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Math 426: Homework 1 Solutions Mary Radcliffe due 9 April 2014

In Bartle: 2B. Show that the Borel algebra B is also generated by the collection of all half-open intervals (a, b] = {x ∈ R | a < x ≤ b}. Also show that B is generated by the collection of all half-rays {x ∈ R | x > a}. Proof. Let S1 = {(a, b) | a < b}, and S2 = {(a, b] | a < b}. By definition, σ(S1 ) = B. To show that σ(S2 ) = B, it suffices to show that S1 ⊂ σ(S2 ) and S2 ⊂ σ(S1 ), as then, since σ(A) is the smallest σ-field containing A, the result follows immediately. Note that if a < b, then (a, b) = ∪n∈Z+ (a, b − n1 ], and (a, b] = ∩n∈Z+ (a, b + 1 n ). Therefore, (a, b) ∈ σ(S2 ) and (a, b] ∈ σ(S1 ), completing the proof. For the second property, let S3 = {(a, ∞) | a ∈ R}. Again, it suffices to show that S3 ⊂ σ(S1 ) and S1 ⊂ σ(S3 ).
Let a < b. Then (a, b) = (a, ∞)\ ∩n∈Z+ (b − n1 , ∞) ∈ σ(S3 ). Moreover, (a, ∞) = ∪n∈Z+ (a, a + n) ∈ σ(S1 ), completing the proof. 2K. Show directly that if f is measurable defined by   f (x) A fA (x) =  −A

and A > 0, then the truncation fA if |f (x)| ≤ A if f (x) > A if f (x) < −A

is measurable. Proof. It suffices to prove that for all α ∈ R, we have fA−1 ((α, ∞)) is measurable. Note that if α > A, then fA−1 ((α, ∞)) = ∅, which is measurable. If α = A, then fA−1 (α) = f −1 ([A, ∞)), which is measurable by Prop 2.4. If −A ≤ α < A, then fA−1 ((α, ∞)) = f −1 ((α, ∞)), which is measurable since f is. Finally, if α < −A, then fA−1 ((α, ∞)) = X, which is measurable. Therefore, fA is measurable. 3E. Let X be an uncountable set and let F be the family of all subsets of X. Define µ on E by requiring that µ(E) = 0 if E is countable, and µ(E) = ∞ if E is uncountable. Show that µ is a measure on F. Proof. Clearly, µ(∅) = 0 and µ(E) ≥ 0 for all E ∈ F. Let {En } ⊂ F be a sequence of pairwise disjoint subsets of F. If all En are countable, then µ(En ) = 0 for all n. Moreover, as the countable union of countably many subsets is countable, we have also that µ(∪En ) = 0. Thus, µ(∪En ) = P µ(En ). If not, there exists some k such that Ek is uncountable. Then ∪En is P uncountable also, so µ(∪En ) = ∞ = µ(Ek ) = µ(En ). 1

Therefore, µ is countably additive, and is thus a measure on F. 3F. Let X = Z+ and let F be the family of all subsets of X. If E is finite, let µ(E) = 0, if E is infinite, let µ(E) = ∞. Is µ a measure on F? Solution. No, µ is not a measure on X. Let En = {n}. Then En is finite for all n, so µ(En ) = 0. However, ∪En = Z+ , so µ(∪En ) = ∞. Thus, µ is not countably additive and thus is not a measure. 3H. Show that Lemma 3.4(b) may fail if the finiteness condition µ(F1 ) < ∞ is dropped. Proof. Let Fn = R\[−n, n], so that µ(Fn ) = ∞ for all n, and Fn ⊃ Fn+1 . Then ∩Fn = ∅, so µ(∩Fn ) = 0. However, lim µ(Fn ) = ∞, and thus the result fails for {Fn }. 3T. Show that the Lebesgue measure of the Cantor set C is zero. Proof. Let E0 = [0, 1], E1 = E0 \( 13 , 23 ), E2 = E1 \( 91 , 29 )\( 79 , 89 ), etc., so that C = ∩En . Then by Prop 3.4(b), since En ⊃ En+1 and λ(E0 ) = 1 < ∞, we have λ(C) = lim λ(En ). Note that by definition, λ((a, b)) = b − a. Moreover, by Prop 3.4(b), we have [a, b] = ∩n∈Z+ (a− n1 , b+ n1 ), and thus λ([a, b]) = lim λ((a− n1 , b+ n1 )) = lim(b−a+ n2 ) = b−a. Therefore, we have λ(En ) = λ(En−1 )− 31 λ(En−1 ) =

n 2 2 n , so λ(C) = lim 23 = 0. 3 λ(En−1 ). Thus, λ(En ) = 3 3U. By varying the construction of the Cantor set, obtain a set of positive Lebesgue measure that contains no nonvoid open interval. Proof. The construction here is called a “Fat Cantor set,” although you certainly may have come up with something else. We begin as with the construction of the Cantor set: Let E0 = [0, 1]. Take E1 to be E0 with the middle third interval deleted, as before: E1 = E0 \( 13 , 23 ). Now, to form E2 , from each interval of E1 , delete an open segment, the sum of whose lengths is 1/6. There are many ways to do this, but, for example, we could 3 5 21 take E1 \( 24 , 24 )\( 19 24 , 24 ), so that the total measure of the removed sets is 1 6 . Similarly, construct Ek from Ek−1 by choosing an interval from each 1 connected set in Ek−1 , the total measure of which will be 3∗2 . Then take P∞ k 1 E = ∩Ek . By Lemma 3.4(b) we have that λ(E) = 1− k=0 3∗2k = 13 > 0, but as we delete from every interval at every step, the resulting intersection contains no nonvoid open intervals. 3V. Suppose that E is a subset of a set N ∈ F with µ(N ) = 0, but E ∈ / F. Then the sequence {fn }, fn = 0 converges µ-a.e. to χE . Hence the almost-everywhere limit of a sequence of measurable functions may not be measurable. Proof. Note that limn→∞ fn (x) = 0 for all x ∈ X. Moreover, χE (x) = 0 for all x ∈ / N . Thus, fn → χE for all x ∈ / N , so {x | fn (x) 6→ χE (x)} has measure 0 under µ. Therefore, fn converges µ-a.e. to χE . Additional Exercises: 1. Let (X, F) be a measurable space.

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(a) Let Λ = {λ | λ is a charge on F}. Prove that Λ is a vector space over R. Proof. As Λ is a subset of all real-valued functions on F, and the set of all real-valued functions is itself a vector space, we need only verify the closure conditions required for a subset of a vector space to be a vector subspace. First, clearly the 0-function is in Λ. Thus, we need only show that if λ1 , λ2 ∈ Λ and c1 , c2 ∈ R, then c1 λ1 + c2 λ2 ∈ Λ. Let λ = c1 λ1 +c2 λ2 . Clearly λ(∅) = 0. Moreover, if {En } are disjoint in F, then λ(∪En )

= = = =

c1 λ1 (∪En ) + c2 λ2 (∪En ) X X c1 λ1 (En ) + c2 λ2 (En ) X (c1 λ1 (En ) + c2 λ2 (En ) X λ(En ),

where we take the convention that ∞ − ∞ = 0. Thus, λ is countably additive, and therefore λ is a charge on F. Thus, Λ is a vector space. (b) Let M = {µ | µ is a measure on F}. Is M a vector space over R? Explain. Proof. No, M is not a vector space. Since measures cannot take on negative values, −µ ∈ / M for all µ ∈ M . However, M is related to a vector space in the sense that any finite linear combination of measures is a measure, provided that the constants used are positive (such a space is called a convex cone). Extra stuff: 2D. Let {An } be a sequence of subsets of a set X. If A consists of all x ∈ X which belong to infinitely many of the sets An , show that A=

∞ [ ∞ \

An .

m=1 n=m

The set A is often called the limit superior of {An } and is denoted by lim sup An . Proof. Let G =

∞ [ ∞ \

An .

m=1 n=m

Suppose xS∈ A. Then for all m, there exists nm > m such that x ∈ Anm . ∞ Thus x ∈ n=m An , and thus x ∈ G. Therefore, A ⊆ G. S∞ On the other hand, suppose x ∈ G. Then for all m, x ∈ n=m An , so for all m there exists nm > m such that x ∈ Anm . Therefore, x is in infinitely many of the sets An , so x ∈ A. Therefore, G ⊆ A. Thus G = A as desired. 2E. Let {An } be a sequence of subsets of a set X. If B consists of all x ∈ X which belong to all but a finite number of the sets An , show that B=

∞ \ ∞ [ m=1 n=m

3

An .

The set B is often called the limit inferior of {An } and is denoted by lim inf An . Proof. Let G =

∞ \ ∞ [

An .

m=1 n=m

Suppose x T∈ B. Then there exists N such that x ∈ An for all n ≥ N . ∞ Thus, x ∈ n=N An , so x ∈ G. Thus, B ⊆ G. On T∞ the other hand, suppose x ∈ G. Then there exists N such that x ∈ n=N An , so x ∈ An for all n ≥ N , and thus there are at most N − 1 sets An for which x ∈ / An . Therefore, x ∈ B, so G ⊆ B. Thus, G = B as desired. 3I. Let (X, F, µ) be a measure space and let {En } ⊂ F. Show that µ(lim inf En ) ≤ lim inf µ(En ). Proof. Let Fn = ∩m≥n Em , so that Fn = Fn+1 ∩ En , and thus Fn ⊂ Fn+1 . Moreover, lim inf En = ∪Fn . Then by Prop 3.4(a), we have µ(lim inf En ) = lim µ(Fn ). As Fn ⊂ En for all n, we have that µ(Fn ) ≤ µ(En ), and thus lim µ(Fn ) ≤ lim inf µ(En ), where we have traded the limit for the limit inferior, since we cannot be assured that lim µ(En ) exists. Therefore, µ(lim inf En ) ≤ lim inf µ(En ), as desired. 3J. Let (X, F, µ) be a measure space and let {En } ⊂ F. Show that lim sup µ(En ) ≤ µ(lim sup En ) when µ(∪En ) < ∞. Show that this inequality may fail if µ(∪En ) = ∞. Proof. As in the previous problem, take Gn = ∪m≥n Em , so that lim sup En = ∩Gn . Note that Gn = Gn+1 ∪ En , so Gn+1 ⊆ Gn , and by hypothesis, µ(G1 ) = µ(∪En ) < ∞. Therefore, by Prop 3.4(b), we have µ(lim sup En ) = µ(∩Gn ) = lim µ(Gn ). Moreover, as En ⊆ Gn , we have that µ(Gn ) ≥ µ(En ), so lim µ(Gn ) ≥ lim sup µ(En ), where again we have traded the limit for the limit superior, in case lim µ(En ) does not exist. Therefore, µ(lim sup En ) ≥ lim sup µ(En ), as desired. Note that if µ(∪En ) = ∞, we find ourselves in a similar situation as with problem 3H. Indeed, the same counterexample applies: take En = R\[−n, n] for all n. Then Fn = En for all n, and λ(Fn ) = λ(En ) = ∞ for all n. On the other hand, lim sup En = ∩Fn = ∅, so λ(lim sup En ) = 0 6≥ lim sup µ(En ) = ∞.

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