Solutions of Homework #3
1. Claim : SE is a σ-algebra on E.
Suppose that B ∈ SE ⇒ ∃A ∈ S such that B = A ∩ E.
∴ Bc = Ac∩ E.
Since S is a σ-algebra ⇒ Ac ∈ S ⇒ Bc ∈ SE.
Suppose Bi ∈ SE ⇒ ∃Ai ∈ S such that Bi = Ai∩ E.
∴T∞
i=1Bi =T∞
i=1(Ai∩ E) = (T∞
i=1Ai) ∩ E.
Since S is a σ-algebra ⇒T∞
i=1Ai ∈ S ⇒T∞
i=1Bi ∈ SE. Therefore SE is a σ-algebra on E.
Claim : ν is a measure on E.
(1) : ν is well-defined.
By the assumption, µ∗(X) = µ∗(E), we have
µ∗(X) = µ∗(A∩X)+µ∗(Ac∩X) = µ∗(A∩E)+µ∗(Ac∩E) = µ∗(E) Since A = A ∩ X ⊇ A ∩ E ⇒ µ∗(A) = µ∗(A ∩ X) ≥ µ∗(A ∩ E), and Ac = Ac∩ XsupseteqAc ∩ E ⇒ µ∗(Ac) = µ∗(Ac∩ X) ≥ µ∗(Ac ∩ E). ∵ µ∗(X) is finite and µ∗(X) = µ∗(E), we have
µ∗(A) = µ∗(A ∩ E) and µ∗(Ac) = µ∗(Ac∩ E).
Similarly, µ∗(B) = µ∗(B ∩ E). Therefore
µ(A) = µ∗(A) = µ∗(A ∩ E) = µ∗(B ∩ E) = µ∗(B) = µ(B).
(2) :ν(∅) = 0
ν(∅) = ν(∅ ∩ E) = ν(∅) = 0.
(3) :A ∩ E ⊇ B ∩ E ⇒ ν(A ∩ E) ≥ ν(B ∩ E).
Since ν is well-defined, ∴ (A ∩ E) ⊇ (B ∩ E) ⇒ µ(A) ≥ µ(B).
Terefore ν(A ∩ E) = µ(A) ≥ µ(B) = ν(B ∩ E).
(4) :Countable additivity.
Let Bi = Ai∩ E ∈ SE, Ai ∈ S and Bi∩ Bj = ∅, ∀i 6= j.
Let C1 = A1, Ci = Ai\ (Si−1
j=1), ∀i ≥ 2.
∴ Ci ∈ S, S∞
i=1Ai =S∞
i=1Ci, Ci disjoint, and Ci∩ E = Bi = Ai∩ E.
Therefore ν(
∞
[
i=1
Bi) = nu(
∞
[
i=1
(Ai∩ E)) = µ(
∞
[
i=1
Ai)
= µ(
∞
[
i=1
Ci) =
∞
X
i=1
µ(Ci)
=
∞
X
i=1
µ(Ai) =
∞
X
i=1
ν(Ai∩ E) =
∞
X
i=1
ν(Bi)
2. (i) :
Suppose B ∈ A, ∴ B = ((a1, b1] ∪ . . . ∪ (an, bn]) ∩ X, where
−∞ ≤ a1 ≤ b1 ≤ a2 ≤ b2 ≤ . . . ≤ an≤ bn≤ ∞.
∴ Bc on X = ((−∞, a1] ∪ (Sn−1
i=1(bi, ai+1]) ∪ (an, ∞]) ∩ X
∴ Bc ∈ A.
Suppose C ∈ A, ∴ C = ((c1, d1] ∪ . . . ∪ (cm, dm]) ∩ X, where
−∞ ≤ c1 ≤ d1 ≤ c2 ≤ d2 ≤ . . . ≤ cm ≤ dm ≤ ∞.
If (a, b] ∩ (c, d] 6= ∅, then (a, b] ∪ (c, d] = (e, f ], for some e, f . Therefore,
B ∪ C = (
n
[
i=1
(ai, bi] ∩ X) ∪ (
m
[
j=1
(cj, dj] ∩ X)
= (
n
[
i=1
(ai, bi] ∪
m
[
j=1
(cj, dj]) ∩ X = (
l
[
k=1
(ek, fk]) ∩ X, for some ek, fk
∴ B ∪ C ∈ A and A is an algebra.
(ii) :
Since 2X is a σ-algebra and A ⊂ 2X ⇒ σ(A) ⊂ 2X. Consider x ∈ Q, since {x} =T∞
n=1((x −1n, x] ∩ X) ⇒ {x} ∈ σ(A).
Suppose S ∈ 2X ⇒ S = {x1, x2, . . .} by the fact that X is count- able.
∴ S =S∞
i=1{xi}, xi ∈ Q, ∀i.
Since {xi} ∈ σ(A) ⇒ S ∈ σ(A) ⇒ 2X ⊂ σ(A).
Hence 2X = σ(A).
(iii) : Let
µ1(S) = 0, if S = ∅
∞, otherwise and µ2(S) = the cardinality of S.
Suppose A ∈ A.
If A = ∅, µ1(∅) = µ2(∅) = µ(∅) = 0.
If A 6= ∅ ⇒ A = {x1, x2, . . .} ⇒ µ1(A) = µ2(A) = µ(A) = ∞.
Therefore µ1 and µ2 are the extension of µ to σ(A).
But µ1({x}) = ∞ 6= 1 = µ({x}).
Hence the extension of µ to σ(A) is not unique.
3. N ote :S is a σ-algebra ,and (X, S, µ) is a measure space, but may not be complete.
(i) By definition, N (µ) = {E ⊂ Xkµ∗(E) = 0}.
10 Let A = {E ⊂ XkE ⊂ N for some N ∈ S with µ(N ) = 0}.
∀E ∈ A, ∃N ∈ S with E ⊂ N such that µ(N ) = 0.
⇒ 0 ≤ µ∗(E) ≤ µ∗(N ) = µ(N ) = 0.
∴ µ∗(E) = 0 ⇒ E ∈ N (µ).
On the other hand, given E ∈ N (µ), ∃ measurable cover B ∈ S of E such that E ⊂ B and µ(B) = µ∗(E) = 0.
∴ E ∈ A.
Hence A = N (µ).
20 By definition, S ∨ N (µ) = the σ-algebra generated by S ∪ N (µ).
Let B = {E ∪ F kE ∈ S and F ∈ N (µ)}.
Obviously, B ⊂ § ∨ N (µ).
Claim:B is a σ-algebra containing S ∪ N (µ).
pf:Since ∅ ∈ S ∩ N (µ), ∀E ∈ S, F ∈ N (µ)
⇒ E = E ∪ ∅ ∈ B and F = ∅ ∪ F ∈ B.
∴ S ∪ N (µ) ⊂ B. Now, we’re going to prove that B is a σ- algebra.
First, it’s easy to see ∅ ∈ B.
Second, given {Bn}∞n=1⊂ B
⇒ Bn= En∪ Fn, for some En∈ S and Fn∈ N (µ). Thus,
∞
[
n=1
Bn=
∞
[
n=1
(En∪ Fn) = (
∞
[
n=1
En) ∪ (
∞
[
n=1
Fn). (1) Since S is a σ-algebra, S∞
n=1En ∈ S.
Now we can choose a measurable cover Nn of Fn, form the problem4, we know that S∞
n=1Nn is also a measurable cover of S∞
n=1Fn. Therefore µ∗(
∞
[
n=1
Fn) ≤ µ(
∞
[
n=1
Nn) ≤
∞
X
n=1
µ(Nn) = 0
⇒ µ∗(S∞
n=1Fn) = 0 ⇒ S∞
n=1Fn ∈ N (µ) (by 10). Hence by (1), S∞
n=1Bn∈ B.
Third, given B ∈ B ⇒ B = E ∪ F , for some E ∈ S and F ∈ N (µ).
⇒ ∃N ∈ S such that F ⊂ N and µ(N ) = 0.
⇒ Bc = (E ∪ F )c = Ec∩ Fc = (Ec∩ Nc) ∪ (N \ F ).
Since (N \ F ) ⊂ N and µ(N ) = 0, (N \ F ) ∈ N (µ).
Therefore Bc ∈ B, since E, N ∈ S, and we complete the proofof the claim.
(ii) : ¯µ is the completion of µ.
⇒ Dom(¯µ) = S ∨ N (µ).
(⇒) : ∀E ⊂ Dom(¯µ).
⇒ E ∈ S∨N (µ) = Sµ∗ = {E ⊂ XkE M B ∈ N (µ), for some B ∈ S}.
⇒ ∃B ∈ S s.t. E M B ∈ N (µ).
⇒ Let C = E ∪ B, A = E ∩ B.
Then A ⊂ E ⊂ C and A, C ∈ Dom(¯µ), and therefore we have 0 ≤ ¯µ(C \ A) = ¯µ(C \ E) + ¯µ(E \ A)
≤ 2¯µ(B M E) = 2µ(B M E) = 0.
Therefore µ(C \ A) = ¯µ(C \ A) = 0 (⇐) : For any E with the following property,
∃A, C ∈ Dom(¯µ) with A ⊂ E ⊂ C s.t. µ(C \ A) = 0.
Since (E \A) (C \A) and µ(C \A) = 0, we have E \A ∈ N (µ).
Consider E = A ∪ (E \ A).
Since A ∈ S and (E \ A) ∈ N (µ), we have E ∈ S ∨ N (µ) = Dom(¯µ).
4. (i) : Let Cj be a measurable cover of Aj. Then Cj is a µ∗-measurable set and
µ∗(Cj) = µ∗(Aj) with Cj ⊃ Aj. Now, let B be a measurable cover of S
jAj, then µ∗([
j
Aj) = µ∗(B) and [
j
Aj ⊂ B.
Set ˜Cj = B ∩ Cj. Thus Aj ⊂ ˜Cj. Moreover, we have µ∗(Aj) ≤ µ∗( ˜Cj) ≤ µ∗(Cj).
∴ ˜Cj is a measurable cover of Aj. Consider µ∗(Cj M ˜Cj) = µ∗(Cj \ ˜Cj).
(a) : If µ∗( ˜Cj) = ∞ for some j
Then µ∗(Aj) = µ∗(Cj) = µ∗( ˜Cj) = ∞
⇒ µ∗(S
jCj) = µ∗(S
jAj) = ∞ and (S
jCj) ⊃ (S
jAj) ThereforeS
jCj is a measurable cover ofS
jAj. (b) : If µ∗( ˜Cj) < ∞, ∀j.
Then µ∗(Cj M ˜Cj) = µ∗(Cj) − µ∗( ˜Cj) = µ∗(Aj) − µ∗(Aj) = 0.
By homework#2, µ∗([
j
Cj) = µ∗( ˜Cj) ≤ µ∗(B) = µ∗([
j
Aj)
Since (S
jAj) ⊂ (S
jCj), we have µ(S
jCj) = µ∗(S
jAj).
ThereforeS
jCj is also a measurable cover ofS
jAj (ii) There are many examples! For example:
(1) X = R, S = {A ⊂ X|A is countable}∪{A ⊂ X|Ac is countable}.
Write A1 = {A ⊂ X|A is countable} and A2 = {A ⊂ X|Ac is countable}.
Define
m(A) = 0, if A ∈ A1 1, if A ∈ A2
Let A1 = (−∞, 0), A2 = (0, ∞) ⇒ µ∗(A1) = µ∗(A2) = 1.
Let C1 = C2 = R, then µ∗(Ai) = µ(Ci) = 1, i = 1, 2.
Thus, µ∗(A1∩ A2) = 0 6= 1 = µ(R) = µ(C1∩ C2).
(2) X = R and m is the counting measure.
Set A1 = (0, 1) and A2 = (1, 2) and C1 = [0, 1] , C2 = [1, 2].
Then µ∗(Ai) = µ∗(Ci) = ∞, i = 1, 2 and µ∗(A1∩ A2) = 0 6= 1 = µ∗(C1∩ C2).
