Solutions of Homework #2 1.

Solutions of Homework #2

1. A = {A ⊂ X|A or A^c is finite}.

(a)

(A is an algebra)

(i) ∵ ∅ = X^c is finite, ∴ ∅, X ∈ A.

(ii) Let A ∈ A.

If A is finite ⇒ (A^c)^c = A is finite ⇒ A^c ∈ A.

If A^c is finite ⇒ A^c ∈ A.

(iii) Let {A_n}^k_n=1 ⊂ A.

If A⁰_ns are finite. ⇒Sk

n=1A_n is finite. ∴Sk

n=1A_n ∈ A.

If there is A_n0 ∈ A such that A^c_n0 is finite.

⇒ (Sk

n=1A_n)^c =Tk

n=1A^c_n ⊂ A^c_n

0, ∴ (Sk

n=1A_n)^c is finite.

∴Sk

n=1A_n∈ A.

(A is not a σ-algebra)

Let X = {x_n}^∞_n=1 and A_k = {x_2k} for k = 1, 2, . . ..

Then S∞

k=1A_k is infinite and (S∞

k=1A_k)^c = {x_2k−1}^∞_k=1 is also in- finite. Therefore S∞

k=1A_k doesn’t belong to A and A is not a σ-algebra.

(b) Let {A_n}^k_n=1 ⊂ A be such that A_n∩ A_m = ∅, ∀m 6= n. Then (i) If A⁰_ns are finite, then Sk

n=1A_n is finite.

⇒ m(Sk

n=1A_n) = 0 =Pk

n=1m(A_n).

(ii) If there is An0 such that (An0)^c is finite, then m(An0) = 1.

⇒ (Sk

n=1A_n)^c =Tk

n=1(A_n)^c ⊂ A_n0 is finite.

∴ m(Sk

n=1A_n) = 1.

On the other hand, ∵ Aⁿ∩ Am = ∅, ∀m 6= n

∴S

n6=n0A_n⊂ (A_n0)^c.

∴S

n6=n0A_n is finite. ⇒ m(S

n6=n0A_n) = 0 =P

n6=n0m(A_n).

Hence m(Sk

n=1A_n) = 1 = m(A_n0)+0 = m(A_n0)+S

n6=n0m(A_n) = Pk

n=1m(A_n).

(c) If X is uncountable, then m can be extended to a countably ad- ditive measure on a σ-algebra.

Claim: m is countably additive on A.

pf: Let {A_n}^∞_n=1⊂ A be such that

An∩ Am = ∅, ∀n 6= m and

∞

[

n=1

An∈ A.

(i) If A_n is finite, ∀n and S∞

n=1A_n ∈ A, then we will claim S∞

n=1A_n is finite.

if not, (S∞

n=1An)^c is finite.

Because X is uncountable, we know thatS∞

n=1A_n is also un- countable.

Then we get a contradiction by the fact that the A⁰_ns are finite.

(ii) If there is A_n0 such that (A_n0)^c is finite, then (S∞

n=1A_n)^c ⊂ (A_n0)^c is finite.

∴ m(S∞

n=1A_n) = 1.

On the other hand, ∵ An∩ A_n0 = ∅, ∀n 6= n₀.

∴S

n6=n0A_n⊂ A^c_n0 ⇒S

n6=n0A_n is finite.

∴ m(S

n6=n0A_n) = 0 =S

n6=n0m(A_n).

So m(S∞

n=1A_n) = 1 = m(A_n0) = P∞

n=1m(A_n).

Since m is countably additive on the algebra A, m can be extended to a countably additive measure on a σ-algebra.

If fact, if we set

S = {A ⊂ X|A or A^c is countable } and

˜

m(A) = 1, if A^c is countable 0, if A is countable

Then we can check that S is a σ-algebra and ˜m is a measure on S such that ˜m = m on A.

2.

µ^∗([

j

Aj) ≤ µ^∗([

j

Aj∪[

j

Bj)

= µ^∗([

j

(A_j ∪ B_j)) = µ^∗([

j

((A_j 4 B_j) ∪ (A_j ∩ B_j)))

= µ^∗(([

j

(A_j 4 B_j)) ∪ ([

j

(A_j∩ B_j)))

≤ µ^∗(([

j

(Aj 4 Bj)) + µ^∗([

j

(Aj ∩ Bj)))

≤X

j

µ^∗(A_j4 B_j) + µ^∗(([

j

A_j) ∩ ([

j

B_j)) = µ^∗(([

j

A_j) ∩ ([

j

B_j))

≤ µ^∗([

j

A_j)

Therefore µ^∗(S

jA_j) = µ^∗((S

jA_j) ∩ (S

jB_j)).

Similarly, µ^∗(S

jB_j) = µ^∗((S

jA_j) ∩ (S

jB_j)).

Hence, µ^∗(S

jA_j) = µ^∗(S

jB_j).

3. Let F be a closed subset of X. We want to show that ∀A ∈ 2^X µ^∗(A) = µ^∗(A ∩ F ) + µ^∗(A \ F ).

Consider A_n= {x ∈ A \ F |d(x, F ) ≥ ¹_n}.

Since ρ(A_n, A ∩ F ) = inf {d(x, y)|x ∈ A_n, y ∈ A ∩ F } > 0, by assump- tion we have

µ^∗(A_n∪ (A ∩ F )) = µ^∗(A_n) + µ^∗(A ∩ F ) Since An∪ (A ∩ F ) ⊂ A, therefore

µ^∗(A_n) + µ^∗(A ∩ F ) ≤ µ^∗(A) (1)

Claim(1): S∞

n=1A_n= A \ F . Obviously,S∞

n=1A_n⊂ A \ F . If not, ∃x ∈ (A \ F ) \S∞

n=1A_n. ⇒ d(x, F ) < ¹_n, ∀n.

Therefore d(x, F ) = 0. Since F is closed, we have x ∈ F , and therefore we get a contradiction.

If we can show µ^∗(S∞

n=1A_n) = lim_n→∞µ^∗(A_n), then let n → ∞ in (1), we have

µ^∗(A \ F ) + µ^∗(A ∩ F ) ≤ µ^∗(A)

Since the opposite inequality is always true, F is µ^∗-measurable.

Claim(2): µ^∗(S∞

n=1A_n) = lim_n→∞µ^∗(A_n) There are two proof of this claim:

pf(a): We will show the general case, that is,

If B_n

∞

[

n=1

, then µ^∗(

∞

[

n=1

B_n) = lim

n→∞µ^∗(B_n)

step1 : If {B_n} are µ^∗-measurable, then the equality holds.(We have showed this in the class.)

step2 : For each n, we can choose a measurable cover ˜C_n of B_n.

Let Cn = T∞

k=nC˜n, then Cn % and measurable. Moreover C˜_k ⊃ B_k ⊃ B_n, ∀k ≥ n. Therefore B_n⊂ C_n.

From the following inequality,

µ^∗(B_n) ≤ µ^∗(C_n) ≤ µ^∗( ˜C_n) = µ^∗(B_n)

we have µ^∗(Cn) = µ^∗(Bn). ie. Cn is also a measurable cover of B_n.

By problem 4 of Homework#3,S∞

n=1C_nis a measurable cover of S∞

n=1Bn.

Then by step1, µ^∗(S∞

n=1C_n) = lim_n→∞µ^∗(C_n).

Hence µ^∗(S∞

n=1B_n) = lim_n→∞µ^∗(B_n).

By Taking A_n= B_n, we proves this claim.

pf(b): (多數同學的證明)

If µ^∗(A) = ∞, then µ^∗(A ∩ F ) + µ^∗(A \ F ) ≥ µ^∗(A) = ∞.

Therefore µ^∗(A) = µ^∗(A ∩ F ) + µ^∗(A \ F ).

If µ^∗(A) < ∞.

Let B1 = A1, Bn= An\ An−1. Then

ρ(B_2k, B_(2(k+1))), ρ(B_2k−1, B_2k+1) > 0, ∀k ∈ N By assumption,

µ^∗(

∞

[

k=1

B_2k) =

∞

X

k=1

µ^∗(B_2k)µ^∗(

∞

[

k=1

B_2k−1) =

∞

X

k=1

µ^∗(B_2k−1)

Since S∞

k=1B_2k,S∞

k=1B_2k−1⊂ A, we have

∞

X

k=1

µ^∗(B_2k),

∞

X

k=1

µ^∗(B_2k−1) < ∞.

Therefore ∀ε > 0, ∃ large integer k₀ such that

∞

X

k=k0

µ^∗(B_2k),

∞

X

k=k0

µ^∗(B_2k−1) < ε 2. Hence ∀2n > k₀,

µ^∗(

∞

[

n=1

A_n) ≤ µ^∗(A_2n) +

∞

X

k=n+1

µ^∗(B_2k) +

∞

X

k=n+1

µ^∗(B_2k−1)

< µ^∗(A_2n) + ε 2+ ε

2 = µ^∗(A_2n) + ε.

Therefore sup_nµ^∗(A_n) ≥ µ^∗(S∞ n=1A_n).

∵ µ^∗(A_n) % and µ^∗(A_n) ≤ µ^∗(A) < ∞,

∴ limn→∞µ^∗(A_n) = sup_nµ^∗(A_n).

µ^∗(S∞

n=1A_n) ≤ lim_n→∞µ^∗(A_n). On the other hand, ∵ An ⊂ S∞

n=1A_n, the opposite inequality is also true.

4. Let E ⊂ X.

If ν^∗(E) = ∞ ⇒ ν^∗(E) ≥ µ^∗(E).

If ν^∗(E) < ∞, given ε > 0, ∃Ai ∈ A such that E ⊂[

i

A_i and X

i

ν(A_i) ≤ ν^∗(E) + ε.

Since µ = ν on A, we haveP

iν(A_i) = P

iµ(A_i) ≥ µ^∗(E). ⇒ µ^∗(E) ≤ ν^∗(E) + ε, ∀ε.

⇒ µ^∗(E) ≤ ν^∗(E), ∀E.

∵ ν is σ-finite, countably additive, σ(A) is the σ-algebra generated by A and µ is the extension of ν. ∴ µ is the measure defined on σ(A) and µ = ν^∗ on σ(A).

Let E ⊂ X.

If µ^∗(E) = ∞ ⇒ µ^∗(E) ≥ ν^∗(E).

If µ^∗(E) < ∞, given ε > 0, ∃B_i ∈ σ(A) such that E ⊂[

i

B_i and X

i

µ(B_i) ≤ µ^∗(E) + ε.

Since µ = ν^∗ on σ(A), we have P

iµ(Bi) =P

iν^∗(Bi) ≤ ν^∗(E).

⇒ ν^∗(E) ≤ µ^∗(E) + ε, ∀ε.

⇒ ν^∗(E) ≤ µ^∗(E), ∀E.

Hence µ^∗ = ν^∗.