61.2. PROBABILITY MEASURES AND TIGHTNESS 1989

Proof: First note every open set is the countable union of closed sets and every closedset is the countable intersection of open sets. Here is why. Let V be an open set and let

Kk ≡{

x ∈V : dist(x,VC)≥ 1/k

}.

Then clearly the union of the Kk equals V. Next, for K closed let

Vk ≡ {x ∈ E : dist(x,K)< 1/k} .

Clearly the intersection of the Vk equals K. Therefore, letting V denote an open set and K aclosed set,

µ (V ) = sup{µ (K) : K ⊆V and K is closed}µ (K) = inf{µ (V ) : V ⊇ K and V is open} .

Also since V is open and K is closed,

µ (V ) = inf{µ (U) : U ⊇V and V is open}µ (K) = sup{µ (L) : L⊆ K and L is closed}

In words, µ is regular on open and closed sets. Let

F ≡{F ∈B (E) such that µ is regular on F} .

Then F contains the open sets. I want to show F is a σ algebra and then it will followF = B (E).

First I will show F is closed with respect to complements. Let F ∈F . Then since µ isfinite and F is inner regular, there exists K⊆F such that µ (F \K)< ε. But KC \FC =F \Kand so µ

(KC \FC

)< ε showning that FC is outer regular. I have just approximated the

measure of FC with the measure of KC, an open set containing FC. A similar argumentworks to show FC is inner regular. You start with V ⊇ F such that µ (V \F) < ε , noteFC \VC = V \ F, and then conclude µ

(FC \VC

)< ε, thus approximating FC with the

closed subset, VC.Next I will show F is closed with respect to taking countable unions. Let {Fk} be a

sequence of sets in F . Then µ is inner regular on each of these so there exist {Kk} suchthat Kk ⊆ Fk and µ (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