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Folland: Real Analysis, Chapter 1
S´ebastien Picard
Problem 1.5
If M is the σ-algebra generated by E, then M is the union of the σ-algebras generated by F as F
ranges over all countable subsets of E. (Hint: Show that the latter object is a σ-algebra.)
Solution: Let N denote the union of the σ-algebras generated by F as F ranges over all count-
able subsets of E.
N = [ M(F):
F⊂E; F countable
When F ⊂ E, the σ-algebra generated by F is contained in the σ-algebra generated by E ie
M(F)⊂M(E)=M,sowehaveN ⊂M.
For the reverse containement, we first observe that E ⊂ N. Indeed, if E = {A } where J
α α∈J
is some index set, then
E ⊂ [M({A })⊂N:
α
α∈J
Next, we will show that N is a σ-algebra. Since M(E) is the smallest σ-algebra containing E, we
will have M = M(E) ⊂ N and this will complete the proof.
c
Let A ∈ N. Then A ∈ M(F) for some countable F ⊂ E. Therefore A ∈ M(F) ⊂ N. Next,
∞ ∞
consider ∪ A where A ∈ N. Then each A ∈ M(F ) for some countable F ⊂ E. Since ∪ F is
i=1 i i i i i j=1 j
∞ ∞ ∞
countable, we have A ∈ M(∪ F ) ⊂ N for each A . Hence ∪ A ∈M(∪ F)⊂N.
i j=1 j i i=1 i j=1 j
Problem 1.12
Let (X;M;µ) be a finite measure space.
a. If E;F ∈ M and µ(E∆F) = 0, then µ(E) = µ(F).
b. Say that E ∼ F if µ(E∆F) = 0; then ∼ is an equivalence relation on M.
c. For E;F ∈ M, define ρ(E;F) = µ(E∆F). Then ρ(E;G) ≤ ρ(E;F)+ρ(F;G), and hence ρ
defines a metric on the space M= ∼ of equivalence classes.
Solution:
(a) First, notice
0 = µ(E∆F)=µ(E\F ∪ F\E)=µ(E\F)+µ(F\E):
Hence µ(E\F) = µ(F\E) = 0. It follows that
µ(E) = µ(E\F)+µ(E∩F)=µ(E∩F);
1
µ(F) = µ(F\E)+µ(F ∩E)=µ(E∩F):
Therefore, µ(E) = µ(F).
(b) It is clear that ∼ is reflexive: µ(E∆E) = µ(∅) = 0. It is also easy to see that ∼ is symmetric: if
E∼F,thenµ(E∆F)=µ(F∆E)=0,soF ∼E.
It only remains to show that ∼ is transitive. Suppose E ∼ F and F ∼ G. Then µ(E ∩ Fc) +
µ(Fc∩E)=0andµ(F ∩Gc)+µ(Fc∩G)=0. Therefore,
µ(E∆G)=µ(E∩Gc)+µ(Ec∩G)
=µ(E∩Gc∩Fc)+µ(Ec∩G∩F)+µ(E∩Gc∩F)+µ(Ec∩G∩Fc)
≤µ(E∩Fc)+µ(Ec∩F)+µ(F ∩Gc)+µ(Fc∩G)
=0+0=0:
Since µ(E∆G) ≥ 0, we must have µ(E∆G) = 0.
(c) We show the triangle inequality. Given E;F;G ∈ M, then
ρ(E;G) = µ(E∆G)
=µ(E∩Gc∩Fc)+µ(E∩Gc∩F)+µ(G∩Ec∩F)+µ(G∩Ec∩Fc)
=(µ(E∩Gc∩Fc)+µ(E∩G∩Fc)+µ(F∩Gc∩Ec)+µ(F ∩G∩Ec))
+(µ(F ∩Gc∩Ec)+µ(F ∩Gc∩E)+µ(G∩Ec∩Fc)+µ(G∩E∩Fc)):
=ρ(E∆F)+µ(F∆G)=ρ(E;F)+ρ(F;G):
Problem 1.16
Let (X;M;µ) be a measure space. A set E ⊂ X is called locally measurable if E ∩ A ∈ M for all
˜ ˜
A∈Msuchthat µ(A)<∞. Let M be the collection of all locally measurable sets. Clearly M ⊂ M;
˜
if M = M, then µ is called saturated.
a. If µ is σ-finite, then µ is saturated.
˜
b. M is a σ-algebra.
˜
c. Define µ˜ on M by µ˜(E) = µ(E) if E ∈ M and µ˜(E) = ∞ otherwise. Then µ˜ is a saturated
˜
measure on M, called the saturation of µ.
d. If µ is complete, so is µ˜
˜
e. Suppose that µ is semifinite. For E ∈ M, define µ(E) = sup{µ(A) : A ∈ M and A ⊂ E}.
Then µ ˜
is a saturated measure on M that extends µ.
f. Let X1;X2 be disjoint uncountable sets, X = X1 ∪ X2, and M the σ-algebra of count-
able or co-countable sets in X. Let µ0 be the counting measure on P(X1), and define µ on M by
˜
µ(E) = µ0(E ∩ X1). Then µ is a measure on M, M = P(X), and in the notation of parts (c) and
(e), µ˜ 6= µ.
2
Solution:
˜ ∞
(a) Suppose µ is σ-finite. Let A ∈ M, and let X = ∪ E where E ∈M and µ(E )<∞. Notice
j=1 j j j
∞ ! ∞
A=A∩ [E =[A∩E:
j j
j=1 j=1
Each E ∩A∈M since A is locally measurable, and since A is a countable union of these sets, we
j
˜
have A ∈ M. Hence M ⊂ M and µ is saturated.
˜
(b) Let E ∈ M. Take any A ∈ M such that µ(A) < ∞. Then
c c c c
E ∩A=A∩(E∩A) =(A ∪(E∩A)) ∈M:
c ˜ ∞ ˜
Hence E ∈ M. Next, consider ∪ E, where each E ∈ M. Then for all A ∈ M such that
i=1 i i
µ(A) < ∞, we have
∞ ! ∞
[E ∩A=[(E ∩A)∈M:
i i
i=1 i=1
˜
Hence M is a σ-algebra.
˜ ˜
(c) We show µ˜ is a measure on M. We have µ˜ : M → [0;∞] and µ˜(∅) = µ(∅) = 0, so it only remains
∞ ˜
to show countable additivity. Let {E } denote a sequence of disjoint sets in M. We partition
j j=1
∞ ∞ ! ∞ !
[E = [A ∪ [F ;
j j j
j=1 j=1 j=1
where A ;F ∈ {E }∞ ∪{∅}, A ∈ M, and F ∈= M or F = ∅. By not selecting the same E twice,
j j j j=1 j j j k
we can make sure that the elements of {A ;A ;:::;F ;F ;:::} are all pairwise disjoint.
1 2 1 2
P P
∞ ∞ ∞ ∞
Case 1: ∪ F =∅. Then µ˜(∪ E)= µ(E ) = µ˜(E ).
j=1 j j=1 j j=1 j j=1 j
∞ S S
Case 2: ∪ F ∈= M. Then some E ∈= M, and µ˜(E ) = ∞. Furthermore, ( A)∪( F)∈= M:
j=1 j k k j j
indeed, if (SA )∪(SF ) ∈ M, then
j j
∞ ∞ ∞ ∞
[F =((([A)∪([F))c∪([A))c∈M:
j j j j
j=1 j=1 j=1 j=1
Therefore, we have
∞ ∞ ! ∞ !! ∞
µ˜([E ) = µ˜ [A ∪ [F =∞=Xµ˜(E):
j j j j
j=1 j=1 j=1 j=1
3
Case 3: SF 6= ∅ and SF ∈ M. In this case we have
j j
∞ ∞ ∞
µ˜([E )=Xµ(A )+µ([F ):
j j j
j=1 j=1 j=1
Hence we need to show that µ(SF ) = ∞. Suppose µ(SF ) < ∞. Choose F such that F 6= ∅.
j j j j
Then since F is locally measurable, we have
j
∞ !
F =F ∩ [F ∈M:
j j j
j=1
This contradiction shows that µ˜(S∞ E ) ≥ µ(SF ) = ∞. Since there is some E ∈= M, we have
j=1 j j k
Pµ˜(E ) = ∞.
j
˜ ˜ ˜
Wenowprove that µ˜ is saturated. Let E ⊂ X be such that E ∩A ∈ M when µ˜(A) < ∞. Choose
˜
A∈Msuchthatµ(A)<∞. Then µ˜(A) <∞, and hence E ∩A ∈ M. But then (E ∩A)∩A ∈ M.
˜
Therefore, E ∩A ∈ M and E ∈ M. It follows that µ˜ is saturated.
˜
(d) Suppose µ is complete. If E ∈ M is such that µ(E) = 0, then µ˜ < ∞ and hence E ∈ M.
˜
Therefore, for all A ⊂ E, we have A ∈ M by completeness of µ. Therefore, A ∈ M and µ˜ is complete.
˜
(e) We prove that µ is a measure on M. Since µ(∅) = µ(∅) = 0, it remains to show countable
˜ S
additivity. Let {E } be a collection of disjoint sets, with E ∈ M. Let A ∈ M be such that A ⊂ E .
j j j
Case 1: µ(A) < ∞. Then
∞ ! ∞ ∞
µ(A) = µ [(A∩E) =Xµ(A∩E)≤Xµ(E):
j j i
j=1 j=1 j=1
˜ ˜
Case 2: suppose µ(A) = ∞. By semifiniteness, for all c > 0 there exists A ⊂ A such that A ∈ M
˜
and µ(A) = c. (Indeed, this is the definition of semifinite in Royden. I do not remember if it is
pointed out directly in Folland, but in any case it is not hard to prove from Folland’s definition.) By
case 1, c ≤ Pµ(E ), so Pµ(E ) = ∞.
i i
Therefore, µ(A) ≤ P∞ µ(E ): Taking the supremum over all such A, we have
j=1 i
∞ ! ∞
µ [E ≤Xµ(E):
j j
j=1 j=1
We now show the reverse inequality. By the definition of the supremum, there exists a sequence
4
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