SOLUTIONS OF HW3
MINGFENG ZHAO
February 20, 2011
1. (Page 32, 18) Let A ⊂ P(X) be an algebra, Aσ the collection of countable unions of sets in A, and
Aσδ the collection of countable intersections of sets in Aσ . Let µ0 be a premeasure on A and µ∗ the
induced outer measure.
a. For any E ⊂ X and > 0, there exists A ∈ Aσ with E ⊂ A and µ∗ (A) ≤ µ∗ (E) + .
b. If µ∗ (E) < ∞, then E is µ∗ -measurable iff there exists B ∈ Aσδ with E ⊂ B and µ∗ (B\E) = 0.
c. If µ0 is σ-finite, the restriction µ∗ (E) in (b) is superfluous.
Proof. Let Ac = {E ⊂ X : E is µ∗ - measurable.}, and σ(A) be the σ-algebra generated by the
algebra A. By Caratheodory’s theorem, in Page 29, 1.13 Proposition in Page 30, and 1.14 Theorem
in Page 30, we know that
A ⊂ Aσ ⊂ Aσδ ⊂ σ(A) ⊂ Ac ,
(1)
and for any F ∈ A, we have µ∗ (F ) = µ(F ).
a. For any E ⊂ X, by the definition of outer measure, we know that for any > 0, there exists
S∞
{Fk }∞
k=1 ⊂ A such that E ⊂
k=1
Fk , and
µ∗ (E) ≥
∞
X
µ(Fk ) − .
k=1
Let A =
S∞
k=1
Fk , then A ∈ Aσ , and E ⊂ A, and
∗
µ (A)
=
∗
µ
∞
[
!
Fk
k=1
≤
∞
X
µ∗ (Fk )
By the subadditivity of the outer measure µ∗
k=1
1
2
MINGFENG ZHAO
=
∞
X
µ(Fk )
By 1.13 Proposition in Page 30
k=1
≤ µ∗ (E) + .
b. Assume µ∗ (E) < ∞.
(=⇒) If E is µ∗ -measurable. By party a, we know for any n, there exists An ∈ Aσ such that
µ∗ (E) ≥ µ∗ (An ) +
1
n,
and E ⊂ An . Let B =
T∞
n=1
An , then B ∈ Aσδ , and E ⊂ B ⊂ An for all
n ≥ 1. By (1), we know that An , B, ∈ Ac , so An , B are µ∗ -measurable. Since µ∗ (E) < ∞, then
lim inf n→∞ µ∗ (An ) < ∞. Hence
µ∗ (B) ≤ lim inf µ∗ (An )
n→∞
1
≤ lim inf µ∗ (E) −
n→∞
n
= µ∗ (E).
So µ∗ (B) = µ∗ (E). Since E ⊂ B, and E, B are µ∗ measurable, then B\E is µ∗ -measurable, and
µ∗ (B\E) = µ∗ (B) − µ∗ (E) = 0.
(⇐=) If there exists B ∈ Aσδ with E ⊂ B and µ∗ (B\E) = 0. By (1), we know that B is µ∗ measurable. Since µ∗ is complete on Ac . Since µ∗ (B\E) = 0, so B\E is µ∗ -measurable. Since E ⊂ B,
then E = B\(B\E) is µ∗ -measurable.
c. Assume µ0 is σ-finite, that is, there exists {Fk }∞
k=1 ⊂ A such that X =
S∞
k=1
Fk , and µ(Fk ) < ∞
for all k ≥ 1.
(=⇒) For any E is µ∗ -measurable, then consider Ek = E
S∞
k=1
T Sk
i=1
S∞
Fk . Since X = k=1 Fk , then
Ek = E. Also we can see that Ek ⊂ Ek+1 , then Ek % E, as k → ∞.
SOLUTIONS OF HW3
3
And Since E, Fk are µ∗ -measurable for all k ≥ 1, then Ek is µ∗ -measurable for all k ≥ 1. Moreover,
since µ(Fk ) < ∞ for all k ≥ 1, then
∗
k
[
∗
µ (Ek ) ≤
µ
!
By the monotonicity of the outer measure µ∗
Fk
i=1
k
X
≤
µ∗ (Fk )
By the subadditivity of the outer measure µ∗
i=1
k
X
=
µ(Fk )
By 1.13 Proposition in Page 30
i=1
∞.
<
By the result of part a, for any k, there exists Ak ∈ Aσ such that Ek ⊂ Ak , and µ∗ (Ak ) ≤
µ∗ (Ek ) + 21k . Since Ek , Ak are µ∗ -measurable, then we can get µ∗ (\Ek ) = µ∗ (Ak ) − µ∗ (Ek ) ≤
B=
∞ [
∞
\
1
.
2k
Let
Al .
k=1 l=k
So B ∈ Aσδ .
Claim I: E ⊂ B.
For any x ∈ E, since Ek % E, as n → ∞, so there exists K ≥ 1, whenever k ≥ K, we have
x ∈ Ek ⊂ Ak , which implies that x ∈ B =
Claim II: For any k ≥ 1, we have B\E ⊂
T∞ S∞
k=1
l=k
Al . By the arbitrary of x ∈ E, we get E ⊂ B.
S∞
l=k (Al \El ).
If not, that is, there exists some k ≥ 1, x ∈ B\E such that x ∈
/
Since x ∈
/
S∞
l=k (Al \El ),
l ≥ k. Since x ∈ B =
S∞
l=k (Al \El ).
then for any x ∈
/ Al \El whenever l ≥ k. Since x ∈
/ E, then x ∈
/ El for all
T∞ S∞
k=1
l=k
Al , then there are infinitely many l ≥ 1 such that x ∈ Al . For such
l ≥ k, we know x ∈ Al , but x ∈
/ El , so x ∈ Al \El , which contradicts with for any x ∈
/ Al \El whenever
l ≥ k. Therefore, for any k ≥ 1, we have B\E ⊂
S∞
l=k (Al \El ).
By Claim II, we know that for any k ≥ 1, we have
∗
∗
µ (B\E) ≤ µ
∞
[
l=k
!
(Al \El )
By the monotonicity of the outer measure µ∗
4
MINGFENG ZHAO
∞
X
≤
µ∗ (Al \El )
By the subadditivity of the outer measure µ∗
l=k
∞
X
1
2l
≤
l=k
1
=
2k−1
.
Take k → ∞, we get µ∗ (B\E) = 0.
(⇐=) If there exists B ∈ Aσδ with E ⊂ B and µ∗ (B\E) = 0. By (1), we know that B is µ∗ measurable. Since µ∗ is complete on Ac . Since µ∗ (B\E) = 0, so B\E is µ∗ -measurable. Since E ⊂ B,
then E = B\(B\E) is µ∗ -measurable.
2. (Page 32, 19) Let µ∗ be an outer measure on X induced from a finite premeasure µ0 . If E ⊂ X,
define the inner measure of E to be µ∗ (E) = µ0 (X) − µ∗ (E c ). Then E is µ∗ -measurable iff
µ∗ (E) = µ∗ (E).
Proof. Let Ac = {E ⊂ X : E is µ∗ - measurable.}, and σ(A) be the σ-algebra generated by the
algebra A. By Caratheodory’s theorem, in Page 29, 1.13 Proposition in Page 30, and 1.14 Theorem
in Page 30, we know that
A ⊂ Aσ ⊂ Aσδ ⊂ σ(A) ⊂ Ac ,
and for any F ∈ A, we have µ∗ (F ) = µ0 (F ).
(=⇒) Assume E is µ∗ -measurable, then E c is µ∗ -measurable. By the additivity of the measure µ∗ ,
we know that
µ∗ (X) = µ∗ (E) + µ∗ (E c ).
By 1.13 Proposition in Page 30, we know µ∗ (X) = µ0 (X). Since µ0 (X) < ∞, then we get
µ∗ (E) = µ∗ (X) − µ∗ (E c ) = µ0 (X) − µ∗ (E c ) = µ∗ (E).
SOLUTIONS OF HW3
5
(⇐=) If µ∗ (E) = µ∗ (E). By 1.13 Proposition in Page 30, we know µ∗ (X) = µ0 (X). By the monotonicity of the outer measure µ∗ , we know µ∗ (E) < ∞, and µ∗ (E c ) < ∞. By the result of part a in Problem
18, for any k ≥ 1, there exists Ak , Bk ∈ Aσ which are µ∗ -measurable, such that E ⊂ Ak , E c ⊂ Bk ,
and µ∗ (Ak ) ≤ µ∗ (E) + k1 , µ∗ (Bk ) ≤ µ∗ (E c ) + k1 .
Let A =
T∞
k=1
Ak , B =
T∞
k=1
Bk , then A, B ∈ Aσδ , so A, B are µ∗ -measurable, and E ⊂ A, and
E c ⊂ B. By the monotonicity of the outer measure µ∗ , we have µ∗ (E) ≤ µ∗ (A), µ∗ (E c ) ≤ µ∗ (B).
Moreover, we have
µ∗ (A) ≤
lim inf µ∗ (Ak )
k→∞
≤
1
lim inf µ∗ (E) −
k→∞
k
=
µ∗ (E).
and
µ∗ (B) ≤
lim inf µ∗ (Bk )
k→∞
≤
1
lim inf µ∗ (E c ) −
k→∞
k
=
µ∗ (E c ).
Since µ∗ (E) = µ0 (X) − µ∗ (E c ) = µ∗ (E), then µ∗ (A) = µ0 (X) − µ∗ (B).
Claim I: µ∗ (A
T
B) = 0.
Since A, B are µ∗ -measurable, then Ac is µ∗ -measurable, and
µ∗ (A) + µ∗ (Ac ) = µ∗ (X) = µ0 (X).
Since µ0 (X) < ∞, and µ∗ (A) = µ0 (X) − µ∗ (B), so µ∗ (Ac ) = µ∗ (B). Since E ⊂ A, then Ac ⊂ E c ⊂
B. By the additivity of measure µ∗ , we can get
µ∗ (B\Ac )
= µ∗ (B) − µ∗ (Ac )
6
MINGFENG ZHAO
=
And B\Ac = B
T
Claim II: A\E ⊂ A
(Ac )c = B
T
T
A, so µ∗ (A
T
0.
B) = 0.
B.
In fact, we get
A\E
= A
\
Ec
⊂ A
\
B
Since E c ⊂ B
By Claim II, we know µ∗ (A\E) = 0. Since A ∈ Aσδ , then by the result of part b in Problem 18,
we know E is µ∗ -measurable.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu