Analisis Real Solutions

8/9/2019 Analisis Real Solutions

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2003

REAL ANALYSIS Spring 2003

SOLUTIONS TO SOME PROBLEMS

Warning:These solutions may contain errors!!

PREPARED BY SULEYMAN ULUSOY

PROBLEM 1. Suppose f n : X → [0, ∞] is measurable for n = 1, 2, 3,...;f 1 ≥ f 2 ≥ f 3 ≥ ... ≥ 0; f n(x) → f (x) as n → ∞, for every x ∈ X .a)Give a counterexample to show that we do not have generally the following result.limn→∞

X

f ndµ = X

f dµ.b) Without changing any other assumptions just add one more assumption and provethat the conclusion is satisfied in this case.

SOLUTION.

a)The standard example is the following.Let f n = 1[n,∞) for n = 1, 2, 3,...,where 1X represents the characteristic function of the set X .Then one can easily show that f = 0

but limn→∞ X f ndµ does not exist.

Consider f n = 1n

1[n,∞) for n = 1, 2, 3,...,where 1X represents the characteristic functionof the set X . Then one can easily checks that f = 0 but limn→∞

X f ndµ does not exist.

Here is another example:In this example we give a counterexample to the case where wehave strict inequality.Let f = 0.Let f n =

1n

if −∞ < x ≤ n and f n = 12 + 1n if n < x f 2 > f 3 > ... > 0 we have gn+1 > gn > 0for all n = 1, 2, 3, .....Also gn(x) → f 1(x) − f (x) as n → ∞.Now we apply the MonotoneConvergence Theorem to the sequence gn and get limn→∞ X gndµ = X limn→∞ gndµ.But this means that

X f 1dµ−limn→∞

f ndµ =

f 1dµ− f dµ.Now since f 1 ∈ L1(X, µ)

we can delete the term

f 1dµ from both sides and get limn→∞ X

f ndµ = X

f dµ.

PROBLEM 2. Suppose µ(X )


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SOLUTION. We first establish that the function f is integrable.For this note that wecan choose N so large that |f N − f | < /2 and for any given > 0.Then we have thefollowing estimate : X

|f |dµ ≤ X

|f N − f |dµ + X

|f N |dµ ≤ (/2)µ(X ) + M µ(X ) (*) where M 0such that

E

|f |dµ < whenever µ(E ) < δ .SOLUTION.Define f n = |f | ∧ n. i.e. f n = |f |, if |f | ≤ n and f n = n ,if |f | > n.Thenf n → |f | as n → ∞. Therefore we can use Monotone Convergence Theorem as f n+1 ≥ f nand f n ≥ 0.Now choose N so that

E

|f − f N | < /2 for given > 0.Then we have thefollowing estimate. E

|f |dµ < E |f − f N |dµ + E |f N |dµ


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1.)µ(∅) = 02.)Countable additivity.i.e.If E 1, E 2, E 3,... is contable collection of sets each of which isin M then µ(∪E i) = Σµ(E i).The first assertion is obvious as ∅ is countable.So by definition µ(∅) = 0.The interesting part is the second assertion.Suppose first that each of E i is countable. In

this case the union of these sets is also countable.So we have µ(∪E i) = 0 by definition.Onthe other hand since each E i is countable µ(E i) = 0 for all i.Thus the second assertionabove holds in this case.Now suppose that ∃ k such that E k is uncountable.Then E ck is countable and by the DeMorgan’s Law used above we again have (∪E i)c ⊂ E ck which shows that µ(∪E i) = 1.If we consider the summation Σµ(E i) we see that it is equal to 1 since the only term thatis nonzero(1) is µ(E k).So we again have the validity of the second assertion.Thus µ defined above is a measure in M .

PROBLEM 5. Let E k be a sequence of measurable sets in X , such that

∞k=1

µ(E k)


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PROBLEM 6. Find a sequence (f n) of Borel measurable functions on R which decresesuniformly to 0 on R , but

f ndm = ∞ for all n.Also,find a sequence (gn) of Borel

measurable functions on [0, 1] such that gn → 0 pointwise but

gndm = 1 for all n.(here m is the Lebesgue measure!)

SOLUTION.For the first part f n = 1/n1[1,∞) works(easy Calculus exercise!!).For thesecond part one sees that gn = n1[0,1/n] satisfies all the assertions(again this is easy toverify).

PROBLEM 7. Show that Monotone Convergence Theorem can be proved as a corollaryof the Fatou’s lemma.

SOLUTION. Applly the Fatou’s lemma to the following sequences (f +f n) and (f −f n).Since f n ↑ f both of these sequences are non-negative. Hence, application of the Fatou’slemma to the sequence (f + f n) gives liminf

f n ≥

f . And application of Fatou’s

lemma to the sequence (f − f n) gives limsup

f n ≤

f .Combination of these twoinequalities proves the Monotone Convergence Theorem.

PROBLEM 8. Let f ∈ L+ and f 0} is σ−finite.SOLUTION.This follows from the following equality.

∪∞n=1{x : f (x) > 1/n} = {x : f (x) > 0}.Each of the sets on the left hand side of this equality is of finite measure , since otherwisef would not have finite integreal.And this shows that the set in question is the union of sets of finite measure.

PROBLEM 9.

a.)If f is nonnegative and integrable on A, then show that

µ({x : x ∈ A, f (x) ≥ c}) ≤ 1/c A

f (x)dµ

b.)If A

|f (x)|dµ = 0, prove that f (x) = 0 a.e.SOLUTION.

a.)This is known as the Chebyshev’s inequality.If A1 = {x : x ∈ A, f (x) ≥ c}, then

A

f (x)dµ = A1

f (x)dµ + A−A1

f (x)dµ

≥ A1f (x)dµ

≥ cµ(A1).

This proves the result.b.)By the Chebyshev’s inequality,

µ({x : x ∈ A, f (x) ≥ 1/n}) ≤ n A

f (x)dµ = 0, ∀n = 1, 2,....

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Therefore,

µ({x : x ∈ A, f (x) = 0}) ≤∞n=1

µ({x : x ∈ A, |f (x)| ≥ 1/n}) = 0.

And this clearly proves the desired result.PROBLEM 10.

a.)Consider a measure space (X, µ) with a finite,positive,finitely additive measure µ.Prove that µ is countably additive if and only if it satisfies the following condition.If An is a decresing sequence of sets with empty intersection then

limn→∞

µ(An) = 0.

b.)Now suppose that X is locally compact Hausdorff space, that Br is the Borel σ−algebra,and that µ is finite, positive, finitely additive measure on Br.Suppose moreover that µ

is regular, that is for each B ∈ Br we have,µ(B) = sup

K {µ(K ) : K ⊆ B, K − compact}

Prove that µ is countably additive.

SOLUTION.

a.)Sufficency:Let (Bn) be countably many measurable sets which are mutually disjoint.Let An = ∪∞i=n+1Bi.Then ∩∞n=1An = ∅.We have

µ(∪∞n=1Bn) = limn→∞

{µ(∪ni=1Bi) + µ(∪∞i=n+1Bi)}

= limn→∞

{∞i=1

µ(Bi) + µ(An)} =∞i=1

µ(Bi).

Therefore µ is a measure.The necessity is obvious.b.) If µ is not countably additive ,by a.) there is a decreasing sequence (An) of measurablesets with empty intersection such that

limn→∞

µ(An) = inf µ(An) > 0.

For each n there exists K n contained in An, such that

µ(An) < µ(K n) + 1/2n+1 inf

iµ(Ai).

Then

µ(An − ∩ni=1K i) ≤n

i=1

µ(Ai − K i) < 1/2inf i

µ(Ai)

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which implies that µ(∩ni=1K i) = 0 and therefore ∩ni=1K i = ∅.Thus {∩ni=1K i : n ∈ N } is a decreasing sequence of nonempty compact subsets in thecompact space K 1.So ∩∞i=1K i = ∅,which contradicts the fact that ∩∞n=1An = ∅.PROBLEM 11. Let λ be Lebesgue measure on R. Show that for any Lebesguemeasurable set E

⊂ R with λ(E ) = 1, there is a Lebesgue measurable set A

⊂ E with

λ(A) = 1/2.

SOLUTION. Define the function f : R → [0, 1] by f (x) = λ(E ∩ (−∞, x]), wherex ∈ R.It is continuous by the following inequality

|f (x) − f (y)| ≤ |x − y|,wherex, y ∈ R.Since limx→−∞ f (x) = 0 and limx→∞ f (x) = 1, there is a point x0 ∈ Rsuch that f (x0) = 1/2.Put A = E ∩ (−∞, x0].PROBLEM 12. Let m be a countably additive measure defined for all sets in aσ−

algebra M .a.)If A and B are two sets in M with A ⊂ B, then show that m(A) ≤ m(B).b.)Let (Bn) be any sequence of sets in M .Then show that m(∪∞n=1Bn) ≤

∞n=1 m(Bn).

SOLUTION.These are almost trivial statements.a.)We have B = A ∪ (B − A) and using countable additivity of m by taking other setsto be empty gives, m(B) = m(A) + m(B − A).But m(B − A) ≥ 0. So the result follows.b.)By set theory we can find a mutually disjoint sequence ( An) such that ∪∞n=1An =∪∞n=1Bn.So we have,

m(∪∞n=1Bn) = m(∪∞n=1An) =∞

n=1

m(An) ≤∞

n=1

m(Bn),

where the second equality follows from the countable additivity of m and the last in-equlity follows from the fact that each term in the sum on the left is less than or equalto the corresponding term on the right.i.e.An ⊆ Bn, ∀n hence by part a.), m(An) ≤m(Bn), ∀n. So the result follows.PROBLEM 13. a.)Let (E n) be an infinite decreasing sequence of Lebesgue measurablesets, that is, a sequence with E n+1 ⊂ E nfor each n. Let m(E 1) be finite, where m is theLebesgue meausre.Then show that m(∩∞i=1E i) = limn→∞ m(E n).b.)Show by acounterexample that we can not omit the condition m(E 1) is finite.

SOLUTION.a.) Let E = ∩∞i=1E i, and let F i = E i − E i+1.Then E 1 − E = ∪∞i=1F i, and the sets F i arepairwise disjoint.Hence,

m(E 1 − E ) =∞i=1

m(F i) =∞i=1

m(E i − E i+1).

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But m(E 1) = m(E 1) + m(E 1 − E ), and m(E i) = m(E i+1) + m(E i − E i+1),since E ⊂ E 1and E i+1 ⊂ E i. Since m(E i) ≤ m(E 1)


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and m(U ) < m(A ∩ (n, n + 1)) + ,where < m(A ∩ (n, n + 1)).There are at most countably many disjoint intervals (a j , b j)

s such that U = ∪ j(a j, b j).Then A ∩ (n, n + 1) = ∪A ∩ (a j , b j).We have

m(A∩(n, n+1)) = j

m(A∩(a j, b j) ≤ j

(b j−a j)/2 = 1/2m(U ) < 1/2(m(A∩(n, n+1))+)

which gives m(A ∩ (n, n + 1)) < , a contradiction.PROBLEM 17. Choose 0 < λ 0. An ⊂ [−1, 2].Thus,

∪∞i=1An ⊂ [−1, 2].Suppose that An ∩ Am = ∅ if n = m. Then

∞

n=1

m(An

) ≤

m([−

1, 2]) = 3,

which contradics m(A) > 0. Therefore there must be some n, m such that An ∩ Am = ∅.Take z ∈ An ∩ Am. Then we can find x, x ∈ A such that

z = x + rn = x + rm.

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Thus x − x = rm − rn.PROBLEM 19. Let f : Rn → R be an arbitrary function having the property that foreach > 0, there is an open set U with λ(U ) < such that f is continuous on Rn − U (inthe relative topology).Prove that f is measurable.

SOLUTION.Let U k be an open set such that λ(U k) < 1/k and f is continuous onRn − U k .Let f k = f 1Rn−U k (where 1A represents the characteristic function of the setA), thenf k is measurable.For any > 0,

m∗({x : |f k − f |(x) ≥ }) = m∗({x ∈ U k : |f (x)| ≥ }) ≤ 1/k.It follows that (f k) converges to f in measure.Since Lebesgue measure is complete f ismeasurable.

PROBLEM 20. Prove or disprove that composition of two Lebesgue integrable func-tions with compact support f , g : R → R is still integable.

SOLUTION.It is not true.For example, letf (x) = 1{0}(x) and g(x) = 1{0,1}(x), where1A represents the characteristic function of the set A.Then f and g are integrable func-tions with compact support. But,since g ◦ f ≡ 1, the function g ◦ f is not integrable.PROBLEM 21. Let (X,M,µ) be a positive measure space with µ(X )


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≤∞

m=0

(m + 1)µ({x : m ≤ f (x) < m + 1}) =∞n=0

µ({x : f (x) ≥ n}) 0

andlimn→∞

(Φ(t)et/n)|t=0 = limn→∞

(et/n/n)|t=0 = 0.

PROBLEM 23. Let f n be a sequence of real-valued functions in L1(R) and suppose

that for some f ∈ L1(R) ∞−∞

|f n(t) − f (t)|dt ≤ 1/n2, n ≥ 1.

Prove that f n → f almost everywhere with respect to Lebesgue measure.SOLUTION. Since

supn

n

k=1 |

f k+1−

f k|dt

≤

∞

k=1

(1/(k + 1)2 + 1/k2) < ∞

by Levi’s theorem there is a measurable set E of measure 0 such that for any t ∈ R − E ,

supn

N k=1

|f k+1 − f k|(t)


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Therefore for any t ∈ R − E,

f n(t) = f 1(t) +n

k=2

(f k − f k−1)(t)

converges. It follows that f n → f almost everywhere.PROBLEM 24. Let m denote the Lebesgue measure on [0, 1] and let (f n) be a sequencein L1(m) and h a non-negative element of L1(m).Suppose thati.)

f ngdm → 0 for each g ∈ C ([0, 1]) andii.) |f n| ≤ h for all n.Show that

A f ndm → 0 for each Borel subset A ⊂ [0, 1].

SOLUTION.For any > 0, there is a δ > 0 such that E

hdm < , whenever m(E ) < δ.For such a δ there are a compact set K and an open set U such that (1)K ⊆ A ⊆ U and (2)m(U − K ) < δ. There is a continuous function g : [0, 1] → R such that(3)0

≤ g

≤ 1, (4)g = 1 on K and (5)g = 0 outside U. Then we have

lim supn→∞

A

f ndm = lim supn→∞

| 10

f n1Adm|

≤ limsupn→∞

| A

f ngdm| + | 10

f n(1A − g)dm|

≤ limsupn→∞

| A

f ngdm| + | 10

h1U −K dm| ≤

It follows that limn→∞

Af ndm = 0.

NOTE : Here 1B

represents the characteristic function of the set B.

PROBLEM 25.

a.)Prove the Lebesgue Dominated Convergence Theorem.b.)Here is a version of Lebesgue Dominated Convergence Theorem which is some kindof extension of it.Prove this.Let (gn) be asequence of integrable functions which converges a.e. to an integarblefunction g.Let (f n) be asequence of measurable functions such that |f n| ≤ gn and (f n)converges to f a.e. If

g = lim

gn, then

f = lim

f n.

c.) Show that under hypotheses of the part b.) we have |f n − f | → 0 as n → ∞.

d.)Let (f n) be asequence of integrable functions such that f n → f a.e. with f is inte-grable.Then show that |f − f n| → 0 as n → ∞ if and only if |f n| → |f | as n → ∞.SOLUTION.

a.) First let us state the theorem properly.Let g be integrable over E and let (f n) be a sequence of measurable functions such that

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|f n| ≤ g on E and for almost all x in E we have f (x) = lim f n(x). Then E

f = lim

E

f n.

Proof :The function g − f n is nonnegative so by Fatou’s Lemma we have E

(g − f n) ≤ lim inf E

(g − f n).

Since |f | ≤ g, f is integrable, and we have E

g − E

f ≤ E

g − limsup E

f n,

whence

E

f ≥

lim sup E

f n.

Similarly, considering g + f n , we get E

f ≤ lim inf E

f n,

and this completes the proof.b.) We will try to use the same idea as in the above proof. Take hn := gn − f n, bynoting that hn ≥ 0 and kn := gn + f n.Applying the Foatou’s lemma to these sequenceswe get the follwing inequalities combination of which proves the result.Application of Fatou’s lemma to hn gives lim sup f n ≤

f and application of the

Fatou’s lemma to kn gives lim inf

f n ≥ f.c.)Take f = 0, gn = |f n| + |f | and g = 2|f | in part b.).Note that now our sequence is|f n − f | which tends to 0 as n → ∞.d.)Suppose

|f −f n| → 0 as n → ∞.We have ||f n|−|f || ≤ |f n−f | and this immediatelygives

|f n| → |f | as n → ∞.Conversely, suppose that

|f n| → |f | as n → ∞.We will use part b.).Take gn =2(|f n| + |f |) and note that gn → 4|f | and note also that |f | is integrable by Fatou’slemma.Now the result follows from part b.) by taking f n = |f n − f | + |f n| − |f |.PROBLEM 26. Evaluate

limn→∞ n0

(1 + x/n)ne−2xdx,

justifying any interchange of limits you use.

SOLUTION. W know that limn→∞(1+x/n)n = ex and (1+x/n)n ≤ (1 +x/(n+1))n+1.

Also we have (1 + x/n)n ≤ ex. Therefore we get (1 + x/n)n ex, which gives that

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(1 + x/n)ne−2x ≤ e−x.Therefore we can apply the Dominated Convergence Theorem tothe function (1 + x/n)nex with the dominating function e−x.An easy computation gives,

limn→∞

n0

(1 + x/n)ne−2xdx = limn→∞

∞0

1[0,n](x)(1 + x/n)ne−2xdx

= ∞0

limn→∞

1[0,n](x)(1 + x/n)ne−2xdx = ∞0

e−xdx = 1.

PROBLEM 27.

a.) Let (an) be a sequence of nonnegative real numbers. Set µ(∅) = 0, and for everynonempty subset A of N (set of natural numbers) set µ(A) =

n∈A an.Show that the

set function µ : P (N ) → [0, ∞] is a measure.b.) Let X be a nonempty set and let f : X → [0, ∞] be a function.Define µ byµ(A) =

a∈A f (x) if A = ∅ and is at most countable, µ(A) = ∞ if A is uncountable,

and µ(∅) = 0.Show that µ is a measure.SOLUTION.

a.) If (An) is a sequence of pairwise disjoint subsets of N and A = ∪∞n=1An, then notethat

µ(A) =k∈An

ak =∞n=1

(k∈An

ak) =∞n=1

µ(An).

This clearly shows that µ is a measure.b.) We need to show the σ−additivity of µ. For that let (An) be asequence of pairwisedisjoint sequence of subsets of X . Set A = ∪∞n=1An. If some An is uncountable then A islikewise uncountable , and hence , in this case µ(A) =

∞n=1 µ(An) = ∞ holds. On the

other hand, if each An is at most countable then A is also at most countable, and so

µ(A) =x∈A

f (x) =

∞n=1

[a∈An

f (x)] =

∞n=1

µ(An)

also holds.Therefore µ is σ−additive and hence it is a measure.PROBLEM 28. Let F be a nonempty collection of subsets of a set X and let f : F →[0, ∞] be a function. Define µ : P (X ) → [0, ∞] by µ(∅) = 0 and

µ(A) = inf {∞n=1

f (An) : (An) ⊆ F, and, A ⊆ ∪∞n=1An}

for each A = ∅, with inf ∅ = ∞. Show that µ is an outer measure.SOLUTION.(1) By definition we have µ(∅) = 0.(2) (Monotonicity) : Let A ⊆ B and let (An) be a sequence in F with B ⊆ ∪∞n=1An.Then A ⊆ ∪∞n=1An, and so µ(A) ≤

∞n=1 f (An). Therefore

µ(A) ≤ inf {∞n=1

f (An) : (An) ⊆ F, and,B ⊆ ∪∞n=1An} = µ(B).

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If there is no sequence (An) with B ⊆ ∪∞n=1An, then µ(B) = ∞, and clearly µ(A) ≤ µ(B).(3) (Subadditivity) : Let (E n) be a sequence of subsets of X and let E = ∪∞n=1E n.If ∞

n=1 µ(E n) = ∞, then µ(E ) ≤∞

n=1 µ(E n) is obvously ture. So, assume∞

n=1 µ(E n) <∞ and let > 0. For each n pick a sequence (Ank) of F with E n ⊆ ∪∞k=1Ank and

∞k=1

f (Ank) < µ(E n) + /2n.

Clearly, E ⊆ ∪∞n=1 ∪∞k=1 Ank holds, and so

µ(E ) ≤∞n=1

∞k=1

f (Ank) <∞n=1

[µ(E n) + /2n] =

∞n=1

µ(E n) + .

Since > 0 is arbitrary , it follows that

µ(E ) = µ(∪∞n=1E n) ≤∞

n=1

µ(E n).

Therefore µ is an outermeasure.

PROBLEM 29. Let f : R → R be a Lebesgue integrable function.Show that

limt→∞

f (x)cos(xt)dλ(x) = lim

t→∞

f (x)sin(xt)dλ(x) = 0

SOLUTION. This is known as the Riemann-Lebesgue lemma. Since simple functionsare dense in integrable functions, it suffices to prove the result for the special casef = 1[a,b) where 1B represents the characteristic function of the set B . So, let f = 1[a,b)where

−∞ < a < b <

∞. In this case, for each t > 0 we have

|

f (x)cos(xt)dλ(x)| = | ba

cos(xt)dx| = |sin(xt)/t|x=bx=a| = |{sin(bt)−sin(at)}/t| ≤ 2/t,

and so limt→∞

f (x)cos(xt)dλ(x) = 0 holds. In a similar fashion, we can show thatlimt→∞

f (x)sin(xt)dλ(x) = 0.

PROBLEM 30. For a sequence (An) of subsets of a set X definelim inf An = ∪∞n=1 ∩∞i=n Ai and lim sup An = ∩∞n=1 ∪∞i=n AiNow let (X,S,µ) be a measure space and let (E n) be a sequence of measurable sets.Showthe following:a.) µ(lim inf E

n) ≤

liminf µ(E n

)b.) If µ(∪∞n=1E n)


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b.) Note that ∪∞i=nE i lim sup E n. Hence,since µ(∪∞n=1E n)


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It follows that E

f dµ +

E c

f dµ = lim inf

E

f ndµ + lim inf

E c

f ndµ,

and from (*) and (**), we see that

lim inf

E

f ndµ =

E

fdµ.

Now let (gn) be a subsequence of (f n). Then ,gn → f a.e. and limn→∞

gndµ =

fdµ. By the above result, we infer that

lim inf

E

gndµ =

E

fdµ,

and so there exists a subsequence (gkn) of the sequence (gk) such that lim E

gkndµ =

E

f dµ. In other words, we have shown that every subsequence of a sequence of realnumbers (

E f ndµ) has a convergent subsequence converging to

E f dµ.This means that

limn→∞ E f ndµ =

E f dµ holds.

PROBLEM 33. Let f : [0, ∞) → R be a continuous function such that f (x+1) = f (x)holds for all x ≥ 0. If g : [0, 1] → R is an arbitrary continuous function, then show that

limn→∞

10

g(x)f (nx)dx = (

10

g(x)dx)(

10

f (x)dx).

SOLUTION. Observe that by induction f (x + k) = f (x) holds for all x ≥ 0 and forall nonnegative integers k.

The change of variables u = nx yields

10

g(x)f (nx)dx = 1/n

n0

g(u/n)f (u)du = 1/nn

i=1

ii−1

g(u/n)f (u)du.

Letting t = u − i + 1, we get ii−1

g(u/n)f (u)du =

10

g((t + i − 1)/n)f (t + i − 1)dt = 10

g((t + i − 1)/n)f (t)dt.

Consequently,

10

g(x)f (nx)dx =

10

[n

i=1

1/ng((t + i − 1)/n)]f (t)dt = 10

hn(t)dt(∗)

where hn(t) = [n

i=1 1/ng((t + i − 1)/n)]f (t) . Clearly, hn is a continuous functiondefined on [0, 1]. In addition, note that if |g(x)| ≤ K and |f (x)| ≤ K hold for each

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x ∈ [0, 1], then hn(t) ≤ K 2 for all t ∈ [0, 1].i.e. the sequence (hn) is uniformly boundedon [0, 1]. Now, note that if 0 ≤ t ≤ 1 then (i − 1)/n ≤ (t + i − 1)/n ≤ i/n.Thus, if mniand M ni denote the minimum and maximum values of g, respectively, on the interval[(i − 1)/n, i/n], then mni ≤ g((t + i − 1)/n) ≤ M ni holds for each 0 ≤ t ≤ 1. LetRn =

ni=1 1/nm

ni and S n =

ni=1 1/nM

ni ,

and note that Rn and S n are two Riemann sums(the smallest and the largest ones)for the function g corresponding to the partition {0, 1/n, 2/n, ..., (n − 1)/n, 1}. Hence,limn→∞ Rn = limn→∞ S n =

10 g(x)dx.From,

|hn(t) − Rn.f (t)| = |[n

i=1

1/ng((t + i − 1)/n)]f (t) − Rn.f (t)|

= |([n

i=1

1/ng((t + i − 1)/n)] − Rn).f (t)| ≤ (S n − Rn)|f (t)|,

we see that limn→∞ hn(t) = f (t) 10 g(x)dx and in fact hn converges uniformly. Now, by(*) and the Lebesgue Dominated Convergence Theorem we obtain

limn→∞

10

g(x)f (nx)dx = limn→∞

10

hn(t)dt

=

10

[ limn→∞

hn(t)]dt =

10

[f (t)

10

g(x)dx]dt = (

10

g(x)dx)(

10

f (x)dx).

PROBLEM 34. Show that ∞0

sin2(x)x2 dx =

π2 .

SOLUTION.Consider the function f (x) = 1,ifx = 0, f (x) = sin2(x)/x2, if 0 < x ≤1, f (x) = 1/x2, ifx > 1.Note that f is Lebesgue integrable over [0, ∞).By the inequality

0 ≤ sin2(x)x2

≤ f (x), we see that the function sin2(x)x2

is Lebesgue integrable over [0, ∞).Now for each r, > 0, we have

r

sin2(x)

x2 dx = −

r

sin2(x)d(1

x)

= −sin2(x)

x |r +

r

2sinxcosx

x dx

=

sin2

− sin2r

r + 2r2

sinx

x dx.

Thus, we see that

∞0

sin(x)

x2 dx = lim

r→∞→0+

r

sin2(x)

x2 dx =

∞0

sin(x)

x dx =

π

2.

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PROBLEM 35.

a.)Let (f n) be a sequence of measurable functions and let f : X → R. Assume that

limn→∞

µ({x : |f n(x) − f (x)| ≥ }) = 0(∗)

holds for every > 0. Show that f is measurable.b.) Assume that (f n) ⊆ M satisfies f n ↑ and f n →µ f (i.e. f n goes to f in measure).Show that

limn→∞

f ndµ =

fdµ.

c.) Assume that (f n) ⊆ M satisfies f n ≥ 0 a.e. and f n →µ f (i.e. f n goes to f inmeasure). Show that f ≥ 0 a.e.SOLUTION.

a.)We will show that there is a subsequence (f nk) of (f n) that converges to f a.e. as thegiven condition (

∗) is equivalent to the condition that f n

→ f in measure.

Pick a sequence (kn) of strictly increasing positive integers such that µ({x : |f n(x) −f (x)| ≥ 1/n}) < 2−n for all k > kn. Set E n := {x : |f n(x) − f (x)| ≥ 1/n} for each n andlet E := ∩∞n=1 ∪∞k=n E k.Then,

µ(E ) ≤ µ(∪∞k=nE k) ≤∞

k=n

µ(E k) ≤ 21−n

holds for all n, and this shows that µ(E ) = 0. Also, if x is not in E , then there exists somen such that x is not in ∪∞k=nE k, and so |f km − f | ≤ 1/m holds for each m ≥ n.Therefore,lim f kn(x) = f (x) for each x ∈ E c, and so f kn → f a.e. holds. Thus, f is measurable aslimit of a sequence of measurable functions is itself measurable.b.)By part a.) there exists a subsequence f nk which converges to f a.e. Since f n ↑, iteasily follows that f n ↑ f .Now apply the Monotone Convergence Theorem to deduce theresult.c.) Again by part a.) there exists a subsequence f nk which converges to f a.e. Thus, wemust have f ≥ 0 a.e.PROBLEM 36. Let g be an integrable function and let (f n) be a sequence of integrablefunctions such that |f n| ≤ g a.e. holds for all n. Suppose that f n →µ f (i.e.f n goes to f in measure), then show that f is an integrable function and lim

|f n − f |dµ = 0.SOLUTION. By the above exercise we know that if f n converges to f in measure then

there exists a subsequence f nk of f n which converges to f a.e.Thus, |f | ≤ g a.e. Andapplication of the Lebesgue Dminated Convergence Theorem gives that f is integrable.Now, assume on the contrary that lim

|f n − f |dµ = 0.Thus, assume that for some > 0 there exists a subsequence (gn) of (f n) such that

|gn − f |dµ ≥ . But we knowthat there exists a subsequence (hn) of (gn) with hn → f a.e. Now Lebesgue Dominated

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SOLUTION. There are various ways to do this. Here is our proof :The result is obvious when f is bounded, since then

| E

f (x)dx| ≤ E

|f |dµ ≤ (supx∈E |f (x)|)µ(E )

In the general case, let

An := {x ∈ A : n ≤ f (x) ≤ n + 1}

BN := ∪N n=0AnC N := A − BN

Then, A

|f (x)|dx = ∞n=0 An |f (x)|dµ.Let N be such that∞

n=N +1

An |

f (x)

|dµ =

C N |f (x)

|dµ <

2.

and let 0 < δ < 2(N +1) . Then µ(E ) < δ implies

| E

f (x)dµ| = E

|f (x)|dµ = E ∩BN

|f (x)|dµ + E ∩C N

|f (x)|dµ

≤ (N + 1)µ(E ) + C N

|f (x)dµ < 2

+

2 = .

PROBLEM 40. Suppose f is integrable function on R(⇔ f ∈ L1(R)).Then, showthat

limt→0

|f (t + x) − f (x)|dx = 0.

SOLUTION. Note that the result is immediate when f is a step function. Now let f be an arbitrary integrable function and let > 0. If f ≈ f 1 + f 2 + f 3 + ...(approximatef by a step function which can be done as step functions are dense in L1), then thereexists n0 ∈ N such that

∞n=n0+1

|f n|dx <

3.

We have |f (x + t) − f (x)|dx ≤

|

n0n=1

f n(x + t) −n0n=1

f n(x)|dx

+∞

n=n0+1

|f n(x + t)|dx +

∞n=n0+1

|f n(x)|dx

20


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PROBLEM 43. Suppose f is an integrable function on X = R p.i) Show that ∀ > 0, there exists a measurable set with finite measure such that f isbounded on A and

(X −A)

|f |dµ < ii) From this deduce that

limµ(E )→0

E |

f |dµ = 0.

SOLUTION.

i)We can assume that f ≥ 0. Consider the following setsA0 := {x : f (x) = 0}, An := {x : 1/n ≤ f (x) ≤ n}, A∞ := {x : f (x) = ∞}.

Clearly, A1 ⊆ A2 ⊆ A3 ⊆ ... and also ∞

n=1(X − An) = A0 ∪ A∞. Note that µ(A∞) = 0as f is integragle. Thus, we have

limn→∞

(X −An)

f dµ =

A0

f dµ +

A∞

f dµ = 0

Therefore ∃n0 such that letting A = An0, (X −A)

fdµ < . Furthermore, f is bounded

on A by n0 and also A has finite measure as 1n0

µ(A) ≤ A fdµ 1. Since A is an algebra, F n ∈ A. Note that F n ⊆ F n+1, so

∞ j=1

E j =∞

j=1

F j ∈ A.

ii) Let m(E ) = n

1 a jµ j(E ). It is clear that m(∅) = 0 as m(∅) = n

1 a jµ j(∅) = 0.Itis also obvious that m(E ) ≥ 0.For the countable additivity, recall the fact that series

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of nonnegative terms can be added and multiplied termwise and rearrenged arbitrarilywithout changing the sum. Thus, m is a measure.

PROBLEM 45.

i) Let (X,M,µ) be a measure space and E , F ∈ M. Show that

µ(E ) + µ(F ) = µ(E ∪ F ) + µ(E ∩ F ).

ii) Given a measure space (X,M,µ) and E ∈ M , define µE (F ) = µ(A ∩ E ). Show thatµE is a measure.

SOLUTION.

i) Write the set as a disjoint union of two sets as follows E = (E − F ) ∪ (E ∩ F ).Nowwrite F as the disjoint union F = (F − E ) ∪ (E ∩ F ) and E ∪ F as the disjoint union(E − F ) ∪ (F − E ) ∪ (E ∩ F ). Then we have,

µ(E ∪ F ) + µ(E ∩ F ) = µ(E − F ) + µ(F − E ) + 2µ(E ∩ F )= µ(E − F ) + µ(E ∩ F ) + µ(F − E ) + µ(E ∩ F )

= µ(E ) + µ(F )

ii) Clearly, µE (A) ≥ 0 and µE (∅) = 0. Countable additivity is also very easy to verify.Suppose An ∩ Am = ∅ when m = n, then

µE (∞n=1

An) = µ(E ∩∞n=1

An) =∞n=1

µ(E ∩ An) =∞n=1

µE (An).

PROBLEM 46. We know that if µ is a measure(in the sense that it is count-ably additive) then it is continuous from below. {i.e. If A1 ⊆ A2 ⊆ A3 ⊆ ...thenµ(

Ai) = lim µ(Ai)}.Similarly, we know that if µ(X ) is finite then µ is continuousfrom above.{i.e. If ... ⊆ A3 ⊆ A2 ⊆ A1 then limn→∞ µ(An) = µ(

An).}. Show that

a finitely additive measure is a measure if and only if it is continuous from below.Nowsuppose that µ(X )


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This clearly shows that µ is countably additive, and hence is a measure.We will try to use previous part. Suppose µ(X )


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PROBLEM 48. Let f : Rn → R be a Lebesgue measurable function such that

m({x : |f (x)| > t}) ≤ ct2

, t > 0.

Prove that there exists a constant C 1 such that for any Borel set E ⊂ Rn of finite andpositive measure

E

|f (x)|dx ≤ C 1

m(E ).

SOLUTION. E

|f |dm = ∞0

m({x : |f | > t} ∩ E )dt

=

√ c|E|

0

m({x : |f | > t} ∩ E )dt + ∞√

c

|E|

m({x : |f | > t} ∩ E )dt

≤ |E

| c

|E | +

∞

√ c|E|c

t2dt =

√ c |E | +

−ct∞√ c|E|

=√

c

|E | + c c|E |

= 2√

c

|E | = C 1

|E |.

Thus C 1 = 2√

c and therefore the assertion is proved.

PROBLEM 49. Let B(m, 1) be m−dimensional ball of radius 1 centered at the originin Rm.a.) Show that there exists a function f : R → [0, 1] such that

m(B(n + 1, 1)) = m(B(n, 1))

[f (t)]

n

dt.

Here m denotes the Lebesgue measure.b.) Show that

[f (t)]ndt → 0 as n → ∞.

c.) Show that for any positive number A, Anm(B(n, 1)) → 0 as n → ∞.SOLUTION.

a.) Let B(n + 1, r) =

x ∈ Rn : x21 + ...x2n+1 = r

.Integrate over t = xn+1 to determinethe volume of B(n + 1, 1). Then we get,

m(B(n + 1, 1)) =

1−1

m(B(n,√

1 − t2))dt

= 2

1−1

m(B(n, 1))√

1 − t2ndt = m(B(n, 1)) ∞−∞

f (t)dt

where f (t) = 1[−1,1]√

1 − t2, here 1A represents the characteristic function of the set A.The first equality follows from the Fubinis theorem, and the second equality follows

25


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from the fact that m(B(n, r)) = rnm(B(n, 1)). This fcat follows from a linear change of variables.b.) Note that [f (t)]n → 0 pointwise if t = 0 and define f n := f n, then f 1 ≥ f 2 ≥... ≥ 0, f 1 is integrable and f n → 0 almost everywhere, so by the first question we have [f (t)]

ndt

→ 0.

c.)Anm(B(n, 1)) = Anm(B(n − 1, 1)

[f (t)n−1dt

= Anm(B(n − 2, 1))

[f (t)]n−2dt

[f (t)]n−1dt

= ... = Anm(B(0, 1))

[f (t)]0dt...

[f (t)]n−1dt

A

[f (t)]0dt

...

A

[f (t)]n−1dt

.

Here we used the fact that m(B(0, 1)) = 1. By part b.), for k sufficiently large,A

[f (t)]kdt < s


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If g ∈ L+ is arbitrary, we can find a sequence (φn)∞1 of nonnegative simple functions thatincreases pointwise to the function g. Then the sequence

φn.f

increases pointwise to

the function g.f . Thus, by the Monotone Convergence Theorem,

gdλ = limn→∞

φndλ = limn→∞

φnfdµ = gfdµ.

PROBLEM 52. Let f (x) = x−1/2 if 0 < x


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for all m ≥ M , so a,b) g2dm = ba g2dm can only be ∞, and this clearly shows that g2can not be integrable.c.) If g = h a.e., we also have

(a,b)

|h|2dm = ∞ for any interval (a, b). Thus, h can notbe bounded on (a, b). Moreover, h is discontinuous at any point x, since otherwise hwould be bounded on some interval containing x.

PROBLEM 53. Compute the following limits and justify the calculations:a.) limn→∞

∞0

1 + xn

−nsin(xn)dx ;

b.) limn→∞ 10

1 + nx2

1 + x2

−ndx;

c.) limn→∞ ∞0 nsin(

xn)

x(1 + x2)]−1dx;d.) limn→∞

∞a

n(1 + n2x2)−1dx , (a ∈ R)SOLUTION.

a.) For n ≥ 2 we have 1 + xn

−nsin(

x

n) ≤ 1 + x

2

−2.

Note also that the function

1 + x2−2

is integrable over [0, ∞). Thus, by the DominatedConvergence Theorem ,we have

limn→∞

∞0

1 +

x

n

−nsin(

x

n)dx =

∞0

limn→∞

1 +

x

n

−nsin(

x

n)dx = 0.

b.)1 + nx21 + x2−n ≤ 1, and 1

0 1dx = 1. Thus by the Dominated Convergence

Theorem, we have

limn→∞

10

1 + nx2

1 + x2

−ndx =

10

limn→∞

1 + nx2

1 + x2

−ndx = 0.

c.)nsin(xn)

x(1 + x2)]−1

≤ (1 + x2)−1 and note that (1 + x2)−1 is integrable over[0, ∞).Hence, again by the Dominated Convergence Theorem, we have,

limn→∞

∞0

nsin(x

n)

x(1+x2)]−1dx =

∞0

limn→∞

nsin(x

n)

x(1+x2)]−1dx =

∞0

(1+x2)−1 = π

2.

d.) limn→∞ ∞a

n(1 + n2x2)−1dx = limn→∞ ∞na

(1 + y2)−1dy = limn→∞ arctan(y)|∞na = 0if a > 0; = π2 if a = 0; = π if a


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so {x : φ ◦ f n φ ◦ f } ⊂ {x : f n f }. Thus, µ({x : φ ◦ f n φ ◦ f }) = 0 asµ({x : f n f }) = 0.Therefore, φ ◦ f n → φ ◦ f a.e.b.) Since φ is uniformly continuous, for any > 0, there exists a δ () > 0 such that|x − y| < δ () implies that |φ(x) − φ(y)| < . Now, if f n → f uniformly, ∀ > 0, thereis M

∈ N such that when n

≥ M , for all x

∈ X ,

|f n(x)

− f (x)

| < δ (), and so that

|φ ◦ f n(x) − φ ◦ f | < . But this shows that φ ◦ f n → φ ◦ f , uniformly.If f n → f almost uniformly, then for any 1, 2 > 0, there is a set E ∈ F (σ − algebra)and a natural number M ∈ N , such that µ(E ) < 1, and when n j ≥ M for x ∈X − E, |f n(x) − f (x)| < δ (2), and so |φ ◦ f n − φ ◦ f | < 2. This clearly shows thatφ ◦ f n → φ ◦ f almost uniformly.If f n → f in measure, then ∀ > 0, µ({x : |f n(x) − f (x)| > δ ()}) →). Since

{x : |φ ◦ f n − φ ◦ f | > } ⊂ {x : |f n − f | > δ ()},

we have µ({x : |φ◦f n−φ◦f | > }) → 0. But this means that φ◦f n → φ ◦f , in measure.c.) A counterexample for a.) is f n(x) =

1n

, f (x) = 0, and φ = 1{0}.

A counterexample for b.) is X = R, f n(x) = x + 1n , f (x) = x and φ(x) = x2.

PROBLEM 55. Suppose f n → f almost uniformly, then show that f n → f a.e. andin measure.

SOLUTION.First let us recall what it means to converge almost uniformly : It means,for all 1, 2 > 0, there is a set E such that µ(E ) < 1 and x ∈ (X − E ) implies|f n(x) − f (x)| < 2.Since f n → f almost uniformly, for any n ∈ N , there is E n ∈ M (σ-algebra such thatµ(E n) <

1n

and f n → f on E cN . Let E = ∞

1 E n, then µ(E ) = 0 and f n → f on

∞1 E

cn = E

c. Thus, f n → f a.e.Since f n

→ f almost uniformly, for every 1, 2 > 0, there is E

∈ M and n1

∈ N

such that µ(E ) < 2 and when n > n1, |f n(x) − f (x)| < 1 for x is not in E, and soµ({x : |f n(x) − f (x)| ≥ 1}) ≤ µ(E ) < 2. Thus,

µ({x : |f n(x) − f (x)| ≥ 1}) → 0.

Therefore, f n → f in measure as 1 and 2 are arbitrary.PROBLEM 56. Show that if f : [a, b] → C is Lebesgue measurable and > 0, thenthere is a set E ⊂ [a, b] such that m(E c) < and f |E is continuous. Moreover, E maybe taken to be compact.

SOLUTION. Since

∞n=1{x : |f (x)| > n} = ∅, there is M ∈ N such that m({x :|f (x)| > M }) < 2 . Let E 1 = {x : |f (x)| ≤ M }, and define h(x) = 1E 1f (x). Now

h ∈ L1[a, b], so we can find a subsequence of a sequence of continuous functions ( gn)which tends to f a.e.Without loss of generality we assume that gn → h a.e. ApplyingEgoroff’s Theorem, we have E 2 ∈ M such that m(E c2) < 2 and gn → h uniformly onE 2. Then, we have that h is continuous on E 2 and so f is continuous on E 1 ∩ E 2 for f

29


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differs from h only on E 1. Now, m((E 1 ∩ E 2)c) = m(E c1 ∪ E c2) < , there is an open setO containing E c1 ∪ E c2 and m(O) < . Let E = Oc, then E is compact, m(E c) < andE ⊂ E ! ∩ E 2, so f is continuous on E .PROBLEM 57. A measure µ is called semi-finite if every set of infinite measurecontains a subset of finite, nonzero measure. Show that every σ

−finite measure is semi-

finite.

SOLUTION. Since µ is σ−finite, we know that

X =∞

j=1

E j, µ(E j) 0µ({x : |f n(x) − f (x)| > }) → 0 as n → ∞. Since µ only takes integer values, thisis equivalent to : ∃N so that µ({x : |f n(x) − f (x)| > }) = 0 for n > N. This says,|f n(x)−f (x)| ≤ for n > N and for all x. But this says f n → f uniformly. The converseis obvious.PROBLEM 59. Prove that for a > 0,

∞−∞

e−x2

cos(ax)dx =√

πe−a2

4 .

SOLUTION. Define,

f n(x) = e−x2

n j=0

(−1) j (ax)2 j

(2 j)! ,

and

g(x) = e−x2

∞ j=0

(−1) j (ax)2 j

(2 j)! = e−x

2

cosh(ax) = e−x2 eax + e−ax

2 .

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Then, g ∈ L1, and |f n(x)| ≤ g(x). Thus, we can use Dominated Convergence Theoremand integrate the series for f term by term. A simple calculation yields,

∞−∞

e−x2 (ax)2n

(2n)! dx = a2n

√ π

4nn!.

Thus,

∞−∞

e−x2

cos(ax)dx =∞n=0

(−1)na2n√

π

4nn! =

√ π

∞n=0

(−a2/4)nn!

=√

πe−a2/4.

PROBLEM 60. Let (q n) be an enumeration of rationals in [0, 1]. Define the functionf on [0, 1] by,

f (x) =

n,so,qn 0.Choose N so large that∞

n=N

2−n < .

Now, choose δ > 0 so small that the interval (x − δ, x + δ ) does not contain any of the q n with n < N. Then for y ∈ (x − δ, x + δ ),

|f (x) − f (y)| <n>N

2−n < .

(b.) The answer is yes. There is a theorem saying, if a function is bounded on abounded interval and it has at most countably many points of discontinuity, thenit is Riemann integrable. Our function satisfies the conditions so it is Riemannintegrable.

31


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(c.) Yes. It is Lebesgue integrable. For this we can use a theorem which says if f isproperly Riemann integrable, then it is Lebesgue integrable.Or we can try to do it directly. The partial sums

f N (x) = n≤N,qn 0 it is not possible to haveB(a, r) ⊂ A for any r > 0. Therefore, for all r > 0, B(a, r) contains points in Ac. Thus,there exists a sequence of points each lying in Ac and converging to a.i.e. There exists(an)∞n=1 such that an ∈ Ac for all n and an → a. But, then limn→∞ f (an) = limn→∞ g(an)since f (an) = g(an) for all n. Since f and g are continuous on A

c the above equalitygives,

f ( limn→∞

an) = g( limn→∞

an)

⇒ f (a) = f ( limn→∞

an) = g( limn→∞

an) = g(a).

Since, a ∈

A is arbitrary we have g(a) = f (a) for all a ∈

A. Since also f (b) = g(b) forall b ∈ Ac we have f (x) = g(x) for any x ∈ Rn.

32

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