1858 CHAPTER 59. BASIC PROBABILITY

Lemma 59.1.8 Let µ be a finite measure defined on B (E) where E is a metric space.Then µ is regular.

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 ∈ X : 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 U 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 (X) such that µ is regular on F} .

Then F contains the open sets and the closed sets.Suppose F ∈F . Then there exists V ⊇ F with µ (V \F)< ε. It follows VC ⊆ FC and

µ(FC \VC)= µ (V \F)< ε.

Thus µ is inner regular on FC. Since F ∈F , there exists K ⊆ F where K is closed andµ (F \K)< ε . Then also KC ⊇ FC and

µ(KC \FC)= µ (F \K)< ε.

Thus if F ∈F so is FC.Suppose now that {Fi} ⊆F , the Fi being disjoint. Is ∪Fi ∈F ? There exists Ki ⊆ Fi

such that µ (Ki)+ ε/2i > µ (Fi) . Then

µ (∪∞i=1Fi) =

∞

∑i=1

µ (Fi)≤ ε +∞

∑i=1

µ (Ki)

< 2ε +N

∑i=1

µ (Ki) = 2ε +µ(∪N

i=1Ki)