let � be the Lebesgue measure and the � the counting measure. Then the diagonal � � � x, y � � X �Y : x � y� is measurable (is an ��, in fact), but its characteristic function fails to satisfy any of the equalities in condition (iii) of the Fubini and Tonelli Theorems.

� ContradictionAgain we will seek an exception of the Fubini and Tonelli Theorems by computing the differanthalves of part (iii) in each result and showing that they are not equal.

Consider �X�Y � �x, y�� . First we wish to compute �Y � �x, y�� , for a particular choice of x.This means that �Y � �x, y�� = �� x�, but � x�y� is only non-zero for the one value where x � y,and so since � is the counting measure, we have that �Y � �x, y�� = 1. So �X �Y � �x, y�� =�X1�� = ��X � � 1.

Now consider �Y�X � �x, y�� . First we start with �X � �x, y�� , for a particular value of y.Which evaluates the integral to �� y�, which again is a set containing only the point where x � y.However since � is the Lesbegue measure this evaluates to 0. So we have that �Y�X � �x, y�� = �Y 0�� 0. Since 0 � 1, we have thus shown that � fails to satisfy the (iii) property of theFubini and Tunelli Theorems. QED.

MATH630-13.nb 6

Problem 12.31

� Let f be a nonnegative measurable function on ��, ��, and let m2 be the two-dimensional Lebesgue measure on �2. Then

m2�� x, y � : 0 � y � f �x�� m2�� x, y � : 0 � y � f �x�� f �x�� x

Let ��t� � m��x : f �x� � t�. Then � is a decreasing function and

�0��t�� t � � f �x�� x.

� Proof

� m2�� x, y � : 0 � y � f �x�� m2�� x, y � : 0 � y � f �x�� f �x�� x

Essentially in the first part we are attempting to show that � f �x�� x is the same thing as the areaunder the curve (with and without its boundary).

Let A � � � x, y � : 0 � y � f �x�� and B � � � x, y � : 0 � y � f �x��. Let � denote the Lesbeguemeasure. Consider � �A�� , since � is �-finite and � is everywhere nonnegative, we mayapply Tonelli Theorem to conclude that � �A�� A�� Ax�� . For anyparticular x we have that Ax is the interval �0, f �x�, and thus that ��Ax� � f �x�. Thus � �A�� = � f �x�� , but the integral with respect to Lesbegue measure is exactly what is intended by thenotation � f �x�� x. Finally we know that � �A�� = ��A� by the definition of an integralof a characteristic function, but �� m2. Hence we have that m2�A = � f �x�� x.

If we then note that B behaves exactly the same as A in the above proof since Bx = �0, f �x��, andhence ��Bx� � ��0, f �x�� = f �x�. Thus the conclusion is equally valid that m2�B � � f �x�� x.

MATH630-13.nb 7

� �0��t�� t � � f �x�� x

While in the above proof we made a construction that showed the area was the same as that underthe curve f �x�, by exploiting a series of vertical bars, in the proof that follows we will attempt todemonstrate that it is equivalent to a series of horizontal bars.

On the question that � is a decreasing function, this follows immediately from that fact thatmA � mB, whenever A ! B, and the observation that if x � �x : f �x� � t1� then x � �x : f �x� � t2�, �t2 � t1. Thus for t1 � t2, we have that �x : f �x� � t1� " �x : f �x� � t2�. So m��x : f �x� � t1� �m��x : f �x� � t2�.

Let � � � x, t � : f �x� � t � 0�

Consider �0��t�� t = �0

�m��x : f �x� � t�� t = �0

�m� t�� t = �0

�� x, t�� x�� t. The last

equality coming from the fact that the measure of a cross-section is the same as the integral of itscharacteristic function. Since characteristic functions are bounded and both measure are taken withrespect to the �-finite real numbers, we may apply Tonelli thoerem to conclude that�0�� x, t�� x�� t = ��

��0�� x, t�� t�� x. From here we need only observe that is the same

as A in the first section of this problem, upto the labelling of variables. Thus from above we mayconclude that ��

��0�� x, t�� t�� x = � f �x�� x. � �0

��t�� t � � f �x�� x. QED.

MATH630-13.nb 8

Documents

Real Analysis Rudin Solution Manual(1-8)(10-13)