ABSTRACT INTEGRATION CHAPTER ONE

Transcription

1 CHAPTER ONE ABSTRACT INTEGRATION Version 1.1 No rights reserved. Any part of this work can be reproduced or transmitted in any form or by any means. Suggestions and errors are invited and can be mailed to 1

2 1 Does there exist an infinite σ-algebra which has only countable many members? Solution: No. Suppose M be a σ-algebra on which has countably infinite members. For each x define B x = x M M M. Since M has countable members, so the intersection is over countable members or less, and so B x belongs M, since M is closed under countable intersection. Define N = {B x x }. So N M. Also we claim if A, B N, with A B, then A B =. Suppose A B, then some x A and same x B, but that would mean A = B = x M M M. Hence N is a collection of disjoint subsets of. Now if cardinality of N is finite say n N, then it would imply cardinality of M is 2 n, which is not the case. So cardinality of N should be at least ℵ 0. If cardinality of N = ℵ 0, then cardinality of M = 2 ℵ 0 = ℵ 1, which is not possible as M has countable many members. Also if cardinality of N ℵ 1, so is the cardinality of M, which again is not possible. So there does not exist an infinite σ-algebra having countable many members. 2

3 2 Prove an analogue of Theorem 1.8 for n functions. Solution: Analogous Theorem would be: Let u 1, u 2,..., u n be real-valued measurable functions on a measurable space, let Φ be a continuous map from R n into topological space Y, and define h(x) = Φ(u 1 (x), u 2 (x),..., u n (x)) for x. Then h : Y is measurable. Proof: Define f : R n such that f(x) = (u 1 (x), u 2 (x),..., u n (x)). So h = Φ f. So using Theorem 1.7, we only need to show f is a measurable function. Consider a cube Q in R n. Q = I 1 I 2 I n, where I i are the intervals in R. So f 1 (Q) = u 1 1 (I 1 ) u 1 2 (I 2 ) u 1 n (I n ) Since each u i is measurable, so f 1 (Q) is measurable for all cubes Q R n. But every open set V in R n is a countable union of such cubes, i.e V = i=1q i, therefore f 1 (V ) = f 1 ( Q i ) = f 1 (Q i ) i=1 Since countable union of measurable sets is measurable, so f 1 (V ) is measurable. Hence f is measurable. i=1 3

4 3 Prove that if f is a real function on a measurable space such that {x : f(x) r} is a measurable for every rational r, then f is measurable. Solution: Let M denotes the σ-algebra of measurable sets in. Let Ω be the collections of all E [, ] such that f 1 (E) M. So for all rationals r, [r, ] Ω. Let α R; we will show (α, ] Ω; hence from Theorem 1.12(c) conclude that f is measurable. Since rationals are dense in R, therefore there exists a sequence of rationals {r i } such that r i > α and r i α. Also (α, ] = 1 [r i, ]. Each [r i, ] Ω and Ω is a σ-algebra (Theorem 1.12(a)) and hence closed under countable union; therefore (α, ] Ω. And so from Theorem 1.12(c), we conclude f is measurable. 4

5 4 Let {a n } and {b n } be sequences in [, ], prove the following assertions: (a) (b) lim sup x ( a n ) = lim inf a n. lim sup (a n + b n ) lim sup a n + lim sup b n provided none of the sums is of the form. (c) If a n b n for all n, then lim sup a n lim sup b n. Show by an example that the strict inequality can hold in (b). Solution: (a) We have for all n N, sup { a i } = inf {a i} i n i n taking limit n, we have desired equality. (b) Again for all n N, we have {a i + b i } sup{a i } + sup{b i } sup i n i n Taking limit n, we have desired inequality. (c) Since a n b n for all n, so for all n we have inf {a i} inf {b i} i n i n Taking limit n, we have desired inequality. For strict inequality in (b), consider a n = ( 1) n and b n = ( 1) n+1. i n 5

6 5 (a) Suppose f : [, ] and g : [, ] are measurable. Prove that the sets {x : f(x) < g(x)}, {x : f(x) = g(x)} are measurable. Solution: Given f, g are measurable, therefore from 1.9(c) we conclude g f is also measurable. But then {x f(x) < g(x)} = (g f) 1 (0, ] is a measurable set by Theorem 1.12(c). Also {x f(x) = g(x)} = (g f) 1 (0) = (g f) 1 ( ( 1 n, 1 n ) = (g f) 1 ( 1 n, 1 n )) Since each (g f) 1 ( 1 n, 1 n) is measurable, so is their countable intersection. Hence {x f(x) = g(x)} is measurable. (b) Prove that the set of points at which a sequence of measurable real-valued functions converges (to a finite limit) is measurable. Solution: Let f i be the sequence of real-measurable functions. Let A denotes the set of points at which f i converges to a finite limit. But then A = n=1 m=1 i,j m {x f i (x) f j (x) < 1 n } = n=1 m=1 i,j m ( (f i f j ) 1 1 n, 1 ) n Since for each i, j, f i f j is measurable, so (f i f j ) 1 ( 1 n, 1 n) is measurable too for all n. Also countable union and intersection of measurable sets is measurable, we conclude A is measurable. 6

7 6 Let be an uncountable set, let M be the collection of all sets E such that either E or E c is at most countable, and define µ(e) = 0 in the first case and µ(e) = 1 in the second. Prove that M is a σ-algebra in and that µ is a measure on M. Describe the corresponding measurable functions and their integrals. Solution: M is a σ-algebra in : M, since c = is countable. Similarly M. Next if A M, then either A or A c is countable, that is either (A c ) c is countable or A c is countable; showing A c M. So M is closed under complement. Finally, we show M is closed under countable union. Suppose A i M for i N, we will show A i also belongs to M. If all A i are countable, so is their countable union, so A i M. But when all A i are not countable means at least one say A j is uncountable. Then A c j is countable. Also ( A i ) c A c j, showing ( A i ) c is countable. So A i M. Hence M is closed under countable union. µ is a measure on M: Since µ takes values 0 and 1, therefore µ(a) [0, ] for all A M. Next we show µ is countable additive. Let A i for i N are disjoint measurable sets. Define A = A i. We will show µ(a) = µ(a i ). If all A i are countable, so is A; therefore µ(a i ) = 0 for all i and µ(a) = 0; and the equation µ(a) = µ(a i ) holds good. But when all A i are not countable means at least one say A j is uncountable. Since A j M, therefore A c j is countable. Also Since all A i are disjoint, so for i j, A i A c j. So µ(a i ) = 0 for i j. Also µ(a j ) = µ(a) = 1 since both are uncountable. Hence µ(a) = µ(a i ). Characterization of measurable functions and their integrals: Assume functions are real valued. First we isolate two class of measurable functions denoted by F and F, defines as: F = {f f is measurable & f 1 ([, α]) is countable for all α R} F = {f f is measurable & f 1 ([α, ]) is countable for all α R} Next we characterize the reaming measurable functions. Since f / F or F, therefore f 1 ([α, ]) is uncountable for some α R. Therefore α f defined as sup{α f 1 ([α, ]) is countable} exists. So if β > α f, then f 1 ([β, ]) is countable. Also if β < α f, then f 1 ([, β]) = f 1 ((β, ]). Since f 1 ((β, ]) is uncountable and belongs to M, therefore f 1 ((β, ]) is countable. And so f 1 (α f ) is uncountable. Also 7

8 f 1 (α f ) M, therefore f 1 (γ) is countable for all γ α f. Thus if f is a measurable function then either f F orf, or there exists α f R such that f 1 (α f ) is uncountable while f 1 (β) is countable for all β α f. Once we have characterization, integrals are easy to describe: if f F f dµ = if f F else α f 8

9 7 Suppose f n : [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, and f 1 L 1 (µ). Prove that then f n dµ = f dµ lim and show that this conclusion does not follow if the condition f 1 L 1 (µ) is omitted. Solution: Take g = f 1 in the Theorem 1.34 to conclude lim f n dµ = f dµ For showing f 1 L 1 (µ) is a necessary condition for the conclusion, take = R and f n = χ [n, ). So we have f(x) = 0 for all x, and therefore f dµ = 0. While f n dµ = for all n. 9

10 8 Put f n = χ E if n is odd, f n = 1 χ E if n is even. What is the relevance of this example to Fatou s lemma? Solution: With the described sequence of f n, strict inequality occurs in Fatou s Lemma (1.28). We have ( ) lim inf f n dµ = 0 While lim inf assuming µ() µ(e). f n dµ = min(µ(e), µ() µ(e)) 0, 10

11 9 Suppose µ is a positive measure on, f : [0, ] is measurable, f dµ = c, where 0 < c <, and α is a constant. Prove that lim n log[1 + (f/n) α ] dµ = if 0 < α < 1, c if α = 1, 0 if 1 < α <. Hint: If α 1, prove that the integrands is dominated are dominated by αf. If α < 1, Fatou s lemma can be applied. Solution: As given in the hint, we consider two cases for α: Case when 0 < α < 1: Define φ n (x) = n log(1 + (f(x)/n) α ). Since φ n : [0, ], therefore Fatou s lemma is applicable. So Also (lim inf lim n log(1 + (f(x)/n)α ) = lim φ n) dµ lim inf Since α < 1 and f dµ <, therefore φ n dµ 1 αf(x) α 1+(f(x)/n) α n α+1 1 = n 2 lim n log(1 + (f(x)/n)α ) = a.e x And hence (lim inf φ n ) dµ =. Therefore lim φ n dµ lim inf φ n dµ = αn1 α f(x)α 1 + (f(x)/n) α Case when α 1: We claim φ n (x) is dominated by αf(x). For a.e. x and α 1, we need to show n log(1 + (f(x)/n) α ) αf(x) for all n ( ( ) α ) f(x) i.e. log 1 + α f(x) n n Define g(λ) = log(1 + λ α ) αλ for λ 0. So if g(λ) 0 for α 1 and λ 0, then (1) follows by taking λ = f(x)/n. So we need show g(λ) 0 for α 1 and λ 0. Computing derivative of g, we have g (λ) = α(1 + λα λ α 1 ) 1 + λ α 11 (1)

12 When 0 λ 1, we have 1 λ α 1 0; while when λ > 1, we have λ α λ α 1 > 0. Thus g (λ) 0. Also g(0) = 0, therefore g(λ) 0 for all λ 0 and α 1. And so for α 1 we have log(1 + (f(x)/n) α ) αf(x) for all n and a.e x. Since αf(x) L 1 (µ), DCT(Theorem 1.34) is applicable. Thus lim n log(1 + (f/n) α ) dµ = lim (n log(1 + (f/n)α )) dµ When α = 1, lim (n log(1 + (f/n) α )) = f(x) (calculating the same as calculated for the case α < 1). And when α > 1, we have lim (n log(1 + (f/n) α )) = 0. And hence lim n log(1 + (f/n) α ) dµ = if 0 < α < 1, c if α = 1, 0 if 1 < α <. 12

13 10 Suppose µ() <, {f n } is a sequence of bounded complex measurable functions on, and f n f uniformly on. Prove that f n dµ = f dµ, lim n= and show that the hypothesis µ() < cannot be omitted. Solution: Let ɛ > 0. Since f n f uniformly, therefore there exists n 0 N such that f n (x) f(x) < ɛ n n 0 Therefore f(x) < f n0 (x) + ɛ. Also f n (x) < f(x) + ɛ. Combining both equations, we get f n (x) < f n0 + 2ɛ n n 0 Define g(x) = max( f 1 (x),, f n0 1(x), f n0 (x) + 2ɛ), then f n (x) g(x) for all n. Also g is bounded. Since µ() <, therefore g L 1 (µ). Now apply DCT(Theorem 1.34) to get lim f n dµ = f dµ To show µ() < is a necessary condition, consider = R with usual measure µ, and f n (x) = 1. We have lim n f n dµ =, while f dµ = 0, since f = 0. f uniformly is also a necessary condi- REMARK : The condition f n tion. 13

14 11 Show that A = n=1 k=n in Theorem 1.41, and hence prove the theorem without any reference to integration. Solution: A is defined as the collections of all x which lie in infinitely many E k. Thus x A x k=n E k n N; and so A = n=1 k=n Now let ɛ > 0. Since k=1 µ(e k) <, therefore there exists n 0 N such that k=n 0 µ(e k ) < ɛ. And µ(a) = µ( Make ɛ 0 to conclude µ(a) = 0. E k E k n=1 k=n E k ) µ( E k ) k=n 0 µ(e k ) k=n 0 < ɛ 14

15 12 Suppose f L 1 (µ). Prove that to each ɛ > 0 there exists a δ > 0 such that f dµ < ɛ whenever µ(e) < δ. E Solution: Let (, M, µ) be the measure space. Suppose the statement is not true. Therefore there exists a ɛ > 0 such that there exists no δ > 0 such that f dµ < ɛ whenever µ(e) < δ. That means for each δ > 0, E there exists a E δ M such that µ(e δ ) < δ, while E δ f dµ > ɛ. By taking δ = 1/2 n, where n N, we construct a sequence of measurable sets {E 1/2 n}, such that µ(e 1/2 n) < 1/2 n for all n and E 1/2 f dµ > ɛ. n Now define A k = n=k E 1/2 n and A = k=1 A k. We have A 1 A 2 A 3, and µ(a 1 ) = µ( n=1 E 1/2 n) n=1 µ(e 1/2 n) < n=1 1/2n <. Therefore from Theorem 1.19(e), we conclude µ(a k ) µ(a). Next define φ : M [0, ] such that φ(e) = f dµ. Clearly, by Theorem 1.29, φ is a measure on M. Therefore, again by Theorem 1.19(e), we E have φ(a k ) φ(a). Since A = k=1 n=k E 1/2n, therefore from previous Exercise, we get µ(a) = 0. Therefore φ(a) = f dµ = 0. While A φ(a k ) = φ( E 1/2 n) φ(e 1/2 k) = n=k E 1/2 k f dµ > ɛ Therefore φ(a k ) φ(a), a contradiction. Hence the result. 15

16 13 Show that proposition 1.24(c) is also true when c =. Solution: We have to show cf dµ = c f dµ, when c = and f 0 We consider two cases: when f dµ = 0 and f dµ > 0. When f dµ = 0, we have from Theorem 1.39(a), f = 0 a.e., therefore, cf = 0 a.e.. And hence cf dµ = 0 = c f dµ While when f dµ > 0, there exist a ɛ > 0 and a measurable set E, such that µ(e) > 0 and f(x) > ɛ whenever x E; because otherwise f(x) < ɛ a.e. for all ɛ > 0; making ɛ 0, we get f(x) = 0 a.e. and hence f dµ = 0, which is not the case. But then cf dµ cf dµ > ɛ c dµ = E E Also c f dµ = Hence the proposition is true for c = too. 16