Set Theory - West Virginia Universitymath.wvu.edu/~chenye/Lecture_notes090224.pdf · Chapter 1 Set Theory ... 1 It is suﬃces to prove that the subset [0,1] of real numbers is not

Chapter 1

Set Theory

This notes has two parts, theorems and definitions that are needed for the proves, the solution of all most

all old entrance exams in WVU math dept. by 2009 and some homework questions in Rudin (v3) as well.

Don’t trust all of them as I made some mistakes when I was doingit. The comments are feedback from Dr.

Lai. You are welcome to discuss with me. —- Ye

1.1 Notes

1.1.1 Ordered pairs:< x, y > is an ordered pair if we distinguish between first and second element. Thus

< x, y >,< y, x > if x , y.

1.1.2 Cartersian product (direct product): X and Y are two sets, the Cartersian product X× Y is the set

x, y of all ordered pairs whose first element is from X and the second element is from Y.

1.1.3 Finite {infinite} sequence:a function whose domain is first n natural numbers{the set N of

natural numbers}.

1.1.4 Sequence:finite or infinite sequence.

1.1.5 Countable set:a set is countable if it is the range of a sequence.

Proposition 1.1.6 The union of a countable collection of countable set is countable.

Proposition 1.1.7 The set of all rational numbers is countable.

Proposition 1.1.8 The set of all real numbers is not countable.

Proof: 1 It is suffices to prove that the subset[0, 1] of real numbers is not countable.

If [0, 1] is countable, let f be an arbitrary function that f: N −→ [0, 1]. Then we can write down[0, 1] as

a sequence ai such that ai = f (i), i ∈ N. Also, we can write down ai = 0.bi1bi2bi3 . . . . Let ci = 1− 0.bii for

all i ∈ N, and c= 0.c1c2c3 . . . . Then f(i) , c for ∀i ∈ N.

Need to add the explanation whyf (i) , c, and need to complete the proof.

Let ci = 9 − bii for all i ∈ N, and c= 0.c1c2c3 . . . . Then f(i) , c for ∀i ∈ N, else if there∃k such that

1

2 CHAPTER 1. SET THEORY

f (k) = ak = 0.bk1bk2 . . .bkk · · · = c, then bkk = ck. But ck = 9 − bkk, that leads to bkk = 4.5, contradicts to

the require that bi j ∈ Z+.

Proof: 2 If R is countable, then it could write down as a sequence{a1, a2, a3, . . . }. And we could find a

covering of R like⋃

n[an − 12n ]. And the length of this cover is at most1, which is a contradiction.

The notation of intervals are not right. It should be Bn = [an, an +12n ] or Bn = [an − 1

2n , an +12n ]. To

complete the proof, need to add the explanation ofR⊆ ∪nBn, and |R| ≤ ∑ |Bn|.

1.1.9 De Morgan’s Laws:

∼ [⋃

A∈CA] =

⋂

A∈CA

∼ [⋂

A∈CA] =

⋃

A∈CA

1.1.10 Algebra of sets:a collectionA of subsets of X is called an algebra or Boolean algebra if

(i) A ∪ B ∈ A , and (ii)∼ A ∈ A for any element A, B inA .

1.1.11 σ-algebra (Borel field): an algebraA of sets is call aσ-algebra if every countable collection of

sets inA is again inA .

1.1.12 Equivalence relation:a relation that is transitive, reflexive, and symmetric on X is said to be an

equivalence relation.

1.1.13 Partial ordering: a relation≺ id said to be partial ordering of a set X if it is transitive andanti-

symmetric on X. (eg.⊂ is a partial ordering onP(X).)

1.1.14 Linear ordering: a set X is said to be linear ordering if for any two elements x and y, we have

either x< y or y< x.

1.1.15 First element:an element a∈ E is the first element if, whenever x∈ E and x, a, we have x< a.

1.1.16 Minimal element:an element a∈ E is the minimal element of E if there is no x∈ E with x, a and

x < a.

1.1.17 Well ordering: a strict linear ordering< on a set X is called a well ordering for X or is said to well

order X if every nonempty subset of X contains a first element.

Well-Ordering Principle: Every setX can be well ordered; that is, there is a relation< well ordersX.

A well-ordered setX is useful for constructing examples.

Chapter 2

The Real Number System

2.1 Notes

2.1.1 Upper bound:If S is a set of real numbers, we say that b is an upper bound for Sif for each x∈ S ,

we have x≤ b.

2.1.2 Least upper bound:A number a is called a least upper bound for S if it is an upper bound for S ,

and for each upper bound b, we have c≤ b. We denote the least upper bound by supS .

2.1.3 Completeness Axiom:Every nonempty set S of real numbers which has an upper bound has a least

upper bound.

2.1.4 Axiom of Archimedes:Given any real number x, there is an integer n such that x< n.

Corollary 2.1.5 Between any two real numbers is a rational; that is if x< y, then there is a rational r with

x < r < y.

Can you show also Corollary 2.1.5 by using the axioms and definitions?

Proof: By the Axiom of Archimedes, for the real number 1/(y − x), there is an integerb such that

1/(y − x) < b. Now consider the set{n} such that{n/b} are rational numbers greater thany. By Axiom of

Archimedes, the real numberyb must have an integern such thatyb < n. That is equal toy < n/b, then the

set{n} is nonempty. By the Completeness Axiom, there must be ak such thatk/b is the smallest one greater

thany. Also, we have (k−1)/b < y ≤ k/b andx = y− (y− x) < k/b−1/b = (k−1)/b < y. Thenr = (k−1)/b

is a rational such thatx < r < y.

2.1.6 Extended real numbers:Extension of real numbers by adding two elements∞ and−∞.

2.1.7 Limit of a sequence:Let {xn} be a sequence of real numbers. A real number l is the limit of{xn} if

for any positiveǫ, there exist an N such that for any n> N we have xn − l < ǫ.

2.1.8 Cauchy sequence:A sequence{xn} is a Cauchy sequence if for anyǫ > 0, there exist an N such that

for any n,m greater than N we have|xn − xm| < ǫ.

3

4 CHAPTER 2. THE REAL NUMBER SYSTEM

Theorem 2.1.9 Cauchy Criterion: A sequence of real numbers converges if and only if it is a Cauchy

sequence.

Proof: ”=⇒”: If a sequence{xn} converges tol, then for anyǫ > 0, there exist anN such that for any

n,m> N, we have|xn − xm| = |xn − l + l − xm| ≤ |xn − l| + |xm− l| < 2ǫ. So{xn} is a Cauchy sequence.

”⇐=”: If {xn} is a Cauchy sequence, then for anyǫ > 0, there exists anN such that for anyn,m> N such

that |xn − xm| < ǫ. Then if we fixm, (To indicate that m is fixed, it is much more clear to letm= N + 1. )

xm− ǫ < xn < xm+ ǫ for anyn > N. That means{xn} is bounded.

Suppose it is bounded by [−M,M], then forǫ1 = 1, at least one intervals [xi − ǫ1, xi + ǫ1], i = 1, 2, . . .

has infinitely many elements from{xn}, let this interval be [a1, b1]. Let ǫ2 = 12, then at least one interval

[xi−ǫ2, xi+ǫ2]∩[a1, b1] contains infinitely many elements from{xi}, let this interval [xi−ǫ2, xi+ǫ2]∩[a1, b1]

be [a2, b2]. Continue such steps, letǫn = 12n , then we can get interval [an, bn] which contains infinitely many

elements from{xi}. Now we havea1 ≤ a2 · · · ≤ an · · · ≤ bn · · · ≤ b2 ≤ b1, and limbn − an = 0, by the nested

intervals principle, there exist only one elementξ ∈ [an, bn] for all n, and that liman = lim bn = ξ.

Since for∀ǫ, there are infinitely many elements inside [ξ − ǫ, ξ + ǫ], then there is a subsequence{xNi }from {xi} converges toξ. That means for∀ǫ > 0,∃N′, such that for anyNi > N′, we have|xNi −ξ| < ǫ. Since

{xi} is a Cauchy sequence, then for∀ǫ > 0, ∃N′′ > 0, such that for anyn,m > N′′, we have|xn − xm| < ǫ.

Let N = max{N′,N′′}, for anyn,Ni > N,

|xn − ξ| = |xn − xNi + xNi − ξ| ≤ |xn − xNi | + |xNi − ξ| < 2ǫ.

Thenxn −→ ξ.

Every claim must have a reason in the proof, and the reason is either a definition, or an axiom, or

some established results. Here you need to add the explanation why there is such a subsequence, and

why there are infinitely many points in the interval.

Suppose it is bounded by [−M,M], M ∈ Z+. Then forǫ1 = 1, at least one interval [d, d + 1], d ∈{−M,−M + 1, . . .M − 1} has infinitely many elements from{xn}. Else if all [d, d + 1] have finitely many

elements from{xn}, since [−M,M] has 2M such intervals which is also finite, then [−M,M] contains only

finite elements from{xn}, which is a contradiction to{xn} is bounded by [−M,M], let this interval be [a1, b1]

and choose one element inside this interval bexN1.

Next, letǫ2 = 1/2, consider the elements in [a1, b1]. Then at least one of the intervals [a1, a1 + 1/2] and

[a1 + 1/2, b1] must have infinitely elements from{xn} since [a1, b1] contains infinitely elements from{xn}.Let this interval be [a2, b2] and choose one element which is not equal toxN1 inside this interval bexN2.

Continue such steps, on the k-th step, letǫk = 1/2k, then at least one interval of [ak−1, ak−1 + 1/2k] and

[ak−1+1/2k, bk−1] has infinitely many elements from{xn}, let this interval be [ak, bk] and also get an element

xNk. Then we havea1 ≤ a2 · · · ≤ an · · · ≤ bn · · · ≤ b2 ≤ b1, and limbn − an = 0, by the nested intervals

principle, there exists only one elementξ ∈ [an, bn] for all n, and that liman = lim bn = ξ.

Also for ǫ = 1/2k, let N′ = k+ 1, then for anyNi > N′, we have|xNi − ξ| < ǫ. That meansxNk → ξ.

Since{xi} is a Cauchy sequence, then for∀ǫ > 0, ∃N′′ > 0, such that for anyn,m > N′′, we have

|xn − xm| < ǫ. Let N = max{N′,N′′}, for anyn,Ni > N,

|xn − ξ| = |xn − xNi + xNi − ξ| ≤ |xn − xNi | + |xNi − ξ| < 2ǫ.

2.1. NOTES 5

Thenxn −→ ξ.

2.1.10 Cluster point: We say that l is a cluster point of{xn} if for any ǫ > 0 and N, there exist an n> N

such that|xn − l| < ǫ.

2.1.11 Limit superior, limit inferior:

lim xn = infn

supk≥n

xk

lim xn = supn

infk≥n

xk

Proposition 2.1.12 lim xn = l iff i) for ∀ǫ > 0, ∃n such that xk < l + ǫ for all k > n;

ii) for ∀ǫ > 0 and any n,∃k ≥ n such that xk > l − ǫ.

Proof: By definition, lim xn = infn

supk≥n

xk. Let sn = supk≥n

xk, then lim xn = infn

sn = l ⇐⇒ i) For ∀ǫ > 0, ∃n

such thatsn < l + ǫ. ii) ⇐⇒ For∀ǫ > 0,∃n such thatxk < l + ǫ for all k > n.

Proof of Prop 2.1.12: The second iff is not correct. Just apply definition of the upper limit.

By definition, lim xn = infn

supk≥n

xk. Let sn = supk≥n

xk, then lim xn = infn

sn = l ⇐⇒ i) For ∀ǫ > 0, ∃n such

thatsn < l + ǫ. That meansxk < l + ǫ for ∀k > n. Also by the definition of limit superior, for∀ǫ > 0 and any

n, ∃k ≥ n such thatxk > l − ǫ. Since if it is not true, there∃ǫ0 > 0 andN ∈ Z+, and for anyk > N, we have

xk < 1− ǫ0, but this leads a contradiction thatlim xn < l − ǫ0 < l.

2.1.13 Open set:A set O of real numbers is called open if for∀x ∈ O, there∃δ > 0 such that each y with

|x− y| < δ belongs to O.

Proposition 2.1.14 The intersection of any finite open set is open.

Proof: It suffices to prove that for any intersection of two open sets is open. Let O1 and O2 be any two

open sets. By definition, for∀a1 ∈ O1 and∀a2 ∈ O2, there∃δ1, δ2 > 0 such that(a1 − δ1, a1 + δ1) ⊂ O1 and

(a2−δ2, a2+δ2) ⊂ O2. Then for∀a ∈ O1∩O2, chooseδ = min{δ1, δ2}, then we have(a−δ, a+δ) ⊂ O1∪O2.

Proposition 2.1.15 The union of any collectionC of open sets is open.

Proof: For ∀a ∈⋃

O∈CO, a must be in at least one open set O. Then there∃δ > 0, such that(a− δ, a+ δ) ⊂

O ⊂⋃

O∈CO. Thus

⋃O∈C

O is open.

Proposition 2.1.16 Every open set of real numbers is the union of a countable collection of disjoint open

intervals.

Proposition 2.1.17 Lindelof: LetC be a collection of open sets of real numbers. Then there is a countable

subcollection{Oi} of C such that⋃

O∈CO =

∞⋃

i=1

Oi .

2.1.18 Point of closure:A real number x is called a point of closure of a set E if for every δ > 0 there is a

y ∈ E such that|x− y| < δ. The closure of E is denotedE.


2.1.19 Closed set:A set F is called closed if F= F.

Proposition 2.1.20 The intersection of any collectionC of closed set is closed.

Theorem 2.1.21 Heine-Borel:Let F be a closed and bounded set of real numbers. Then each open cover-

ing of F has a finite subcovering.

2.1.22 continuous at one point:We say that f is continuous at the point x in E if for∀ǫ > 0, there∃δ > 0

such that for any y in E with|x− y| < δ, we have| f (x) − f (y)| < ǫ.

2.1.23 Continuous:The function f is said to be continuous on a subset A of E if it iscontinuous at each

point of A.

Proposition 2.1.24 Let f be a real valued function defined and continuous on a closed and bound set F.

Then f is bounded on F and assumes its maximum and minimum on F.

Proof: By Heine-Borel open covering theorem.

2.1.25 (May, 2009)(Apr, 2005)

Let f(x) be monotone increasing on[0, 1] with f(0) = 0 and f(1) = 1. If the set{ f (x); x ∈ [0, 1]} is dense

in [0, 1], show that f is a continuous function on[0, 1]. Is it absolutely continuous on[0, 1]? Prove your

conclusion.

Proof: It is continuous on [0, 1]. Since f (x) is monotone increasing on [0, 1] with f (0) = 0 and f (1) = 1,

the set{ f (x); x ∈ [0, 1]} ⊂ [0, 1]. For an arbitraryy ∈ (0, 1), since{ f (x); x ∈ [0, 1]} is dense in [0, 1], then for

ǫ = 1/n > 0, there is af (xn) ∈ (y, y+ ǫ) and f (x′n) ∈ (y− ǫ, y). Since f (x) is monotone increasing, then for

∀ f (x) = y ∈ (0, 1), let δ = min{|xn − x|, |x′n − x|}, such that for any|x′ − x| < δ, we have| f (x′) − f (x)| < ǫ.

Then f (x) is continuous at (0, 1).

Also for x = f (x) = 0 and∀ǫ = 1/n > 0, there is ayn = f (xn) ∈ (0, ǫ), let δ = |xn − x|, then for any

|x′ − x| < δ, we have| f (x′) − f (x)| < ǫ. Then f (x) is continuous atx = 0. The same forx = 1.

I want to use your proof for 2.1.25 as an example.

The current proof has the following problems:

(1) Confusing notations: x is used to mean a general variable in Line -10, butx also mean a special

point in lines -7, -8, and -9.

(2) Lack of sufficient details: the monotone property must be used in the proof, and so the argu-

ment whenx < x′ and whenx′ < x are different, even though they are similar.

(3) When working on a proof, ask the question what do I need to prove? In this exercise, we need

to prove, by definition of continuity, that f (x) is continuous at every pointa ∈ [0, 1]. (Usea instead of

x to avoid notational confusion).

I am going to show you an example of this proof: (What are in parentheses are not part of the

proof).

Proof of 2.1.25: It suffices to show thatf (x) is continuous at every pointa ∈ [0, 1]. (This sentence

tells the reader what you are going to do in the proof).

2.1. NOTES 7

(To make you easy to understand the writing of the proof, I argue on the cases when0 < a < 1 and

whena ∈ {0, 1}. These cases are not necessary and the proofs can be combinedby using slightly more

complicated notations).

First we assume thata ∈ (0, 1). Let b = f (a) (this is again a better notation than the confusing

y = f (x)). Since f (0) = 0 and f (1) = 1, and since f (x) is increasing (reason: assumption), we conclude

that 0 ≤ b ≤ 1 (conclusion).

For any ǫ > 0, since0 < b < 1, and since{ f (x) : x ∈ [0, 1]} is dense in[0, 1] (reason: assumption

and established facts) , there must be anx′ ∈ [0, 1] such that y′ = f (x′) ∈ (b, b + ǫ), and an x′′ ∈ [0, 1]

such that y′′ = f (x′′) ∈ (b− ǫ, b) (conclusion). Sincef is increasing (reason: assumption),x′ > a > x′′

(conclusion).

Let δ = min{a − x′′, x′ − a}. Sincex′ > a > x′′ (reason: established fact),δ > 0 (conclusion). For

any x ∈ [0, 1] with |x − a| < δ, either x > a − δ ≥ a − (a − x′′) = x′′, or x < a + δ ≤ a + (x′ − a) = x′

(establishing the fact by the definition of absolute values).

Since f is increasing, and by the definition ofδ, (reason: assumption and established facts), when

x < x′, we have f (x) ≤ f (x′) = y′ < b + ǫ; and when x > x′′, we have f (x) ≥ f (x′′) = y′′ > b − ǫ(conclusion). It follows by the definition of absolute values that | f (x) − b| < ǫ. Hence by the definition

of continuity, f (x) is continuous ata.

(The cases whena = 0 or a = 1 can be done by using help intervals. You can also combine the

arguments, but separating the proofs makes the proof easierto be understood).

General Comments: Most of the other proofs in Chapter 1 (I am completing your Chapter 1

today) have the similar problems (confusing notations and lack of sufficient details). Need to train

yourself to be a master of these.

2.1.26 Let f be a function defined by setting

f (x) =

x if x irrational

psin 1q if x = p

q in lowest terms.

At what points is f continuous?

Proof: The function f is continuous on irrationals and discontinuous on rationals.

First prove the case for irrationals. Since limx→0

sinaa = lim

q→∞

sin 1q

1q= 1. That means for∀ǫ > 0, there exists

an sufficient largeM, such that for anyq > M, we have∣∣∣∣sin 1

q1q− 1

∣∣∣∣ < ǫ. Let x be an irrational number, for

∀ǫ > 0, let δ1 = ǫ, δ2 = minq≤M|x − p

q |, let δ = min{δ1, δ2}. Then for any rationaly = pq with |x − y| < δ, we

have| f (x) − f (y)| = |x− psin 1q | = |x−

pq +

pq − psin 1

q | ≤ |x−pq | + |

pq − psin 1

q | ≤ 2ǫ.

Also we have for∀x ∈ [0, 1] and∀ǫ > 0, letδ = ǫ > 0 such that for any irrationaly with |x− y| < δ, we

have that| f (x) − f (y)| = |x− y| < ǫ.Then we can get the conclusion thatf (x) is continuous on [0, 1]. It is easy to prove it is true for any

[N,N + 1], whereN is an integer.

If x = pq , let ǫ0 = 1

2 p| sin 1q −

1q |, for any 0< δ < ǫ0, there exists an irrationaly with |x− y| < δ, such that

2ǫ0 = |psin 1q −

pq | ≤ |psin 1

q − y| + |y− pq |, implies | f (x) − f (y)| = |psin 1

q − y| ≥ ǫ0.


2.1.27 (Apr, 2004)(May, 1996)

Let f be a function in(0, 1) defined by setting

f (x) =

0 if x irrational1q if x = p

q in lowest terms.

Determine all the points where f is continuous. Prove your conclusion.

Proof: The set where f is continuous is the irrationals in(0, 1). Since for∀ǫ = 1n > 0, the sequence{xi}of

rational number such that for each xi =pq ∈ (0, 1) and q≤ n is finite, then letδ = min

i|x− xi |, such that for

any y with|x− y| < δ, we have| f (x) − f (y)| < ǫ.For any rational number x= p

q , there∃ǫ0 = 12q, such that for anyδ > 0, there exists an irrational number y

with |x− y| < δ, such that| f (x) − f (y)| = |1q − 0| > ǫ0.

2.1.28 (Aug, 2004)

Let f be a function in(0, 1) defined by setting

f (x) =

0 if x irrational

sin 1q if x = p

q in lowest terms.

Find the set C of points where f is continuous and the set D of points where f is discontinuous. Justify your

conclusion.

Is function f Riemann integrable on(0, 1)? Lebesgue integrable on(0, 1)? Justify your conclusion.

Proof: The function f is continuous at irrationals. Since for any irrational x and∀ǫ = 1n > 0, let

δ = minp≤q≤n

|x− pq |, such that for any y= p

q with |x− y| < δ, we have that| f (x) − f (y)| = | sin 1q | < |

1q | < ǫ.

2.1.29 (Apr, 2004)(Apr, 2002)

Let f be an real function defined on R with f(0) = 1. If the set A= {x : f (x) > 0} is both open and close,

prove that f(x) > 0 everywhere on R.

Proof: Since f(0) = 1, we have0 ∈ A. If A , R, then there must be some real numbers not in A, then

consider the smallest interval containing0. At least one side of this interval must be a real number, if this

side is open, then this will lead a contradiction to A is closed, for the same reason, it could not be closed.

So A must be R.

2.1.30 (Aug, 2003)

Let fn : (0, 1] → [0,∞) be a decreasing sequence of continuous function convergingpointwise to a zero

functionθ. Must fn converge uniformly?

?

2.1.30: The answer is NO. When saying no, we need to constructa sequence of decreasing func-

tions that converge point-wise to 0.

(How do we find such an example? Think this way: If the domain isa closed interval (or more

generally, a compact set), then the answer will be a YES, as the point-wise convergence will turn into

a finite number of subintervals, and the decreasing propertyassures that we only need to consider one

function on each subinterval. Therefore, the problem must be the open interval.)

2.1. NOTES 9

Define, for eachn > 2,

fn(x) =

1 if x < 1n

−nx+ 2 if 1n ≤ x ≤ 2

n

0 if 1 ≥ x > 2n

(Then we need to verified that fn is decreasing, point-wise convergent to 0, and not uniformly

convergent. )

Proof: For x = 1/mand anyǫ > 0, there existsN = 3msuch that for anyn > N, since fn is a decreasing

sequence,fn(x) = fn(1/m) ≤ fN(1/m) = f3m(1/m) = ǫ. And for anyy ≥ x, we havefn(y) = fn(x) = 0 < ǫ.

Sincex is arbitrary, fn(x) 9 0 point wise.

But fn(x) 9 0 uniformly. There existsǫ0 = 1/2, for anyN, there existsx = 1/2N andn = N such that

fn(x) = fN(1/2N) = 1 > ǫ0. Thereforefn(x) 9 0 uniformly.

Additional Comment to this exercise: You should be able to prove, by using Theorem 2.1.21, the

following:

2.1.30’: Let fn : [0, 1] 7→ [0,∞) be a sequence of decreasing and continuous functions point-wisely

convergent to a zero function. Show thatfn must converge uniformly.

Hint: Use assumption to showfn(x) ≥ 0. Given ǫ > 0, for eacha ∈ [0, 1], there must be aNa such

that n ≥ Na, | fn(a)| < ǫ. Since fn’s are continuous,δa > 0 such that when x ∈ Ia = (a − δa, a + δa),

0 ≤ fn(x) ≤ fNa(x)| < ǫ. Then use the compactness to show that there are only finitelymany suchIa’s

that covers [0, 1], and choose the rightδ.

Proof: Since{ fn} is convergent pointwise to zero function, forxi ∈ [0, 1], ∀ǫ, there∃Ni ∈ Z+ such

that for anyn ≥ Ni , we have| fn(x) − 0| = fn(x) < ǫ. Since fn is continuous, there∃δi > 0 such that for

∀y ∈ (x1−δi , x1+δi), we have| fNi (xi)− fNi (y)| < ǫ. That is fNi (y) < fNi (xi)+ǫ ≤ 2ǫ. Since{ fn} is decreasing,

we have for anyy ∈ (xi − δi , xi + δi) and anyn > Ni that fn(y) < fN1(y) < 2ǫ.

Since [0, 1] is compact, we could use finitely many (xi − δi , xi + δi) to cover it. Then [0, 1] ⊆ ⋃mi=1(xi −

δi , xi + δi). Let N = max{N1,N2, . . . ,Nm}. Then for anyx ∈ [0, 1] andn > N, x must be in one of the

(xi − δi , xi + δi), therefore| fn(x) − f (x)| = fn(x) < 2ǫ.

2.1.31 (May, 1996)

Let { fn : R→ R} be a sequence of continuous function and let f be a point wise limit of fn. Show that for

every a< b, f−1([a, b]) is a Gδ set.

?

A Gδ set is a countable intersection of open sets. Here is a hint:

First show that

f −1[a, b] = ∩∞m=1 ∩∞n=1 f −1

n (a− 1m, b+

1m

).

by showing each point on one side is also a point on the other side.

Then quoting the definition of continuous functions to show that each f −1n (a− 1

m, b+1m) is an open

set.

Since [a, b] = ∩∞m=1(a− 1/m, b + 1/m), I can understand the part∩∞m=1. But for the part∩∞n=1 f −1n , there

is an example:

Let fn(x) = x2 − 1/n. Then fn(x) → f (x) = x2 point wise. Consider the setf −1n ([0, 1]). We have

f −11 ([0, 1]) = [−2,−1] ∪ [1, 2], f −1([0, 1]) = [0, 1]. Then f −1([0, 1]) * f −1

1 ([0, 1])


2.1.32 (Apr, 2005)

Let fn(x) be a sequence of continuous functions on[0, 1] and fn(x) ≥ fn+1(x) n = 1, 2, 3 . . . . For every

x ∈ [0, 1], limn→∞

fn(x) < 0. Determine and prove if there is aδ > 0 such that

limn→∞

fn(x) ≤ −δ ∀x ∈ [0, 1].

Proof: For x1 ∈ [0, 1], let y1 = limn→∞

fn(x1). Then forǫ1 = −1/3y1, there∃N1 > 0, such that for any n≥ N1,

we have

| fn(x1) − y1| < ǫ1. (2.1)

Also, fn(x) is a sequence of continuous functions, then there∃δ1 > 0 such that for any x∈ (x1− δ1, x1+ δ1),

we have

| fN1(x) − fN1(x1)| < ǫ1. (2.2)

By 2.1, we have fN1(x1) < ǫ1 + y1, and by 2.2, fN1(x) < ǫ1 + fN1(x1). Since fn(x) ≥ fn+1(x), we have

fn(x) ≤ fN1(x) < ǫ + fN1(x1) < 1/3y1 for any x ∈ (x1 − δ1, x1 + δ1) and n ≥ N1. Then we can find∞⋃

nk=1(xi − δi , xi + δi) to cover[0, 1], and since[0, 1] is closed and bounded, then we can find a finite sub-

coveringm⋃

nk=1(xnk−δnk, xnk+δnk) of [0, 1]. Then letδ = min{−1/3ynk}, we havelim

n→∞fn(x) ≤ −δ for ∀x ∈ [0, 1].

2.1.33 (Apr, 2002)

Show that if a Cauchy sequence{ fn} has a subsequence converges to l, then fn converges to l.

Proof: See the proof of Cauchy criteria.

Chapter 3

Lebesgue Measure

3.1 Notes

3.1.1 Outer measure:For each set A, define m∗(A) = infA⊂∪In

∑l(In), where{In} is the countable collection

of open intervals that cover A.

Proposition 3.1.2 Let {An} be countable collection of sets of real numbers. Then

m∗(∪An) ≤∑

m∗(An).

3.1.3 A set E is measurable if for every set A, we have m∗A = M∗(A∩ E) +m∗(A∩ E).

Since A= (A ∩ E) ∪ (A ∩ E), we have m∗A ≤ m∗(A ∩ E) + m∗(A ∩ E). So we only need to prove m∗A ≥m∗(A∩ E) +m∗(A∩ E).

Theorem 3.1.4 The collection of measurable set is aσ − algebra; that is, the complement of a measur-

able set is measurable, the union (intersection) of a countable collection of measurable set is measurable.

Moreover, every set with outer measure 0 is measurable.

Theorem 3.1.5 Every Borel set is measurable. In particular each open set and each closed set is measur-

able.

If E is a measurable set, define the Lebesgue measuremE to be the outer measure ofE.

Proposition 3.1.6 Let {Ei} be a sequence of measurable set. Then

m(∪Ei) ≤∑

mEi .

If the sets Ei are pairwise disjoint, then

m(∪Ei) =∑

mEi .

Proposition 3.1.7 Let E be a given set. Then the following five statements are equivalent:

(1) E is measurable.

(2) Givenǫ > 0, there is an open set O⊃ E with m∗(O ∼ E) < ǫ.

11

12 CHAPTER 3. LEBESGUE MEASURE

(3) Givenǫ > 0, there is a closed set F⊂ E with m∗(E ∼ F) < ǫ.

(4) There is a G in Gδ with E ⊂ G, m∗(G ∼ E) = 0.

(5) There is an F in Fσ with F ⊂ E, m∗(E ∼ F) = 0.

If m∗E is finite, the above statements are equivalent to:

(6) Givenǫ > 0, there is a finite union U of open intervals such that M∗(U△E) < ǫ.

3.1.8 An extended real-valued function f is said to be measurable if its domain is measurable, and for each

real number r, the set{x : f (x) > r} is measurable.

3.2 Exercises

3.2.1 (Apr, 2005)

For a bounded set E, define

m∗(E) = b− a−m∗([a, b] \ E),

where [a, b] is an interval containing E, and m∗ denotes the usual outer measure. Prove the following

statements.

(a)If E be the set of all irrational numbers in[0, 1], then m∗(E) = 1.

(b)m∗(E) is independent of the choice of[a, b], as long as it contains E.

(c)m∗(E) ≤ m∗(E).

Proof: (a)If E is the set of all irrational numbers in[0, 1], then [0, 1] \ E is all the rational numbers in

[0, 1], hence m∗([0, 1] \ E) = 0 and m∗(E) = 1 − 0 − 0 = 1. The conclusion ”hencem∗(0, 1] − E) = 0”

needs a reason. Exercise: IfE is countable, thenm∗(E) = 0. (See 3.2.3.)

(b)Assume that[a, b] is the smallest interval that contains E. Then any[c, d] contains E, we have c≤ a

and d≥ b. Since the set[a, b] is measurable, for the set[c, d] \ E we have

m∗([c, d] \ E) = m∗([c, d] \ E ∩ [a, b]) +m∗([c, d] \ E ∩ [a, b])

= m∗([a, b] \ E) +m∗([c, a] ∪ [b, d]).

Then

m∗([c, d]) −m∗([c, d] \ E) = m∗([c, d]) −m∗([a, d] \ E) −m∗([c, a] ∪ [b, d])

= m∗([a, b]) −m∗([a, b] \ E)

= m∗(E).

We can get the conclusion that m∗(E) is independent of the choice of[a, b]. The equality needs the as-

sumption E ⊆ [a, b]. (It is in the beginning that: assume that[a, b] is the smallest interval that contains

E.)

(c)Since[a, b] \ E ∪ E = [a, b], we have

m∗([a, b]) ≤ m∗([a, b] \ E) +m∗(E),

that is

m∗(E) ≥ m∗([a, b]) −m∗([a, b] \ E) = b− a−m∗([a, b] \ E) = m∗(E).

3.2. EXERCISES 13

3.2.2 (Apr, 2005)(Aug, 2001)

Let E be a measurable set in [0,1] with mE= c(12 < c < 1). Let E1 = E + E = x+ y; x, y ∈ E. Show that

there exists a measurable set E2 ⊂ E1 such that mE2 = 1.

Let a be the greatest element inE, thenE ∩ E + a has at most one element, thenm(E ∪ E + a) =

m(E) +m(E + a) = 2m(E) > 1.

Define f (x) = m([0, x] ∩ E1), 0 ≤ x ≤ 2a. Then f (0) = 0, f (2a) > 1. Now we want to provef is a

continuous function.

f (x+ ∆x) = m([0, x+ ∆x] ∩ E)

= m(([0, x] ∩ E) ∪ [x, x+ ∆x] ∩ E)

≤ f (x) + ∆x

That is also true for∆x ≤ 0. So | f (x + ∆x) − f (x)| ≤ |∆x|. f is a continuous function. By mean value

theorem,∃y′ ∈ [0, 2a], such thatf (y) = m([0, y] ∩ E) = 1. And [0, y] ∩ E is the setE2 we want.

SinceE2 is the intersection of two measurable sets, then it is also measurable.

Line 2: SinceE∩ E = E, E∩ E+ a = E+ a. Why it has at most one element? Also, why this would

imply that m(E ∪ E + a) > 1?

Let a be the greatest element inE. Then for anyx ∈ E, we have 0≤ x ≤ a, that isa ≤ x+ a ≤ 2a. So

E∩E+a ⊆ {a}. If E∩E+a =, by the measurability ofE andE+a, we havem(E∪E+a) = mE+m(E+a).

If E ∩ E + a = {a}, thenE ∪ E + a = E \ {a} ∪ {a} ∪ E + a \ {a}. Since the set{a} has measurable

0, it is measurable. AlsoE \ {a} = E ∩ {a}, thenE \ {a} is measurable, andm(E) = m(E \ {a} ∪ {a}) =m(E\ {a})+m({a}) = m(E\ {a})+0 = m(E\ {a}). And for the same reason we havem(E+a) = m(E+a\ {a})is measurable. Since the three setsE \ a, {a} andE + a \ {a} are disjoint,

m(E ∪ E + a) = m(E \ {a}) +m({a}) +m(E + a \ {a}) = m(E) +m(E + a) = 2m(E) > 1.

English: Sentences starting with ”Let ..” should end with a period ”.”, not a ”,”. Also, the next

sentence should start with ”Then”, not ”then”.

3.2.3 (Aug, 2004)

Use the definition of Lebesgue measure show that the set of allrational numbers in[0, 1] is a Lebesgue

measurable set.

Because the rational numbers in [0, 1] is a countable set, denoted as A, it can be written down as

x1, x2, . . .. Then for∀ǫ,A ⊂⋃∞

i=1(xi − ǫ2i+1 , xi +

ǫ2i+1 ). So

m∗A ≤ m∗(∞⋃

i=1

(xi −ǫ

2i+1, xi +

ǫ

2i+1))

≤∞∑

i=1

2iǫ

= ǫ.

This holds for anyǫ, som∗A = 0.


Exercise 3.2.3: You need to carefully read the problem before you prove it. This exercise asks us to

prove that A, the set of all rational numbers in [0,1], is a Lebesgue measurable set, and you prove that

it has outer measure zero. This will be a BIG mistake. I was told that in the analysis exam, you also

made similar errors and caused at least one problem with zeropoints. This is something we MUST

correct.

Hint 1: To prove A is a measurable set, we need to verify Definition 3.1.3. (Remember what is the

Purpose of doing exercise?)

Hint 2: You can also add to the current proof, a property that every subsetA of the real numbers

with m∗(A) = 0 is Lebesgue measurable., by verifying Definition 3.1.3.

By the definition of measurable set, we need to show that for any setE, mE= m(E∩A)+m(E∩ A). That

means we only need to show thatmE≥ m(E ∩ A) +m(E ∩ A). SinceE ∩ A ⊆ A, we havem(E ∩ A) ≤ mA.

Then m(E ∩ A) = 0. And sinceE ∩ A ⊆ E, m(E ∩ A) ≤ mE. Then we can get the conclusion that

mE≥ 0+m(E ∩ A) = m(E ∩ A) +m(E ∩ A).

3.2.4 (Aug, 2004)(Aug, 1997)

Let E ⊂ [0, 1] be closed and with no interior point. Is it true that the measure m(E) = 0? Justify your

conclusion.

Proof: No. Consider the generalized Cantor set.

Letα be an fixed number in(0, 1), we construct the generalized Cantor set as follows: First delete the middle13α open interval. Second, for the remaining two intervals, delete each interval the middle19α open interval.

In the n-th step, delete the middleα 13n open interval from each of the remaining2n−1 intervals. Let n→ ∞,

the measure of the remaining set E is1− α∞∑

i=1

13i = 1− α.

In each step, we delete the open intervals, then E is a closed set. If E has an interior point x, then there

exists aδ > 0 with (x − δ, x + δ) ⊂ E. But each time we delete the middle13n open intervals, there must

be an m, such that13m < 2δ. Then in the m-th step, we will delete part of(x − δ, x + δ), contradicting to

(x− δ, x+ δ) ⊂ E.

The set E⊂ [0, 1] is closed and with no interior point and mE= 1− α , 0.

Cantor set is a nowhere dense set.

3.2.5 (Aug, 2003)

Let E1 ⊃ E2 ⊃ . . . be an infinite sequence of measurable subest ofR and assume that∞⋂

n=1En = ∅.

(a)Show that if m(E1) < ∞, then limn→∞m(En) = 0

(b)Give an example showing that the conclusion of (a) may be false when m(E1) = ∞.

Proof: Let Fi = Ei \ Ei+1, then we have Fi ∩ F j = ∅ if i , j, and that E1 =⋃i

Fi ∪⋂i

Ei . Since the family

of measurable sets is an algebra of sets, then⋃i

Fi and⋂i

Ei are measurable and that

m∗(E1) −m∗(⋂

i

Ei) = m∗(⋃

i

Fi)

= m∗(E1) −m∗(E2) +m∗(E2) −m∗(E3) . . .

= m∗(E1) − limi

Ei .

3.2. EXERCISES 15

Thenlimi

Ei = m∗(⋂i

Ei) = m∗(∅) = 0.

(2)Let Ei = [i,∞], then∞⋂

n=1En = ∅, but lim Ei , 0.

3.2.5: This is a well-written exercise.

English:

(1) Avoid ”Let ..., then ...”, which is incorrect English. Use ”Let.... Then ...” (This occurs very

often, and almost everywhere in your writing.)

(2) ”... and that” should be ”that ... and that ...” This does not fit in both uses in your writing. You

can delete the two ”that” in your writing here.

3.2.6 (Apr, 2005)

Let { fn} be a sequence of measurable functions on(a, b), and

E = {x ∈ (a, b) : fn(x) is convergent}.

Show that E is measurable.

Since for a fixedx, if { fn(x)} convergent, then it is a Cauchy sequence. Forǫk = 1/k, there∃Nk > 0,

such that for anyn,m > Nk, we have| fn(x) − fm(x)| < ǫk. Then the setE where fn(x) is convergent could

write down as follows:

E =∞⋂

k=1

Fk,

Fk =

⋃

n,m>Nk

{x : | fn(x) − fm(x)| < ǫk}.

SinceFk is the union of countable collection of measurable set, thenFk is measurable, andE is measurable.

3.2.6:

Mathematical Logic: Need to quote a reason whyFk is measurable. (Hint: use definition of mea-

surable functions).

A General Rule: You can always check your proof and see if all the conditions in the problems are

applied. Reading your proof, you may ask yourself where did Iuse the assumption that the functions

are measurable functions? If there is a condition not used inthe proof, check the proof again.

English: Avoid ”Since ...., then ...” This is Chinese English. See Bondy and Murty’s book to see

”Since ..., (without then)...”

Now we want to showFk is measurable. We can write downFk like this

Fk =⋃

n,m>Nk{x : | fn(x) − fm(x)| < ǫk}

=⋃

n,m>Nk{x : −ǫk + fm(x) < fn(x) < fm(x) + ǫk}

=⋃

n>Nk

⋃m>Nk{x : fn(x) > fm(x) − ǫk} ∩ {x : fn(x) < fm(x) + ǫk}

.

Since eachfn(x) is measurable, by the definition of measurable function,the set{x : fn(x) > fm(x) − ǫk}and{x : fn(x) < fm(x) + ǫk} are measurable. Since the collection of measurable set is aσ−algebra,Fk is

measurable and henceE is measurable.


3.2.7 (Apr, 2002)

Prove that function f(x) defined on R is measurable if for any rational number r, the setE = {x : f (x) < r}is measurable.

Proof: For each x∈ R, there is a sequence of rational numbers rn ↓ x, then{x : f (x) < x} = ⋂n{x : f (x) <

rn}. Since the countable intersection of measurable set is measurable, then{x : f (x) < x} is measurable,

and therefore f(x) is measurable.

3.2.7: Good math. Same English problem as above.

3.2.8 (Aug, 2001)

Suppose E1,E2, . . . ,En are n measurable sets in[0, 1], and every x∈ [0, 1] belongs to at least q of these

sets. Show that, there is at least an Ek such that mEk ≥ q/n.

Proof: Since every x∈ [0, 1] belongs to at least q sets from{E1,E2, . . . ,En}, then⋃k

Ek covers[0, 1] q

times, and that

q = m(n⋃

k=1

Ek) ≤ mE1 +mE2 + · · · +mEn.

Then there is at least an Ek such that mEk ≥ q/n. 3.2.8: Good.

3.2.9 (Apr, 2001)

(a) Let{ fn} be a sequence of measurable functions. Show that if fn→ f a.e., then f is measurable.

(b) Let f be a differentiable function. Show that f′ is measurable.

(a)The functionf is measurable if for eachr ∈ R, the setEr = {x : f (x) > r} is measurable. Since

fn → f a.e., the set wherefn 9 f is measure 0 hence a measurable set, and any subset of it is also of

measure 0 and hence measurable, then{x : f (x) > r, fn(x) 9 f (x)} is measurable. And we only need to

consider the part thatfn(x)→ f (x). And for fn(x)→ f (x), we havelim fn(x) = lim fn(x) = f (x) and:

{x : lim fn(x) = f (x) > r} = {x : lim fn(x) > r}

= {x : infk

supn≥k

fn(x) > r}

= {x :∞⋂

k=1

∞⋃

n=k

fn(x) > r}.

Each fn(x) is measurable and the collection of measurable set is aσ-algebra, sof (x) is measurable.

(b) Let fn(x) =f (x+ 1

n )− f (x)1n

. Then f ′(x) = limn→∞

fn(x). Since the set{x : fn(x) > r} is measurable,

{x : f ′(x) > r} =⋂n{x : fn(x) > r} is also measurable.

Page 13:

Do you think you need to justify the equalities of the sets?I don’t understand.

English:

(1) Avoid ”Let ..., then ...”, which is incorrect English. Use ”Let.... Then ...” (This occurs very

often, and almost everywhere in your writing.)

3.2.10 (Apr, 2001)(Aug, 1996)

Let { fn} be a sequence of measurable function on[0, 1] such that{ fn(x)} is a bounded sequence for each x.

Define the set E= {x ∈ [0, 1] : limn→∞

fn(x) exists.} Show that E is a measurable set.

3.2. EXERCISES 17

See 3.2.6.3.2.10: Why fn’s must be bounded? The condition thatfn is bounded has nothing to do

with the problem. Since the condition thatE = {x ∈ [0, 1] : limn→∞

fn(x) exists.} only asks us to consider that

set thatfn convergent to a real number smaller than∞.

3.2.11 (Aug, 2000)(Jan, 2001)(Apr, 2001)

For each statement below, either prove it (if true) or give a counter example (if false).

(a) E is measurable if and only if m∗(P∪ Q) = m∗(P) +m∗(Q) for any P⊂ E and any Q⊂ E.

(b) If E is a countable set, then E is measurable and mE= 0.

(c) If E is a measurable set with mE= 0, then E is a countable set.

(d) If f is measurable, then so is| f |.(e) If | f | is measurable, then so is f .

(f) If f ≥ 0 is measurable on E and∫

Ef = 0, then f= 0 a.e. on E.

(g) If f is continuous on[a, b], then f is of bounded variation on[a, b].

Proof: (a) True. ”=⇒” is obvious. ”⇐=”: For any set A, let P= A∩ E, Q= A∩ E, then A= P∪ Q, and

m∗(A) = m∗(P∪ Q) = m∗(P) +m∗(Q) = m∗(A∩ E) +m∗(A∩ E).

(b) True. If E is countable, the we could write down E as{x1, x2, . . . }, for ∀ǫ > 0, the set∞⋃

n=1(xn− ǫ

2n , xn+ǫ2n )

covers E, and its length isǫ, then mE< ǫ. Sinceǫ is arbitrary, mE= 0.

For any set A, m(A∩E) = 0. Also A⊃ A∩ E, and so mA≤ m(A∩ E) = m(A∩E)+m(A∩ E), then therefore

E is measurable.

(c) False. Cantor set is measurable but not countable. It hasa one to one correspondence with[0, 1].

(d) True. If f is measurable, then both{x; f (x) > r} and{x; f (x) < −r} are measurable for∀r ∈ R. Therefore

{x; | f (x)| > r} = {x; f (x) > r} ∪ {x; f (x) < −r} is measurable.

(e) False. Define

f (x) =

1 x < S

−1 x ∈ S ,

where S is a nonmeasurable set. Then f is nonmeasurable but| f | is measurable.

(f) Let En = {x : f (x) > 1/n}, then∫

Ef ≥ mE1 + 1/2m(E2 \ E1) + 1/3m(E3 \ E2) + . . . , and E =

E1 ∪ (E2 \ E1) ∪ (E3 \ E2) ∪ . . . . If one of the En \ En−1 has measure greater than 0, then we will have∫E

f (x) > 0 which is a contradiction. Therefore mE= 0.

(g) Consider the function f(x) =

sin 1

x x ∈ (0, 1]

0 x = 0on [0, 1]. For any M∈ Z+, there is a subdivision0 =

x1 <1

2Mπ+ 3π2< 1

2Mπ+ π2< 1

2(M−1)π+ 3π2< 1

2(M−1)π+ π2< · · · < x2M+2 = 1 such that

2M+2∑k=1| f (xk) − f (xk−1)| > 4M.

Then f(x) is not of bounded variation on[0, 1].

3.2.11:

(a) Never use the phrases like ”obvious”, ”it is clear” ... Can you quote a definition, or a theorem

to justify your conclusion?

English: (a), (f): Avoid ”Let ..., then ...”,

3.2.12 (Jan, 2001)(Aug, 1997)

(a) State the definition of a measurable function.

(b) Use the definition to deduce that, if f is measurable on a measurable set E, then for everyα ∈ R, the set


Eα = {x ∈ E : f (x) = α} is measurable.

(c) Construct a function f on E= (0, 1) to show the converse of (b) is not true.

(a) The functionf is measurable if for anyr ∈ R, the set{x : f (x) > r} is measurable.

(b) If {x : f (x) > r} is measurable, then its complement{x : f (x) ≤ r} is also measurable. Since the

set{x : f (x) < r} =∞⋃

n=1{x : f (x) ≤ r − 1

n}. The countable collection of measurable set is aσ-algebra, then

˜{x : f (x) < r} ∩ ˜{x : f (x) > r} = {x : f (x) = r} is measurable.

(c) LetS be a nonmeasurable set in (0, 1), definef (x) =

0 x ∈ E \ S

x x∈ S. Then the set{x : f (x) > 0} =

S which is also nonmeasurable.

3.2.12:

English: (a) ”The function...” should be ”A function...”

Math logic: (b) The proof is an example of a bad logic. Here is how you would think:

Want to prove {x : f (x) = r} is measurable? It suffices to show that both{x : f (x) < r} and

{x : f (x) > r} are measurable.

Definition says {x : f (x) > r} is measurable. How do I prove{x : f (x) < r} is measurable? As

{x : f (x) < r} = {x : − f (x) > −r}, can I used property or definition to show that − f (x) is also

measurable?

(c) The example is correct.

English: Let.... Define... (Two sentences).

Do you need to explain whyf satisfies the converse of (b)? It is better for you to write down clearly

what the converse of (b) is before you display your example.

(b) If {x : f (x) > r} is measurable, then its complement{x : f (x) ≤ r} is also measurable. Since the set

{x : f (x) < r} =∞⋃

n=1{x : f (x) ≤ r − 1

n} =∞⋃

n=1

˜{x : f (x) > r − 1n}. The countable collection of measurable set

is aσ-algebra, then{x : f (x) < r} is measurable. Also, ˜{x : f (x) < r} ∩ ˜{x : f (x) > r} = {x : f (x) = r} is

measurable.

(c) The converse of (b) is: If for anyα ∈ R, the setEα = {x : f (x) = α} is measurable, thenf

is measurable. LetS be a nonmeasurable set in (0, 1). Define f (x) =

0 x ∈ E \ S

x x∈ S. Then the set

{x : f (x) > 0} = S which is also nonmeasurable. Thereforef is nonmeasurable.

3.2.13 (May, 1996)

Let the set E⊂ (0, 2) be defined as follows:

E =

{1j+

sinkk

; j, k = 1, 2, 3 . . .

}.

Is there a finite collection of open intervals{Ik, k = 1, 2, 3, . . . } such that E⊂ ∪nk=1Ik and

∑nk=1 l(Ik) ≤ 1

2?

Prove your conclusion.

We can order the setE as we order the rational numbers, thenE is a countable set, hencemE = 0. By

the definition, we havemE= infE⊂In

∑n

l(In). Then there must be an open covering{In} of E with∑n

l(In) = 1/2.

Also the setE is bounded and closed, then any open covering has a finite sub-covering.

3.2. EXERCISES 19

3.12.13

English: We ... Then ... (two sentences)

It is much better to explicitly show the open cover then saying ”there must be...” Saying ”there

must be ...” in many cases is equivalent to say nothing. List elements inE by a1, a2, ... and cover them

with In = (an − 1/2n+1, an + 1/2n+1).

We can order the setE as we order the rational numbers. ThenE is a countable set andE = {a1, a2, a3 . . . },hencemE = 0. By the definition, we havemE = inf

E⊂In

∑n

l(In). There is an open covering{In} = (an −

1/2n+1, an + 1/2n+1) of E with∑n

l(In) = 1/2. Also the setE is bounded and closed, then any open covering

has a finite sub-coveringO ⊂∞⋃

n=1In, andmO≤ l(In) = 1/2.

How to proveE is closed?

3.2.14 (may, 1996)

Is there a monotone function on[0, 1] which is discontinuous at all rational points? Prove your conclusion.

Proof: No? 3.2.14:

I will think it over before answering it.


New begin from July 16

Proposition 3.2.15 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 giveǫ, δ > 0, there is a subset A⊂ E with mA< δ and an

integer N such that for∀x ∈ E \ A and∀n ≥ N,

| fn(x) − f (x)| < ǫ

Proof: Let

Gn = {x : | fn(x) − f (x)| ≥ ǫ},

and

En = {∞⋃

k=n

Gk} = {x : | fk(x) − f (x)| ≥ ǫ for somek ≥ n}.

ThenE1 ⊇ E2 ⊇ E3 . . . .

Since fn → f (x) a.e., the setF that fn 9 f (x) has measure 0, and⋂

En = F. By 3.2.5, we have

lim mEn = mF = 0. Then forδ > 0, there exists anN, such that for anyn ≥ N, we havemEn ≤ δ. Let

A = EN. Then we have

A = E \ A = {x : | fn(x) − f (x)| < ǫ for anyn ≥ N}.

�

Theorem 3.2.16 Egoroff’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 giveη > 0, there is a subset A⊂ E

with mA< η such that fn converges to f uniformly on E\ A.

Proof: By 3.2.15, we have forǫk = 1/k andδk = 1/2kη, there exists a setAk with mAk < δk andNk,

such that for anyx ∈ Ak andn ≥ Nk, we have| fn(x) − f (x)| < ǫk. Let A =⋃

Ak. ThenmA ≤ ∑Ak <

η. And for ǫ = 1/k, there existsN = k, such that for anyn > N and anyx ∈ A, | fn(x) − f (x)| < ǫ.

�

Chapter 4

The Lebesgue Integral

4.1 Notes

The functionχE defined by

χE(x) =

1 x ∈ E

0 x < E

is called the characteristic functionof E. A linear combination

ϕ =

n∑

i=1

aiχEi (x)

is called a simple functionif all Ei are measurable. Note that we don’t requireEi to be disjoint.

Proposition 4.1.1 Let f be a bounded function defined on a measurable set with finite measure. In order

that

infψ≥ f

∫

Eψ(x)dx= sup

f≥ϕ

∫

Eϕ(x)dx

for all simple functionψ andϕ, it is necessary and sufficient that f be measurable.

Proof: Long proof, omit.

�

4.1.2 If f is a bounded measurable function defined on a measurable set E with mE finite, we define the

(Lebesgue) integralof f over E by ∫

Ef (x)dx= inf

∫

Eψ(x)dx

for all simple functionsψ ≥ f .

Proposition 4.1.3 Let f be a bounded function defined on[a, b]. If f is Riemann integrable on[a, b], then

it is measurable and

R∫ b

af (x)dx=

∫ b

af (x)dx.

21

22 CHAPTER 4. THE LEBESGUE INTEGRAL

Proof: Since a step function is also a simple function,

R∫ b

a

f (x) ≤ supf≥ϕ

∫

Eϕ(x)dx≤ inf

ψ≥ f

∫

Eψ(x)dx≤ R

∫ b

af (x).

Since f is Riemann integrable and by 4.1.1, we havef is measurable. By 4.1.2,

∫ b

af (x)dx = inf

ψ≥ f

∫

Eψ(x)dx= R

∫ b

af (x)dx.

�

Theorem 4.1.4 Bounded Convergence Theorem:Let { fn} be a sequence of measurable function defined

on a set E of finite measure, and suppose that there is a real number M such that| fn(x)| < M for all n and

all x ∈ E. If fn(x)→ f (x) pointwise for all x∈ E, then

∫

Ef = lim

∫

Efn.

Proof: By 3.2.16, we have for∀ǫ > 0, there exists a setA with mA< ǫ/2M, such that for anyx ∈ A, we

have fn→ f uniformly. Then there exists anN, such that for anyn > N, we have| fn(x) − f (x)| < ǫ/mE for

all x ∈ A. Thus for alln > N, we have

|∫

Efn −

∫E

f | = |∫

Efn − f |

≤∫

E| fn − f |

=

∫E\A | fn − f | +

∫A| fn − f |

≤ mE∗ ǫ/mE+ 2M ∗ ǫ/2M

= 2ǫ.

Then∫

Ef = lim

∫E

fn.

�

Proposition 4.1.5 A bounded function f on[a, b] is Riemann integrable if and only if the set of points at

which f discontinuous has measure zero.

If f is a nonnegative measurable function defined on a measurablesetE, we define

∫

Ef = sup

h≤ f

∫

Eh,

whereh is a bounded measurable function such thatm(x : h(x) , 0) is finite.

Theorem 4.1.6 Fatou’s Lemma:If { fn(x)} is a sequence of nonnegative measurable functions and fn(x)→f (x) a.e. on a set E, then ∫

Ef ≤ lim

∫

Efn.

4.1. NOTES 23

Proof: Since fn(x)→ f (x) a.e., the integral on a measure 0 set is 0. Then we can assume that fn(x)→ f (x)

everywhere. Leth be a bounded measurable function no greater thanf which and vanishes outside a setE′

with finite measure. Definehn(x) = min{h(x), fn(x)}.Now we want to show thathn(x) → h(x) on E′. Since fn(x) → f (x), for any ǫ > 0, there exists an

N, such that for anyn > N, we have| fn(x) − f (x)| < ǫ. Also, for anyn > N, if hn(x) = h(x), we have

|hn(x) − h(x)| = 0 < ǫ. If hn(x) = fn(x), that meansfn(x) ≤ h(x) ≤ f (x), then|hn(x) − h(x)| = | fn(x) − h(x)| ≤| fn(x) − f (x)| < ǫ. Thereforehn(x)→ h(x). And by Bounded Convergence Theorem, we have

∫

Eh =

∫

E′h = lim

∫

E′h ≤ lim

∫

E′fn.

Taking the supremum overh, we have∫

Ef ≤ lim

∫E

fn.

�

Theorem 4.1.7 Monotone Convergence Theorem:Let { fn} be an increasing sequence of nonnegative

measurable functions, and let fn(x)→ f (x) a.e. Then∫

f = lim∫

fn.

Proof: By Fatou’s Lemma, we have ∫f ≤ lim

∫fn.

Since{ fn} is an increasing sequence of functions, we havefn(x) ≤ f (x) a.e., and so∫

fn(x) ≤∫

f (x). Then

lim∫

fn ≤∫

f .

Since lim∫

fn ≥ lim∫

fn, we have ∫f = lim

∫fn.

�

4.1.8 A nonnegative measurable function f is called integrableover the measurable set E if

∫

Ef < ∞.

Proposition 4.1.9 Let f be a nonnegative function which is integrable over a setE. Then givenǫ > 0, there

is aδ > 0 such that for any subset A⊂ E with mA< δ, we have∫

Af < ǫ.

Proof: If f is bounded byM. Then givenǫ > 0, for any subsetA ⊂ E with mA< ǫ/M, we have∫

Af < ǫ.

If f is unbounded, define

fn =

f (x) if f (x) ≤ n

n if f (x) > n.


Then eachfn is bounded,fn is increasing and converges tof pint wise. LetA be any subset ofE. By

Monotone Convergence Theorem, we have for anyǫ > 0, there exist anN such that∫

AfN−

∫A

f =∫

AfN− f ≤

ǫ. Let δ = ǫ/N. Then for anyA ⊆ E with mA< δ, we have

∫A

f ≤∫

A[ f (x) − fN(x)] +

∫A

fN(x)

≤ ǫ + N ∗ ǫ/N= 2ǫ.

�

Let f +(x) = max{ f (x), 0} and f −(x) = max{− f (x), 0}, we havef = f + − f −.

4.1.10 A measurable function f is said to be integrable over E if f+ and f− are both integrable over E. In

thus case we define ∫

Ef =

∫

Ef + +

∫

Ef −.

Theorem 4.1.11 Lebesgue Convergence Theorem:Let g be an integrable over E and let{ fn(x)} be a

sequence of measurable function such that| fn| ≤ g on E and fn(x)→ f (x) a.e. Then

∫

Ef = lim

∫

Efn.

Proof: Since| fn| ≤ g and fn(x) → f (x) a.e.,g− fn is nonnegative andg(x) − fn(x) → g(x) − f (x) a.e., by

Fatou’s Lemma, ∫

Eg(x) − f (x) ≤ lim

∫

Eg(x) − fn(x).

Since fn(x)→ f (x) a.e., by 3.2.9,f (x) is measurable. Since| fn| ≤ g,∫

f ≤∫

g < ∞, so f is integrable.

Then we have ∫g−

∫f ≤

∫g− lim

∫fn,

that is equal to ∫f ≥ lim

∫fn.

Again by Fatou’s Lemma,∫

f ≤ lim∫

fn. So∫

Ef = lim

∫E

fn.

�

Chapter 4.

General Comment: Looks like you are good in understanding and applying the definition of inte-

gral, using simple functions. You are a lot better in using the epsilon argument this time.

Theorems 4.1.4, 4.1.6, and 4.1.7 are the most important, andcommonly used tools. Need to re-

member them all. One way to remember them is to find more applications of them. When you work

on an exercise related to convergence, ask yourself: can I apply one of them?

4.2. EXERCISES 25

4.2 Exercises

4.2.1 (Aug, 2005)

Determine if each of the following is true or false:

(1) If f is continuous on[0, 1], then it is uniformly continuous on[0, 1].

(2) If { fn} is a sequence in L1(R), and fn(x)→ 0 uniformly for all x∈ R, then∫R

fn→ 0.

(3) For measurable sets En ⊆ [0, 1] with lim mEn = 1, there holdslim∫

Enf =

∫ 10 f for every integrable

function f on[0, 1].

Proof: (1) True. For anyx ∈ [0, 1], and anyǫ > 0, there exists aδx > 0, such that for anyy ∈ (x− δx, x+ δx),

we have|xk − y| < ǫ. Let Ox be (x − 1/2δx, x + 1/2δ). Now we can use{Ox : x ∈ [0, 1]} to cover

[0, 1]. Since [0, 1] is a closed bounded interval, by Heine-Borel Open Covering Theorem, there is a finite

sub-covering∪nk=0Ok ⊃ [0, 1]. Let δ = 1/2 min{δ1, δ2, . . . , δn}, for any |y − z| < δ, the elementy must in

(x−1/2δxk , x+1/2δxk) for somek. Since|y−z| < δ ≤ 1/2δxk , |z− x| ≤ |z−y|+ |y− x| ≤ 1/2δxk +1/2δxk = δxk.

That meansz∈ (xk − δxk , xk + δxk). Therefore| f (xk) − f (y)| < ǫ and| f (xk) − f (z)| < ǫ. Also

| f (y) − f (z)| ≤ | f (xk) − f (y)| + | f (xk) − f (z)| < 2ǫ.

Then f is uniformly continuous on [0, 1].

(2) False. Definefn(x) = 1nχ([0, n]). Then for anyx ∈ R andǫ = 1/k > 0, there exists anN = k, such

that for anyn > N, we have| fn(x) − 0| ≤ 1/n < 1/k = ǫ. Then fn(x)→ 0 uniformly on [0, 1]. But∫R

fn = 1

for all n.

(3) True. By the definition of Lebesgue integral,∫ 10 f = sup

ϕ≤ f

∫ 10 ϕ. That means for anyǫ > 0, there exist

a simple functionϕ ≤ f , such that∫ 10 f −

∫ 10 ϕ ≤ ǫ. Without generality, assumeϕ =

∑ni=1 aiχ(Fi) and the

setsFi are disjoint. Then∫ 10 ϕ =

∑ni=1 aimFi. Let a = max{a1, a2, . . . , an}.

Since limmEn = 1 andEn ⊆ [0, 1], then there exists anN, such that for anyn ≥ N, we have 1−mEn <

ǫ/a. Let F′i = EN∩Fi . SinceFi are disjoint measurable sets, we have the setsF′i are disjoint and measurable.

Also,∫ 10 ϕ ≥

∫ENϕ. Then

|∫ 10 ϕ −

∫ENϕ| =

∫ 10 ϕ −

∫ENϕ

=

n∑i=1

aimFi −n∑

i=1aimF′i

≤n∑

i=1amFi −

n∑i=1

amF′i

= a(n∑

i=1mFi −mF′i )

= a(m([0, 1]) −mEN)

≤ aǫa= ǫ

.

Now we have∫ 10 ϕ −

∫E′ϕ ≤ ǫ, combine with

∫ 10 f −

∫ 10 ϕ ≤ ǫ, we have

∫ 10 f −

∫ENϕ ≤ 2ǫ. Since

∫ENϕ ≤

∫EN

f ≤∫ 10 f , we have|

∫ 10 f −

∫EN

f | ≤ 2ǫ. Since for anyn > N, m([0, 1]) −mEn ≤ ǫ, we can also derive

such inequalities. Therefore lim∫

Enf =

∫ 10 f .

2nd Proof for (3).


Since f is integrable over [0, 1], we have bothf + and f − are integrable. By 4.1.9, we have for anyǫ,

there exists aδ > 0, such that for anyA ⊆ E with mA< δ, we have∫

Af + < ǫ

∫

Af − < ǫ.

Since limmEn = 1, then there exists anN, such that for anyn > N, we have 1−mEn < δ. So∫

[0,1]\En

f + < ǫ∫

[0,1]\En

f − < ǫ.

Then ∫

[0,1]f −

∫

En

f =∫

[0,1]\En

f =∫

[0,1]\En

f + −∫

[0,1]\En

f − < ǫ.

Then lim∫

Enf =

∫ 10 f .

Ex 4.2.1(3) The first proof uses definition, which is OK. Can wehave a better (here is simpler)

proof?

The second (another proof) has a gap. The last line of Page 25 does not imply the first line on Page

26. Check the definition of a limit, and understand absolute values.

The might be a third proof, which will train you how to apply a T heorem. Suppose thatf ≥ 0 (or

f ≥ 0 almost every where). Definefn = f when restricted to En. Then fn ≤ f , and fn is increasing.

Therefore, Theorem 4.1.7 can give you the answer.

How about if we do not have f ≥ 0 almost everywhere? Thenf + and f − will. You may need to use

the ”Twin Sister” theorem of Theorem 4.1.7 when the sequencefn is decreasing.

3rd Proof for (3).

Since f is integrable over [0, 1], we have bothf + and f − are integrable. Letf +n and f −n be restricted

on En. Then bothf +n and f −n are increasing sequences. SinceEn ⊆ [0, 1] with lim mEn = 1, we havef +nconverges tof + a.e. andf −n converges tof − a.e.. By Monotone Convergence Theorem, we have

∫

Ef + = lim

∫

En

f +n = lim∫

En

f + and∫

Ef − = lim

∫

En

f −n = lim∫

En

f −.

And so∫

Ef =

∫

E( f + − f −) =

∫

Ef + −

∫

Ef − = lim

∫

En

f + − lim∫

Ef − = lim

∫

En

( f + − f −) = lim∫

En

f .

�

4.2.2 (Aug, 2005)

Let E be a measurable set in[0, 1] with mE> 0. Then there are two numbers x and y in E(x , y) such that

x− y is a rational number.

Proof: By way of contradiction. Assume that any two elementsx, y in E, x − y is not a rational number.

Since the rational number is a countable set, let the rational number in [0, 1] be{r1, r2, r3, . . . }. Then for any

two different rational numbersr i andr j , if the setE+ r i ∩ E+ r j is not empty, then there exists an elementz

inside, alsoz− r i , z− r j andz− r i , z− r j ∈ E, butz− r i − (z− r j) is a rational number, contradicts to our

assumption. Then we assume thatE + r i ∩ E + r j for all i , j is empty.

4.2. EXERCISES 27

SinceE is a subset of [0, 1], we haveE + rk is in [0, 2] for k = 0, 1, 2, . . . . And so∞⋃

k=1Ek ⊆ [0, 2].

We have∑

mEk ≤ m([0, 2]) = 2. But mE = mEk > 0, the summation∑

mEk should be∞, which is a

contradiction to∑

mEk ≤ m([0, 2]) = 2. So there must be two numbersx andy in E(x , y) such thatx− y

is a rational number.

�

4.2.3 (Aug, 2005)

let { fn} be a sequence of integrable functions on[0, 1] such that fn→ f a.e. in[0, 1] with f integrable. Then

lim∫ 1

0| fn − f | = 0 ⇐⇒ lim

∫ 1

0| fn| =

∫ 1

0| f |.

Proof:” =⇒ ”:

If lim∫ 10 | fn− f | = 0, then for anyǫ > 0, there exists anN, such that for anyn > N, we have

∫ 10 | fn− f | < ǫ.

Since∫ 10 (| fn| − | f |) ≤

∫ 10 | fn − f |, we also have

∫ 10 (| fn| − | f |) =

∫ 10 | fn| −

∫ 10 | f | < ǫ. Then lim

∫ 10 | fn| =

∫ 10 | f |.

4.2.3 ” =⇒ ” : same gap in absolute values. You need to show|∫ 10 (| fn| − | f |)| < ǫ, and so your

arguments need to be used twice.

If lim∫ 10 | fn− f | = 0, then for anyǫ > 0, there exists anN, such that for anyn > N, we have

∫ 10 | fn− f | < ǫ.

Since∫ 10 (| fn| − | f |) ≤

∫ 10 | fn − f | and

∫ 10 (| f | − | fn|) ≤

∫ 10 | fn − f |, we have|

∫ 10 (| fn| − | f |)| ≤

∫ 10 | fn − f | < ǫ.

Then lim∫ 10 | fn| =

∫ 10 | f |.

” ⇐= ”:

Since f is integrable in [0, 1], by 4.1.9, for anyǫ > 0, there exists aδ > 0, and anyE ∈ [0, 1] with

mE< δ, we have∫

Ef < ǫ.

By Egoroff’s Theorem, if fn → f a.e. on a bounded measurable set [0, 1], for suchδ > 0, there exists

a setA ⊆ [0, 1] with mA< δ, such thatfn → f uniformly on A. Then there exists anN1, such that for any

n > N1 andx ∈ A, | fn(x) − f (x)| < ǫ.Since lim

∫ 1

0| fn| =

∫ 1

0| f |, there exists anN2, such that for anyn > N2, |

∫ 1

0| fn| −

∫ 1

0| f || < ǫ. Then we

have that∫ 1

0| fn| <

∫ 1

0| f | + ǫ. And so for the setA ⊂ [0, 1] with mA< δ and anyn > N = max{N1,N2}, we

have ∫ 10 | fn| =

∫A| fn| +

∫A| fn|

<∫

A| f | +

∫A| f | + ǫ

.

Then we have that ∫A| fn| <

∫A| f | +

∫A| f | −

∫A| fn| + ǫ

≤∫

A| f | +

∫A| f − fn| + ǫ

< ǫ + ǫ + ǫ

= 3ǫ

Now we have that for any anyA ⊂ [0, 1] with mA< δ and anyn > N = max{N1,N2},∫ 10 | fn − f | ≤

∫A| fn − f | +

∫A| fn − f |

< ǫ +∫

A| fn| +

∫A| f |

< 5ǫ

.

Therefore lim∫ 10 | fn − f | = 0.


4.2.4 (Real Analysis, Royden, Page 94, Problem 17)(Apr,2004)(May, 1996)

(a) Let f be integrable over R. Then∫

f (x)dx =∫

f (x+ t)dx.

(b) Let g be a bouned measurable function. Then

limt→0

∫ ∞

−∞|g(x)[ f (x) − f (x+ t)]| = 0.

Proof: (a)By the definition of Lebesgue integral, we have∫

f (x)dx= infϕ≥ f

∫ϕ(x)dx.

So for anyǫ > 0, there exists a simple functionϕ(x) such that∣∣∣∫

f (x)dx−∫ϕ(x)dx

∣∣∣ < ǫ, and therefore∣∣∣∫R

f (x+ t)dx−∫Rϕ(x+ t)dx

∣∣∣ =∣∣∣∫R

f (x+ t)d(x+ t) −∫Rϕ(x+ t)d(x+ t)

∣∣∣ =∣∣∣∫R

f (y)dy−∫Rϕ(y)dy

∣∣∣ < ǫ, for

y = x+ t.

For such simple functionϕ, we have

∫ϕ(x)dx =

∑nk=1 aimEi

=∑n

k=1 aim{Ei + t}=

∫ϕ(x+ t)dx.

Then ∣∣∣∫

f (x)dx−∫

f (x+ t)dx∣∣∣ ≤

∣∣∣∫


∣∣∣ +∣∣∣∫ϕ(x)dx−

∫ϕ(x+ t)dx

∣∣∣+

∣∣∣∫ϕ(x+ t)dx−

∫f (x+ t)dx

∣∣∣< 2ǫ.

Sinceǫ is arbitrary, we have∫

f (x)dx=∫

f (x+ t)dx.

(b) Assume that|g| is bounded byM. Since f (x) is integrable overR, givenǫ > 0, there is a continuous

functionh defined on a closed interval [−N,N] and vanishes outside, such that∫ ∞

−∞| f − h| < ǫ/M.

Thenh is a continuous function defined on [−N,N], and so it is uniformly continuous. That means for

any x ∈ R there exists aδ > 0, such that for any|t| < δ, we have|h(x) − h(x + t)| < ǫ2MN . And so∫ N

−N|h(x) − h(x+ t)| < ǫ

M . Then we have for anyǫ > 0, there existsδ > 0, such that for any|t| < δ, we have

∫ ∞−∞ |g(x)[ f (x) − f (x+ t)]| =

∫ ∞−∞ |g(x)[ f (x) − f (x+ t)]|dx

=

∫ ∞−∞ |g(x)| |[ f (x) − f (x+ t)]| dx

≤ M∫ ∞−∞ |[ f (x) − f (x+ t)]|dx

≤ M∫ N

−N|[ f (x) − h(x)| + |h(x) − h(x+ t)| + |h(x+ t) − f (x+ t)]| dx

< 3ǫ.

Therefore limt→0

∫ ∞−∞ |g(x)[ f (x) − f (x+ t)]| = 0.

�

4.2. EXERCISES 29

4.2.4 (a) Over what set the integration takes place? Withoutit, the equality might be false. I think

it should be overR, the real number line.

Why do you think you do not need a reason for this ”therefore” conclusion:

such that∣∣∣∫


∣∣∣ < ǫ, and therefore∣∣∣∫

f (x+ t)dx−∫ϕ(x+ t)dx

∣∣∣ < ǫ.(You presented the reason after the conclusion, without mentioning the relation between the two).

4.2.5 (Aug, 2005)

For f ∈ L1(R) and g∈ L∞(R), define

F(t) =∫

Rg(x) f (t − x)dx for t ∈ R.

Then F is uniformly continuous in R.

Proof: The condition thatg ∈ L∞(R) meansg is bounded except for a measure 0 set, so it is the same with

4.2.4.

4.2.6 (May, 2009)(Apr, 2005)(Jan, 2001)

Let f ∈ L1(R1) and define

F(t) =∫ ∞

−∞f (x)sin(xt)dx.

Prove F(t) is continuous in R. Is F(t) uniformly continuous in R? Prove your conclusion.

Proof: Since f ∈ L1(R1), for anyǫ > 0, there exists a step functionϕ(x) =∑n

i=1 aiχ(Ei) vanishes outside

[−N,N] such that∫ ∞−∞ | f (x) − ϕ(x)| < ǫ. Let a = max{a1, a2, . . . , an}.

Now consider sinxt. For anyǫ > 0, there exists aδ = ǫ4aN2 , such that for any|△t| < δ, we have

| sinxt − sin[x(t + △t)]| = |2 cos(xt+ xt + x△t) sin(xt − xt− x△t)|≤ 2| sin(−x△t)| = 2| − sin(x△t)|= 2| sin(x△t)| ≤ 2|x△t|= 2|x||△t| ≤ 2N ǫ

4aN2

=ǫ

2aN.

Then we have

|F(t) − F(t + △t)| = |∫ ∞−∞( f (x)sin(xt) − f (x)sin(x(t + △t)))dx|

= |∫ ∞−∞ f (x)[sin(xt) − sin(x(t + △t))]dx|

≤ |∫ ∞−∞[ f (x) − ϕ(x)][ sin(xt) − sin(x(t + △t))]dx| + |

∫ ∞−∞ ϕ(x)[sin(xt) − sin(x(t + △t))]dx|

≤∫ ∞−∞ | f (x) − ϕ(x)|dx+

∫ N

−N|ϕ(x)[sin(xt) − sin[x(t + △t)]] |dx

≤ ǫ +∫ N

−N|a ǫ

2aN |dx

≤ 2ǫ.

ThereforeF(t) is uniformly continuous onR.

�

4.2.6: You should indicate that youra depends on the given epsilon. Some of the inequalities are

not correct.


See a correction below (note that|[sin(xt) − sin(x(t + △t))]| ≤ |sin(xt)| + |sin(x(t + △t)) ≤ 2).

|F(t) − F(t + △t)| = |∫ ∞

−∞( f (x)sin(xt) − f (x)sin(x(t + △t)))dx|

= |∫ ∞

−∞f (x)[sin(xt) − sin(x(t + △t))]dx|

≤ |∫ ∞

−∞[ f (x) − ϕ(x)][ sin(xt) − sin(x(t + △t))]dx| + |

∫ ∞

−∞ϕ(x)[sin(xt) − sin(x(t + △t))]dx|

≤ 2∫ ∞

−∞| f (x) − ϕ(x)|dx+

∫ N

−N|ϕ(x)[sin(xt) − sin[x(t + △t)]] |dx

≤ 2ǫ +∫ N

−N|a ǫ

2aN|dx

≤ 3ǫ.

4.2.7 (Aug, 2004)(Apr, 2004)(May, 1996)

If f (x) is integrable over R, then

limλ→∞

∫

Rf (x) cos(λx)dx = 0.

Proof: Since f (x) is integrable overR, for anyǫ > 0, there exists a step functionϕ(x) =∑n

i=1 aiχ(Ei) defined

on [−M,M] and vanishes outside with that∫R| f (x) − ϕ(x)| < ǫ. Let a = max{a1, a2, . . . , an} andN =

[aǫ

].

Then for anyλ > N, we have∫R

f (x) cos(λx)dx =∫R[ f (x) − ϕ(x)] cos(λx)dx+

∫Rϕ(x) cos(λx)dx

≤∫R[ f (x) − ϕ(x)]dx+

∫ M

−M

∑ni=1 aiχ(Ei) cos(λx)dx

≤∫R| f (x) − ϕ(x)|dx+

∫ M

Macos(λx)dx

< ǫ + aλ

sinλx|M−M

≤ ǫ + 2aλ

≤ 3ǫ.

Therefore limλ→∞∫R

f (x) cos(λx)dx= 0.

�

4.2.8 (Apr, 2005)

(a) State Fatou’s Lemma.

(b) Show by an example that the strict inequality in Fatou’s Lemma is possible.

(c) Show that Fatou’s Lemma can be derived from the Monotone Convergence Theorem.

Proof: (a) See 4.1.6.

(b) Define fn = nχ([0, 1n]). Then for anyǫ = 1

k > 0, and anyx ∈ [ǫ, 1], let N = k, we have| fn(x) − 0| = 0

for any n > N. Sinceǫ is arbitrary, we havefn(x) → f (x) = 0 a.e. on [0, 1]. But∫

fn = 1 for any real

numbern, and∫

f = 0.

(c) Definehn(x) = inf i≥n{ fi(x)}. Thenhn is an increasing sequence of functions withhn(x) → f (x) a.e.

By the Monotone Convergence Theorem,∫

f = lim∫

hn ≤ lim∫

fn.

�

4.2. EXERCISES 31

4.2.9 (Aug, 2003)

Is the product of two integrable functions from R to R integrable?

Proof: It’s not. Let f (x) = x−12 with x ∈ (0, 1). Sincef (x) is measurable,f (x) ≥ 0 and

∫ 10 f (x)dx = 2, we

have f (x) is integrable. But∫ 10 f (x)2

=

∫ 10 x−1 >

∞∑k=1

1k = ∞, so the product is not an integrable function.

�

4.2.10 (Aug, 2003)

Show that

limn→∞

∫ 1

0

xn+ 1

xn + 2

exists and find its value.

Proof: Since∫ 10

xn+1

xn+2 =∫ 10

xn+2−1

xn+2 = 1 −∫ 10

1xn+2 and fn(x) = 1

xn+2 with x ∈ [0, 1] is an increasing se-

quence of nonnegative measurable function bounded by 1/2, by Monotone Convergence Theorem, we have

lim∫ 10

1xn+2 =

∫ 10 lim 1

xn+2 = 1/2. And so

limn→∞

∫ 1

0

xn+ 1

xn + 2= 1− lim

n→∞

∫ 1

0

1xn + 2

= 1/2.

�

4.2.11 (Aug, 2003)

Let { fn} be a sequence of measurable functions on[0, 1]. Describe the three concept of convergence stated

in (1)-(3) and give any implications between them. The implication must be proved. The lack of implication

must be supported by a counter example.

(1) fn→ 0 in measure as n→ ∞.

(2) fn→ 0 a.e. as n→ ∞.

(3) || fn||1→ 0 as n→ ∞.

Proof:

(1) fn → 0 in measure asn→ ∞ means for anyǫ > 0, there exists anN, such that for eachn > N, we

havem{x : | fn(x) − 0| ≥ ǫ} < ǫ.(2) fn→ 0 a.e. asn→ ∞means the set of points inE which is not converges to 0 has measure 0.

(3) || fn||1 → 0 asn → ∞ means for anyǫ > 0, there exists anN, such that for eachn > N, we have

||| fn(x)|| − 0| < ǫ.(1) ; (2)

Let n = k + 2m with 0 ≤ k < 2m andm = 1, 2, 3 · · · . Thenn = 1, 2, 3, · · · . Define fn(x) = 1 when

n = k+ 2m andx ∈ (k2−m, (k+ 1)2−m]. Otherwisefn(x) = 0.

Then for any 1> ǫ > 0, there exists 2−m0 < ǫ, let N = 2m0, then we havem{x : | fn(x) − 0| ≥ ǫ} < ǫ for

anyn > N. So fn→ 0 in measure.

But for any x ∈ (0, 1], let ǫ0 = 1/2, for anyN > 0, there exists 2m1 > N, such thatx is in one of the

interval (k2−m1, (k + 1)2−m1] with 0 ≤ k < 2m1. Assume that this interval is (k02−m1, (k0 + 1)2−m1] and so

| fk0+2m1 − 0| = 1 > ǫ0. Then fn 9 0 anywhere in (0, 1].


4.2.11

(1) ; (2): You need to show, by definition, thatm{ fn 9 0} > 0. In this example, you have

m{ fn 9 0} = 2−m→ 0.

Here is a question: how do we describe the setA = { fn 9 0}?x ∈ A iff there exists a1/m > 0, and for any N > 0, there exists ann ≥ N such that | fn(x)| ≥ 1/m.

This might suggests that:

Can we show that (you may need to modify it)

A = ∪∞m=1 ∪∞N=1 ∪n≥N{| fn(x)| ≥ 1/m}?

Then, ask the question (the same idea of thinking by definitions), can we show thatm(A) = 0? If

not, how do we construct a correct example from the proof? Remember, you need an example with

m(A) > 0.

Since the setA = { fnn→ 0} = {}(2)⇒ (1)

If fn→ 0 a.e., by 3.2.16, we have for anyǫ > 0 andδ > 0, there exists anN andA ⊂ [0, 1] with mA< δ,

such that for anyn > N andx ∈ A, we have| fn(x)| < ǫ. Let δ = ǫ. Then{x : | fn(x)| ≥ ǫ} ⊆ A, that means

m{x : | fn(x)| ≥ ǫ} < ǫ.(2); (3)

Define fn(x) = nχ([0, 1/n]), by 4.2.8, fn→ 0 a.e., but|| fn||1→ 1.

(3); (2)

We could use the example given in (1); (2). We have|| fn||1→ 0 but fn 9 0 a.e..

(1); (3)

We could use the example given in (2); (3). We havefn → 0 a.e. hencefn → 0 in measure, but

|| fn||1→ 1.

(3)⇒ (1)

By way of contradiction, assume thatfn 9 0 in measure, then there exists anǫ0 > 0, such that for any

N, there exists ann > N with m{x; | fn(x)| ≥ ǫ0} ≥ ǫ0. But then|| fn||1 ≥ ǫ20 > 0, contradicts to|| fn||1→ 0.

�

4.2.12 (Aug, 2003)

Suppose that f is a continuous function on[0, 1] for which

∫

[0,1]f (t)tndt = 0, ∀n = 1, 2, 3, . . .

Show that f is the zero function.

Proof: Since f is a continuous function on [0, 1], assume that| f | is bounded byM. And for anyǫ > 0,

there exists a polynomialp(x) such that| f (x) − p(x)| < ǫ/M for any x ∈ [0, 1]. And since∫[0,1] f (t)tndt = 0,

for ∀n = 1, 2, 3, . . . , we have∫

[0,1]f (x)p(x) = 0 =

∫

[0,1]f (x)[p(x) − f (x)] +

∫

[0,1]f (x)2.

4.2. EXERCISES 33

Since|∫[0,1] f (x)[p(x)− f (x)]| < ǫ, we have|

∫[0,1] f (x)2| < ǫ. Then f (x)2

= 0 a.e., and sof (x) = 0 a.e..Since

f is a continuous function withf (x) = 0 a.e.,f (x) is the zero function.

�

4.2.12: Need a reason why there exists such a poly. I told you this solution before. You need an

approximation theorem here:

And for any ǫ > 0, there exists a polynomialp(x) such that | f (x) − p(x)| < ǫ/M for any x ∈ [0, 1].

4.2.13 (Apr, 2002)

Let {gn} be a sequence of nonnegative integrable function which converges to an integrable function g. Let

fn be a sequence of measurable functions such that| fn| < gn and fn→ f a.e.. if

∫ 1

0g(x)dx = lim

∫ 1

0gn(x)dx,

then ∫ 1

0f (x)dx= lim

∫ 1

0fn(x)dx.

Proof: Since Let fn is a sequence of measurable functions with| fn| < gn, we have∫

fn ≤∫

gn < ∞, so fnis integrable. And sof is integrable since

∫f ≤ lim fn < ∞. And sogn − fn is nonnegative. By Fatou’s

Lemma, we have ∫g− f ≤ lim

∫gn − fn.

Then ∫g−

∫f ≤ lim

∫gn − lim

∫fn.

That is ∫f ≥ lim

∫fn.

Again by Fatou’s Lemma, sincefn + gn is nonnegative, we have

∫g+ f ≤ lim

∫gn + fn.

Then ∫f ≤ lim

∫fn.

So∫ 1

0 f (x)dx = lim∫ 1

0 fn(x)dx.

�

4.2.14 (Aug, 2001)

Let f be nonnegative and measurable on E, and En = {x ∈ E : f (x) ≥ n}. Show that if∑∞

n=1 n ·mEn < ∞,

then f is integrable on E, but the converse is not true.

Proof: Let f (x) = 1/2 with E = R, that ismE = ∞, thenEn = ∅ for all n, so∑∞

n=1 n ·mEn = 0 < ∞. But∫R

f = ∞, so f is not integrable onR.


If mE< ∞. Let E0 = E andFn = En \ En+1 with n = 0, 1, 2 . . . . ThenFn = {x ∈ E : n ≤ f (x) < n+ 1}.Then ∫

f ≤ mF0 + 2mF1 + 3mF2 + . . .

= mE−mE1 + 2(mE1 −mE2) + 3(mE2 −mE3) + . . .

= mE+mE1 +mE2 +mE3 + . . .

≤ mE+∑∞

n=1 n ·mEn

< ∞.

So f is integrable onE.

For the converse part, letf (x) = x−1/2. Then∫(0,1)

f (x)dx = 2, so f is integrable over (0, 1). But the set

En = (0, 1/n2) for n = 1, 2, 3 . . . . Then we have∑∞

n=1 n ·mEn =∑∞

n=1 1/n = ∞.

�

4.2.15 (Apr, 2001)

Let fn be a sequence of integrable functions over E such that0 ≤ fn+1 ≤ fn hold almost everywhere for each

n. Show that fn ↓ 0 holds almost everywhere if and only if∫

fndx ↓ 0.

Proof: ⇒: Since f1 is integrable,fn ↓ 0 a.e. and 0≤ fn+1 ≤ fn, by Lebesgue Convergence Theorem, we

have∫

fndx ↓ 0.

⇐:

For any fixedǫ > 0, let En = {x : | fn| < ǫ} with n = 1, 2, 3 · · · . Then we haveE1 ⊆ E2 ⊆ E3 . . . . If

mEn 9 mE, then there exists anǫ0 > 0, for anyN, there exists ann > N, such that

mE−mEn = m(E \ En) = m{x : | fn(x)| ≥ ǫ} > ǫ0,

then∫

Efn ≥ ǫ · ǫ0 > 0, which is a contradiction to

∫fndx ↓ 0.

SomEn→ mE. That means for anyδ > 0, there exists anN, such that for anyn > N, we have

mE−mEn = m{x : | fn(x)| ≥ ǫ} < δ.

Sinceǫ andδ are arbitrary, andE1 ⊆ E2 ⊆ E3 · · · , we have if for anyy < {x : | fN(x)| ≥ ǫ}, | fn(y)| < ǫ for any

n > N. And since the set{x : | fn(x)| ≥ ǫ} is smaller than anyδ, we havefn ↓ 0 a.e..

�

4.2.16 (Aug, 2000)

Suppose{ fn} is a sequence of nonnegative measurable functions, and∑∞

n=1 fn converges a.e. on E. Show

that ∫

E

∞∑

n=1

fn =∞∑

n=1

∫

Efn.

Proof: Let∑∞

n=1 fn converges tof a.e. onE. Since{ fn} is nonnegative and measurable, thengn =∑n

k=1 fnis a increasing sequence of nonnegative measurable function with gn → f a.e.. By Monotone Convergence

Theorem, we have

∫

E

∞∑

n=1

fn =∫

Ef = lim

∫

Egn = lim

∫

E

n∑

k=1

fn = limn∑

k=1

∫fn =

∞∑

k=1

∫fn.

4.2. EXERCISES 35

�

4.2.16: Need a reason whyf exists:

Since{ fn} is nonnegative and measurable, thengn =∑n

k=1 fn is a increasing sequence of nonnegative

measurable function with gn→ f a.e..

4.2.17 (Aug, 1997) Suppose{ fn} and f are measurable functions and fn→ f a.e. in E with mE< ∞. Show

that there exists a sequence of measurable sets{Ek}∞k=1 such that

∞⋃

k=0

Ek = E, mE0 = 0, and fn→ f uniformly on each Ek for k = 1, 2, · · · .

Proof: Since fn → f a.e., by Egoroff’s Theorem, we have forδk = 1/k, there exists a measurable set

Fk ⊆ E with mFk < δk such thatfn converges tof uniformly on Fk = Ek. Let E0 = ∩∞k=1Fk. ThenmE0 = 0

and

E =∞⋂

k=1

Fk ∪∞⋂

k=1

Fk = E0 ∪∞⋃

k=1

Fk =

∞⋃

k=0

Ek.

�

4.2.18 (May, 1996) Give one example of Lebesgue integrable function f(x) on [0, 1] which is not Riemann

integrable. Explain why.

Proof: Define f (x) =

1 x < [0, 1] ∩ Q

0 x ∈ [0, 1] ∩ Q. Then f (x) is discontinuous at every point in [0, 1], so it

is not Riemann integrable. But it is Lebesgue integrable, since f can write down as a simple function

f (x) = c1χ(E1) + c2χ(E2) with c1 = 1, c2 = 0, E1 = [0, 1] \ Q andE2 = [0, 1] ∩ Q, whereE1 andE2 are

measurable. So∫

f = c1mE1 + c2mE2 = 1 < ∞. Thereforef (x) is Lebesgue integrable.

�

4.2.19 (Aug, 2005)

If E is a measurable set with m(E) < ∞, and{ fn} is a sequence of measurable function on E, then

fn→ 0 in measure⇐⇒∫

E

| fn|1+ | fn|

→ 0.

Proof: =⇒: Since fn → 0 in measure, we have for anyǫ > 0, there exists anN, such that for anyn > N,

the setAn = {x : | fn(x) − 0| ≥ ǫ/mE} is with measure smaller thanǫ/mE. Now we have

∫E| fn|

1+| fn| =∫

An

| fn|1+| fn| +

∫E\An

| fn|1+| fn|

≤∫

An1+

∫E\An

ǫ/mE1+| fn|

≤∫

An1+

∫E\An

ǫ/mE

< 2ǫ.

So∫

E| fn|

1+| fn| → 0.


⇐=: By way of contradiction. Assume thatfn 9 0 in measure, then there exists anǫ0 > 0, for anyN,

there exists ann > N such that the setA = {x : | fn(x) − 0| ≥ ǫ0} is with measuremA≥ ǫ0. Then we have∫E| fn|

1+| fn| =∫

E\A| fn|

1+| fn| +∫

A| fn|

1+| fn| ≥∫

Aǫ0

1+| fn|???

4.2.19:⇐=:

Your using the definition is the right direction. What is missing is a little bit calculus:

The function h(x) = x/(1 + x) is an increasing function (compute the derivative to seeh′(x) =

1/(1 + x)2 > 0). So the proof goes like this: (I am using your wording. Also pay attention to English:

where did I change your wording?)

By way of contradiction. Assume that fn 9 0 in measure. Then (not then) there exists anǫ0 > 0,

such that for any N, there exists ann > N such that the setA = {x : | fn(x)| ≥ ǫ0} is with measure

mA≥ ǫ0.

Then we have∫

E| fn|

1+| fn| =∫

E\A| fn|

1+| fn| +∫

A| fn|

1+| fn| ≥∫

Aǫ0

1+ǫ0≥ 0. (you add the details).

4.2.20 (Apr, 2005)

Suppose f is a non-negative integrable function on[0, 1]. If

∫ 1

0f n=

∫ 1

0f for all n = 1, 2, . . . ,

then f(x) must be the characteristic function of some measurable set E⊂ [0, 1] except for a measure 0 set.

If there exists anǫ > 0, the set{x : f (x) > 1+ ǫ} has positive measure, thenf n → ∞ and so∫ 10 f n → ∞, a

contradiction to∫ 1

0 f n=

∫ 1

0 f and f is integrable.

Also, if the set{x : ǫ < f (x) < 1 − ǫ} has positive measure, thenf n → 0 and so∫ 1

0f n → 0. Then

contradicts to∫ 1

0f n=

∫ 1

0f for anyn = 1, 2, . . . .

Since f is a non-negative integrable function on [0, 1], it is measurable and only takes 0 and 1 except for

a measure 0 set, sof (x) must be the characteristic function of some measurable setE ⊂ [0, 1] except for a

measure 0 set.

�

New begin from Aug 1st

Chapter 5

Differentiation and Integration

5.1 Notes

Theorem 5.1.1 Let f be an increasing real-valued function defined in the interval [a, b]. Then f is differ-

entiable a.e., the derivative f′ is measurable, and∫ b

af ′(x)dx≤ f (b) − f (a).

Let f be a real-valued function defined on the interval [a, b], and leta = x0 < x1 < · · · < xk = b be any

subdivision of [a, b]. Define

p =k∑

i=1

[ f (xi) − f (xi−1)]+

t =k∑

i=1

[ f (xi) − f (xi−1)]−

t = p+ n =k∑

i=1

| f (xi) − f (xi−1)|,

where we user+ to denoter, if r ≥ 0 and 0, ifr < 0, and setr− = |r | − r+. We havef (b) − f (a) = p− n. Set

P = supp,

N = supn,

T = supt,

where we take the suprema over all possible subdivisions of [a, b]. We call P, N, T the positive negative,

and total variation off over [a, b]. We sometimes writeTba( f ) to denote the total variation off on [a, b]. If

T < ∞, we sayf is of bounded variation over [a, b] or abbreviated byf ∈ BV.

Lemma 5.1.2 If f is of bounded variation on[a, b], then

Tba = Pb

a + Nba

and

Pba − Tb

a = f (b) − f (a).

37

38 CHAPTER 5. DIFFERENTIATION AND INTEGRATION

Proof: Since for any subdivision, we havep = n + f (b) − f (a), taking suprema of both sides, we have

Pba − Nb

a = f (b) − f (a).

Sincet = p+ n = 2n+ f (b) − f (a), taking suprema of both sides, we haveTba = 2Nb

a + f (b) − f (a), and

by what we have proved,Nba = − f (b) + f (a) + Pb

a. So

Tba = 2Nb

a + f (b) − f (a) = Nba + f (b) − f (a) − f (b) + f (a) + Pb

a = Pba + Nb

a.

Theorem 5.1.3 A function f is of bounded variation on[a, b] if and only if it is the difference of two mono-

tone real-valued functions on[a, b].

Proof: ”⇒: ” Define g(x) = Pxa( f ) andh(x) = Nx

a( f ). Theng andh are monotone increasing functions on

[a, b]. By 5.1.2 and sincePxa + f (a) is a monotone increasing function, we havef (x) = Px

a − Nxa + f (a), the

difference of two monotone increasing functions.

”⇐: ” If f = g − h whereg and h are two monotone increasing functions on [a, b], then for any

subdivision, we have

∑| f (xi) − f (xi−1)| ≤

∑g(xi) − g(xi−1) +

∑h(xi ) + h(xi−1) = g(b) − g(a) + h(b) + h(a) < ∞.

SoTba < g(b) − g(a) + h(b) + h(a) < ∞, then f is of bounded variation.

Corollary 5.1.4 If f is of bounded variation on[a, b], then f′(x) exists a.e. in[a, b].

Theorem 5.1.5 Let f be an integrable function in[a, b], and suppose that

F(x) = F(a) +∫ x

af (t)dt.

Then F′(x) = f (x) a.e. in[a, b].

A real-valued functionf defined on [a, b] is said to be absolutely continuous on [a, b] if for any ǫ > 0,

there exists aδ > 0 such thatn∑

i=1

| f (x′i ) − f (xi)| < ǫ

for every finite collection{(x′i , xi)} of nonoverlapping intervals with

n∑

i=1

|x′i − xi | < δ.

Theorem 5.1.6 If f is absolutely continuous on[a, b], then it is of bounded variation on[a, b].

Theorem 5.1.7 If f is absolutely continuous on[a, b] and f′(x) = 0 a.e., then f is constant.

Theorem 5.1.8 A function F is an indefinite integral if and only if F is absolutely continuous.

5.2. EXERCISES 39

5.2 Exercises

5.2.1 (Aug, 2005)

Determine if the following is true or false:

If f is absolutely continuous in[0, 1], then f is of bounded variation in[0, 1].

Proof: By the definition of absolutely continuous, we have forǫ = 1, there exists aδ > 0, such that

n∑

i=1

| f (x′i ) − f (xi)| < 1

for every finite collection{(x′i , xi)} of nonoverlapping intervals with

n∑

i=1

|x′i − xi | < δ.

Then for any subdivision of [0, 1], we can split them (by adding new division point) intoK =[

1δ

]+ 1 sets of

intervals with each length smaller thanδ. SoT10( f ) ≤ K, f is of bounded variation on [0, 1].

�

5.2.2 (May, 2009)(Apr, 2005)

Let f(x) be monotone increasing on[0, 1] with f(0) = 0 and f(1) = 1. If the set{ f (x); x ∈ [0, 1]} is dense

in [0, 1], show that f is a continuous function on[0, 1]. Is it absolutely continuous on[0, 1]? Prove your

conclusion.

Proof: We have provedf is continuous in 2.1.25. Letf be the Cantor ternary function in [0, 1], then f

is monotone increasing and continuous. Alsof ′(x) = 0 a.e.. If it is absolutely continuous, by 5.1.8, we

have∫ 1

0f ′(x) = f (1) − f (0). But

∫ 1

0f ′(x) = 0 not equal tof (1) − f (0) = 1. So f may not be absolutely

continuous on [0, 1].

�

5.2.3 (Aug, 2004)(Apr, 2004)

Let f , g be two absolutely continuous functions on[0, 1]. Prove that f g is absolutely continuous on[0, 1].

Is it also true if the interval is replaced by(−∞,∞)? Justify your conclusion.

Proof: Since f , g are absolutely continuous [0, 1], it is continuous and so bounded. Assume thatf is

bounded byK, g is bounded byM. By the definition of absolutely continuous, we have for anyǫ, there

existsδ1 andδ2, such that

n∑

i=1

| f (x′i ) − f (xi)| < ǫ/Mn∑

i=1

|g(y′i ) − g(yi )|/K

for any nonoverlapping intervals{(x′i , xi)} and{(y′i , yi)} with

n∑

i=1

|x′i − xi | < δ1

n∑

i=1

|y′i − yi | < δ.


Then letδ = min{δ1, δ2}. We have

n∑i=1| f (x′i )g(x′i ) − f (xi)g(xi )| ≤

n∑i=1| f (x′i )g(x′i ) − f (x′i )g(xi)| +

n∑i=1| f (x′i )g(xi) − f (xi)g(xi)|

≤n∑

i=1K|g(x′i ) − g(xi)| +

n∑i=1

M| f (x′i ) − f (xi)|

< 2ǫ

for any nonoverlapping intervals{(x′i , xi)} withn∑

i=1|x′i − xi | < δ.

It is not true if the interval is replaced by (−∞,∞). Let f (x) = g(x) = x. Then for anyǫ > 0, letδ = ǫ,

for any finite nonoverlapping intervals{(x′i , xi)} with∑n

i=1 |x′i − xi | < δ, we have that∑n

i=1 | f (x′i ) − f (xi)| =∑ni=1 |x′i − xi | < ǫ.

But f (x)g(x) = x2 is not absolutely continuous. Since forǫ0 = 1 and anyδ > 0, there existsx ∈ R with

|1/x| < δ such that|(x+ 1/x)2 − x2| = 2+ 1/x2 > ǫ0.

�

5.2.4 (Aug,2004)

Let f(x) = xcosπx for 0 < x ≤ 1 and f(0) = 0.

(1) Is f continuous on[0, 1]?

(2) Is f uniformly continuous on[0, 1]?

(3) Is f absolutely continuous on[0, 1]?

Justify your conclusion.

Proof: (1) For a fixedx ∈ (0, 1] and anyǫ, let δ1 =12 x, δ2 =

x2πǫ, δ3 = ǫ andδ = min{δ1, δ2, δ3}, for any

y ∈ (x− δ, x+ δ), we have

| cosπx − cosπy | = |2 sin 1/2(πx +πy) sin 1/2(πx −

πy)|

≤ 2| sin 1/2(πx −πy)|

< 2|1/2(πx −πy)|

= π|x−y|

xy

≤ 2π |x−y|x2

< ǫx.

And so| f (x) − f (y)| = |xcosπx − ycosπy |

≤ |xcosπx − xcosπy | + |xcosπy − ycosπy |= |x|| cosπx − cosπy | + | cosπy ||x− y|< ǫ + ǫ.

Then f is continuous on [0, 1].

(2) Since [0, 1] is a closed bounded interval, we havef is also uniformly continuous on [0, 1].

(3) f is not absolutely continuous on [0, 1]. By 5.2.1, we want to provef < BV. Since∞∑

k=1

1k = ∞, for any

M > 0, there existsN > 0, such that for anyn > N, we haveN∑

k=1

1k > M. Let 0< 1/N < 1/(N − 1) < · · · < 1

5.2. EXERCISES 41

be a subdivision of [0, 1], then the total variation for this subdivision is

| f (0)− f (1/N)| + | f (1/N) − f (1/(N − 1)| + · · · + | f (1/2)− f (1)|> | f (1/N) − f (1/(N − 1))| + · · · + | f (1/2)− f (1)|= |1/N cos π

1/N − 1/(N − 1) cos π1/(N−1) | + |1/(N − 1) cos π

1/(N−1) − 1/(N − 2) cos π1/(N−2) |

+ · · · + |1/2 cos π1/2 − cosπ|

> 1/N + 1/(N − 1)+ · · · + 1

> M.

Then f < BV.

�

5.2.5 (Apr, 2002)

Prove that a function F is an indefinite integral⇐⇒ it is absolutely continuous.

⇒: By 4.1.9.

⇐: If f is absolutely continuous, it is also of bounded variation, then

F(x) = F1(x) − F2(x)

whereF1 andF2 are monotone increasing functions. ThenF′ = F′1 − F′2 exists a.e. and so∫

F′(x) =∫

F′1(x) − F′2(x) ≤∫

F1(x) + F2(x) ≤ F1(b) + F2(b) − F1(a) − F2(a) < ∞.

ThenF′ is integrable. LetG(x) =∫ x

aF′(t)dt, thenG is absolutely continuous and so isf = F −G.

???

5.2.6 (Aug, 2001)

Let f(x, y) be a bounded function on the unit square Q= (0, 1) × (0, 1). Suppose for each y, that f is a

measurable function of x. For each(x, y) ∈ Q, let the partial derivative∂ f∂x exists. Under the assumption

that ∂ f∂y is bounded in Q, prove that

ddy

∫ 1

0f (x, y)dx=

∫ 1

0

∂ f (x, y)∂y

dx.

Proof: Since ∫ 10 f (x, y+ h)dx−

∫ 10 f (x, y)dx

h=

∫ 1

0

f (x, y+ h) − f (x, y)h

dx,

andf (x, y+ h) − f (x, y)

h→ ∂ f (x, y)

∂yas h→ 0.

If ∂ f∂y is bounded, by bounded convergence theorem, we have

ddy

∫ 1

0f (x, y)dx=

∫ 1

0

∂ f (x, y)∂y

dx.

�


5.2.7 (Aug, 2001)(Aug, 1997)(May, 1996)

Prove:

(1) If f is absolutely continuous on[a, b], then for any set E⊂ [a, b] with mE= 0, there holds m( f (E)) =

0.

(2) For a continuous and increasing function f on[a, b], if m( f (E)) = 0 for every E in[a, b] with

mE= 0, then f is absolutely continuous on[a, b].

Proof: (1) SinceE ⊂ [a, b] with mE = 0, we have for anyδ > 0, there exists a open coveringO = ∪Oi

with mO < δ to coverE in [a, b] and andf (O) covers f (E). WLOG, Oi are disjoint. By the definition of

absolutely continuous, for anyǫ > 0, there exists aδ > 0, such that∑n

i=1 m f(Oi) < ǫ for any finiten with∑n

i=1 mOi < δ. So it also holds forn = ∞.

(2)???

5.2.8 (Jan, 2001)

(1) Prove that if f is absolutely continuous on[a, b], then f is continuous and of bounded variation on

[a, b].

(2) Let f(x) = xsin 1x for 0 < x ≤ 1 and f(0) = 0. Show that f is continuous on[0, 1] but not of bounded

variation on[0, 1].

Proof: (1) See 5.2.1.

(2) See 5.2.4.

5.2.9 (Aug, 2000)

Show that a function satisfying a Lipschitz condition on[a, b] is absolutely continuous. (Note: A function f

is said to satisfying a Lipschitz condition on[a, b] if there is a constant M such that| f (x) − f (y)| ≤ M|x− y|for all x, y in [a, b].)

Proof: If satisfying a Lipschitz condition on [a, b], then for anyǫ > 0, letδ = ǫ/M, we have

n∑

i=1

| f (x′i ) − f (xi)| ≤n∑

i=1

M|x′i − xi | < ǫ

for any nonoverlapping interval{(x′i , xi)} with∑n

i=1 |x′i − xi | < δ. Then f is absolutely continuous in [a, b].

�

5.2.10 (may, 1996)

Is there a monotone function on[0, 1] which is discontinuous at all rational points? Prove your conclusion.

Proof: Define f (x) = 1 whenx is a rational number in [0, 1], and be 0 otherwise. And letg(x) be the

total variation of f (x), that meansg(x) = Tx0 f (t). Theng(x) is a monotone function on [0, 1] which is

discontinuous at all rational points.

Here is a monotone functionf on [0, 1] which is discontinuous at every rational points.

Let p1, p2, ..., pn, ... be an infinite sequence of positive numbers such that∑∞

n=1 pn < ∞ is convergent

(for example, pn = 2−n), and let a1, a2, ..., an, ... be a sequence that listing all the rational numbers in

[0, 1].

5.2. EXERCISES 43

Define f : (−∞,∞) 7→ (−∞,∞) as follows: f (x) = 0 for all x < 0. For any x ≥ 0, let A(x) = {ai : ai ≤x}, and define f (x) =

∑ai∈A(x) pi . Sincepn > 0 for all n ≥ 1, f (x) is an increasing function. If x = an is a

rational number in [0, 1], then

limx→a+n

f (x) − limx→a−n

f (x) = pn > 0,

and so f (x) is not continuous atx = an.


Chapter 6

The Classical Banach Space

6.1 Notes

A measurable function defined on [0, 1] is said to belong to the spaceLp= Lp(E) if

∫E| f |p < ∞, 0 < p < ∞.

For a functionf ∈ Lp, we define

|| f ||p =(∫

E| f |p

)1/p

.

And we defineL∞ the space of all bounded measurable functions on [0, 1], and

|| f ||∞ = inf {M : m(t : f (t) > M) = 0}.

Two important inequalities:

If f andg are inLp with 1 ≤ p <≤ ∞, then so isf + g and

|| f + g||p ≤ || f ||p + ||g||p.

If p andq are nonnegative extended real numbers with that1p +

1q = 1, and if f ∈ Lp andg ∈ Lq, then

f · g ∈ L1 and ∫| f g| ≤ || f ||p||g||q.

A sequence{ fn} in a normed linear linear space is said to converge to an element f in the space if given

ǫ > 0, there is anN such that for alln > N, we have|| fn − f || < ǫ. And we denoted it byfn→ f .

A normed linear space is calledcompleteif every Cauchy sequence{ fn} in the space, there is an element

f such thatfn→ f . A complete normed linear space is calledBanach space.

6.2 Excercises

6.2.1 (Aug, 2004)

Let { fn} be a sequence of real Lebesgue measurable functions on[0, 1]. If for any real g(x) ∈ L2[0, 1], the

sequence of real numbers ∫ 1

0g(x) fn(x)dx

converges. Does fn(x) converges to a function f(x) in L2[0, 1]? Justify your conclusion.

45

46 CHAPTER 6. THE CLASSICAL BANACH SPACE

Proof: Define fn(x) =√

nχ[0,1/n] . Then for anyg(x) ∈ L2[0, 1], we have for anyǫ andδ, there exists a

A ⊂ [0, 1] with mA < δ such that∫

A|g|2 < ǫ2. And there exists anN, such that for anyn > N, we have

1/n < δ, then ∫ 1

0g(x) fn(x)dx =

∫ 1/n

0g(x) fn(x)dx

≤(∫ 1/n

0 |g(x)|2)1/2 (∫ 1/n

0 | f(x)|2)1/2

< ǫ.

Now suppose thatfn→ f in L2, then for anyǫ, there exists anN such that for anyn ≥ N, we have

|| fN(x) − f ||2 <=∫ 1

0| fN − f |2 =

∫ 1/N

0| fN − f |2 +

∫ 1

1/N| f |2 < ǫ.

So∫ 1/N0 | fN − f |2 < ǫ, that meansfN → f a.e. inR. But for n = 2N, we can also get thatf2N → f which is

a contradiction. So suchf does not exist.

6.2.2 (Apr, 2004)

Let f ∈ L∞[0, 1], show that

limn→∞|| f ||Ln[0,1] = || f ||L∞[0,1].

Proof: Assume that|| f ||L∞[0,1] = M, then f is bounded byM except on a measure 0 setE. Then

limn→∞ || f ||Ln[0,1] = limn→∞(|| f ||Ln[E] + || f ||Ln[E])

= 0+ limn→∞ || f ||Ln[E]

≤ (∫ 10 |M|

n)1/n

≤ M.

For anyǫ, the setE = {x : | f (x)| > M − ǫ} has measure greatera > 0. Then we have(∫ 1

0| f |n

)1/n=

(∫E| f |n +

∫E| f |n

)1/n

≥(∫

E| f |n

)1/n

≥ ((M − ǫ)nmE)1/n

= (M − ǫ)mE1/n.

Let n→ 0, we have

limn→∞|| f ||Ln[0,1] ≥ M − ǫ.

So limn→∞ || f ||Ln[0,1] = || f ||L∞[0,1].

6.2.3 (Aug, 2000)

Suppose{ fn} and f are function in Lp[0, 1](p ≥ 1), and fn → f a.e.. Show that{ fn} converges to f in Lp

⇐⇒ || fn||p→ || f ||p.

Proof: ⇒: If { fn} converges tof in Lp, then for anyepsilon, there exists anN such that for anyn > N, we

have|| f − fn||p < ǫ. Then∣∣∣|| f ||p − || fn||p

∣∣∣ ≤∣∣∣|| f − fn + fn||p − || fn||p

∣∣∣ ≤∣∣∣|| f − fn||p + || fn||p − || fn||p

∣∣∣ = || f − fn||p < ǫ.

⇐: ???

6.2. EXCERCISES 47

6.2.4 (Aug, 1997)

Use the Holder inequality to establish the generalized Holder inequality: Let pi > 1 with∑m

i=11pi= 1. Then

||m∏

i=1fi ||1 ≤

m∏i=1|| fi ||pi for any fi ∈ Lpi (0, 1).

Proof: We prove it by induction. Since it is true forn = 2, suppose it is true for anyn ≤ k, for n = k+ 1, we

have∑k+1

i=11pi=

∑ki=1

1pi+

1pk+1= 1. Let

∑ki=1

1pi=

1t , then

||k∏

i=1

fi ||1 · || fk+1||1 ≤ ||k∏

i=1

fi ||t · || fk+1||pk+1.

Since∑k

i=1tpi=

∑ki=1

1pi/t= 1, we have

||k∏

i=1

fi ||t =

k∏

i=1

∫ 1

0| fi |t

1/t

≤

k∏

i=1

|| f ti ||pi/t

1/t

=

k∏

i=1

|| fi ||pi .

So ||m∏

i=1fi ||1 ≤

k∏i=1|| fi ||pi · || fi ||pk+1 =

k+1∏i=1|| fi ||pi .

6.2.5 (Aug, 2001)

For f ∈ Lp[a, b](p > 1), set f= 0 outside of[a, b] and define

fh(x) =12h

∫ x+h

x−hf (t)dt for h> 0.

Show that

|| fh||p ≤ || f ||p and limh→0+

|| fh − f ||p = 0.

(Note: You can use the fact, without giving its proof, that for integrable φ, there holds∫ b

a|φh(x)|dx ≤∫ b

a|φ(x)|dx.)

Proof: Sincep > 1, there existsq ≥ 1 such that 1/p+ 1/q = 1. Then

|| fh||pp =∫ b

a

∣∣∣∣∣∣12h

∫ x+h

x−hf (t)dt

∣∣∣∣∣∣

p

dx≤∫ b

a

∣∣∣∣∣12h

(2h)1/q|| f (t)||p∣∣∣∣∣p

dx=∫ b

a

(∫ x+h

x−h| f (t)|dt

)dx

6.2.6 (Apr, 2001)

(1) Assume that1 ≤ p < q ≤ ∞. If f is in Lq[0, 1], show that|| f ||p ≤ || f ||q.

(2) Let f be in L2[0, 1]. Show thatlim p→1+ || f ||p = || f ||1.

(3) Let f be a bounded measurable function on[0, 1]. Show thatlim p→∞ || f ||p = || f ||∞.

6.2.7 (May, 1996)

Let{ fn} be a sequence of functions in L2(0, 1), which converges almost everywhere to0 and satisfies|| fn||L2 ≤1. Can we conclude that∀g ∈ L2(0, 1),

limn→∞

∫g(x) fn(x)dx= 0?

Prove your conclusion.

Proof: Use the example offn in 6.2.1, and letg(x) = 1 in (0, 1). Then fn→ 0 a.e., but

limn→∞

∫g(x) fn(x)dx= 1.

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