12.1. OUTER MEASURES 281

Next I will show F is closed with respect to taking countable unions. Let {Fk} bea sequence of sets in F . Then since Fk ∈ F , there exist {Kk} such that Kk ⊆ Fk andµ (Fk \Kk)< ε/2k+1. First choose m large enough that

µ ((∪∞k=1Fk)\ (∪m

k=1Fk))<ε

2.

Then

µ ((∪mk=1Fk)\ (∪m

k=1Kk))≤m

∑k=1

ε

2k+1 <ε

2

and so

µ ((∪∞k=1Fk)\ (∪m

k=1Kk)) ≤ µ ((∪∞k=1Fk)\ (∪m

k=1Fk))

+µ ((∪mk=1Fk)\ (∪m

k=1Kk))

<ε

2+

ε

2= ε

Since µ is outer regular on Fk, there exists Vk such that µ (Vk \Fk)< ε/2k. Then

µ ((∪∞k=1Vk)\ (∪∞

k=1Fk)) ≤∞

∑k=1

µ (Vk \Fk)

<∞

∑k=1

ε

2k = ε

and this completes the demonstration that F is a σ algebra. This proves the lemma.

Theorem 12.1.12 Let µ be a finite measure defined on B (E) where E is a closed subsetof Rn. Then µ is regular.

Proof: From Lemma 12.1.11 µ is outer regular. Now let F ∈B (E). Then since µ isfinite, there exists K ⊆ F such that K is closed, K ⊆ F, and

µ (F)< µ (K)+ ε.

Then let Kk ≡ K ∩ B(0,k). Thus Kk is a closed and bounded, hence compact set and∪∞

k=1Kk = K. Therefore, for all k large enough,

µ (F)

< µ (Kk)+ ε

< sup{µ (K) : K ⊆ F and K compact}+ ε

≤ µ (F)+ ε

Since ε was arbitrary, it follows

sup{µ (K) : K ⊆ F and K compact}= µ (F) .

This proves the theorem.