Math Math 426: Homew Homework ork 1 Soluti Solutions ons Mary Radcliffe due 9 April 2014
In Bartle: Bartle: 2B. Show that the Borel algebra B is also generated by the collection of all half-open intervals (a, (a, b] = {x ∈ R | a < x ≤ b}. Also Also show show that that B is generated by the collection of all half-rays { x ∈ R | x > a}. Proof. Let S Let S 1 = { (a, b) | a < b}, and S 2 = { (a, b] | a < b}. By definition,
σ(S 1 ) = B . To show that σ that σ (S 2 ) = B , it suffices to show that S 1 ⊂ σ( σ (S 2 ) ⊂ and S and S 2 σ( σ (S 1 ), as then, since σ (A) is the smallest σ -field containing A, the result follows immediately. Note that if a (a, b) = ∪ n∈Z+ (a, b − n1 ], and (a, (a, b] = ∩n∈Z+ (a, b + a < b, b , then (a, 1 ). Therefore, (a, (a, b) ∈ σ( σ (S 2 ) and (a, (a, b] ∈ σ( σ (S 1 ), completing completing the proof. n For the second property, let S 3 = { (a, ∞) | a ∈ show that S that S 3 ⊂ σ( σ (S 1 ) and S and S 1 ⊂ σ( σ (S 3 ).
R}.
Again, Aga in, it suffices suffices to
Let a < b. Then Then (a, (a, b) = (a, ∞)\ ∩n∈Z+ (b − n1 , ∞) ∈ σ( σ (S 3 ). Moreov Moreover, er, + (a, ∞) = ∪ n∈Z (a, a + n) ∈ σ( σ (S 1 ), completing completing the proof.
2K. Show Show directly that that if f is f is measurable and A and A > 0, then the truncation f A defined by f (x) if | | f ( f (x)| ≤ A ( ) = A if f ( f (x) > A f A x −A if f ( f (x) < − A is measurable.
Proof. It suffices to prove that for all α ∈ R , we have f have f A−1 ((α, ((α, ∞)) is mea-
surable. surable. Note that that if α > A, then f then f A−1 ((α, ((α, ∞)) = ∅ , which is measurable. −1 −1 If α = α = A A,, then f then f A (α) = f ([A, ([A, ∞)), which is measurable by Prop 2.4. If −A ≤ α < A, then f then f A−1 ((α, ((α, ∞)) = f −1 ((α, ((α, ∞)), which is measurable since −1 f f is. Finally Finally,, if α < −A, then f A ((α, ((α, ∞)) = X , which is measurable. Therefore, f Therefore, f A is measurable. 3E. Let X Let X be be an uncountable set and let F be the family of all subsets of X of X .. Define µ Define µ on on E E by by requiring that µ that µ((E ) = 0 if E E is is countable, and µ and µ((E ) = ∞ if E is E is uncountable. Show that µ is a measure on F . Proof. Clearly, µ(∅) = 0 and µ(E ) ≥ 0 for all E ∈ F . Let {E n } ⊂ F be
a sequence of pairwise disjoint subsets of F . If all E all E n are countable, then µ(E n ) = 0 for all n. n . Moreover, as the countable union of countably many subsets is countable, we have also that µ(∪E n ) = 0. Thu Thus, µ(∪E n ) = µ(E n ).
If not, there exists some k such that E k is is uncounta uncountable. ble. Then ∪E n is uncountable also, so µ so µ((∪E n) = ∞ = µ = µ((E k ) = µ(E n ).
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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 E n = { n}. Then E n is finite
for all n, so µ(E n ) = 0. However, ∪ E n = Z+ , so µ(∪E n ) = ∞. 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(F 1 ) < ∞ is dropped. Proof. Let F n = R \[−n, n], so that µ(F n ) = ∞ for all n, and F n ⊃ F n+1 .
Then ∩F n = ∅, so µ(∩F n ) = 0. However, lim µ(F n ) = ∞, and thus the result fails for { F n }. 3T. Show that the Lebesgue measure of the Cantor set C is zero. Proof. Let E 0 = [0, 1], E 1 = E 0 \( 13 , 32 ), E 2 = E 1 \( 19 , 92 )\( 79 , 98 ), etc., so
that C = ∩ E n . Then by Prop 3.4(b), since E n ⊃ E n+1 and λ(E 0 ) = 1 < ∞, we have λ(C ) = lim λ(E n ).
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+ 2n ) = b − a. Therefore, we have λ(E n ) = λ(E n−1 ) − 13 λ(E n−1 ) = n n 2 λ(E n−1 ). Thus, λ(E n ) = 23 , so λ(C ) = lim 23 = 0. 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 E 0 = [0, 1]. Take E 1 to be E 0 with the middle third interval deleted, as before: E 1 = E 0 \( 13 , 32 ). Now, to form E 2 , from each interval of E 1 , 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 take E 1 \( 24 , 24 )\( 19 , 21 ), so that the total measure of the removed sets is 24 24 1 . Similarly, construct E k from E k−1 by choosing an interval from each 6 connected set in E k−1 , the total measure of which will be 3∗12 . Then take 1 E = ∩ E k . By Lemma 3.4(b) we have that λ(E ) = 1 − ∞ = 31 > 0, k =0 3∗2 but as we delete from every interval at every step, the resulting intersection contains no nonvoid open intervals. k
k
3V. Suppose that E is a subset of a set N ∈ F with µ(N ) = 0, but E ∈ / F . Then the sequence {f n }, f n = 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→∞ f n (x) = 0 for all x ∈ X . Moreover, χE (x) = 0
→ χ E (x)} has for all x ∈ / N . Thus, f n → χ E for all x ∈ / N , so { x | f n (x) measure 0 under µ. Therefore, f n converges µ-a.e. to χ E . 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 c 1 , c2 ∈ R , then c 1 λ1 + c2 λ2 ∈ Λ. Let λ = c 1 λ1 +c2 λ2 . Clearly λ(∅) = 0. Moreover, if {E n } are disjoint in F , then λ(∪E n ) = c1 λ1 (∪E n ) + c2 λ2 (∪E n )
(E ) + (E ) λ c λ (c λ (E ) + c λ (E )
= c1 = =
1
1
n
2
1
n
2
2
2
n
n
λ(E n ),
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 Explain.
R?
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 A n , show that ∞
A =
∞
An .
m=1 n=m
The set A is often called the limit superior of {An } and is denoted by limsup An . ∞
Proof. Let G =
∞
An .
m=1 n=m
Suppose x ∈ A. Then for all m, there exists n m > m such that x ∈ A n . Thus x ∈ ∞ An , and thus x ∈ G. Therefore, A ⊆ G. n=m m
∞ On the other hand, suppose x ∈ G. Then for all m, x ∈ n=m An , so for all m there exists n m > m such that x ∈ A n . Therefore, x is in infinitely many of the sets A n , so x ∈ A. Therefore, G ⊆ A. m
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 A n , show that ∞
B =
∞
m=1 n=m
3
An .
The set B is often called the limit inferior of {An } and is denoted by liminf An . ∞
Proof. Let G =
∞
An .
m=1 n=m
Suppose x ∈ B. Then there exists N such that x ∈ An for all n ≥ N . ∞ Thus, x ∈ n=N A n , so x ∈ G. Thus, B ⊆ G.
On the other hand, suppose x ∈ G. Then there exists N such that x ∈ ∞ A n , so x ∈ A n for all n ≥ N , and thus there are at most N − 1 sets n=N An for which x ∈ / A n. Therefore, x ∈ B, so G ⊆ B.
Thus, G = B as desired. 3I. Let (X, F , µ) be a measure space and let { E n } ⊂ F . Show that µ(liminf E n ) ≤ liminf µ(E n ). Proof. Let F n = ∩m≥n E m , so that F n = F n+1 ∩ E n , and thus F n ⊂
F n+1 . Moreover, lim inf E n = ∪F n . Then by Prop 3.4(a), we have µ(lim inf E n ) = lim µ(F n). As F n ⊂ E n for all n, we have that µ(F n) ≤ µ(E n ), and thus lim µ(F n ) ≤ liminf µ(E n ), where we have traded the limit for the limit inferior, since we cannot be assured that lim µ(E n ) exists. Therefore, µ(lim inf E n ) ≤ lim inf µ(E n ), as desired. 3J. Let (X, F , µ) be a measure space and let { E n } ⊂ F . Show that limsup µ(E n ) ≤ µ(lim sup E n ) when µ(∪E n ) < ∞. Show that this inequality may fail if µ(∪E n ) = ∞. Proof. As in the previous problem, take Gn = ∪ m≥n E m , so that lim sup E n =
∩Gn . Note that Gn = Gn+1 ∪ E n , so Gn+1 ⊆ Gn , and by hypothesis, µ(G1 ) = µ(∪E n ) < ∞. Therefore, by Prop 3.4(b), we have µ(lim sup E n ) = µ(∩Gn ) = lim µ(Gn ). Moreover, as E n ⊆ Gn , we have that µ(Gn ) ≥ µ(E n ), so lim µ(Gn ) ≥ limsup µ(E n ), where again we have traded the limit for the limit superior, in case lim µ(E n ) does not exist. Therefore, µ(lim sup E n ) ≥ limsup µ(E n ), as desired. Note that if µ(∪E n ) = ∞, we find ourselves in a similar situation as with problem 3H. Indeed, the same counterexample applies: take E n = R\[−n, n] for all n. Then F n = E n for all n, and λ(F n ) = λ(E n ) = ∞ for all n. On the other hand, limsup E n = ∩ F n = ∅ , so λ(lim sup E n ) = 0 ≥ limsup µ(E n ) = ∞ .
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