1990 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

showing µ is inner regular on ∪∞k=1Fk. 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 shows µ is outer regular on ∪∞k=1Fk and this proves the lemma.

Lemma 61.2.3 Let µ be a finite measure on B (E) , the Borel sets of E, a separable com-plete metric space. Then if C is a closed set,

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

Proof: Let {ak} be a countable dense subset of C. Thus ∪∞k=1B

(ak,

1n

)⊇C. Therefore,

there exists mn such that

µ

(C \∪mn

k=1B(

ak,1n

))≡ µ (C \Cn)<

ε

2n .

Now let K =C∩ (∩∞n=1Cn) . Then K is a subset of Cn for each n and so for each ε > 0 there

exists an ε net for K since Cn has a 1/n net, namely a1, · · · ,amn . Since K is closed, it iscomplete and so it is also compact. Now

µ (C \K) = µ (∪∞n=1 (C \Cn))<

∞

∑n=1

ε

2n = ε.

Thus µ (C) can be approximated by µ (K) for K a compact subset of C. This proves thelemma.

This shows that for a finite measure on the Borel sets of a separable metric space, theabove definition of regular coincides with the earlier one.

61.3 Tight MeasuresNow here is a definition of what it means for a set of measures to be tight.

Definition 61.3.1 Let Λ be a set of probability measures defined on the Borel sets of atopological space. Then Λ is “tight” if for all ε > 0 there exists a compact set, Kε such that

µ ([x /∈ Kε ])< ε

for all µ ∈ Λ.

Lemma 61.2.3 implies a single probability measure on the Borel sets of a separablemetric space is tight. The proof of that lemma generalizes slightly to give a simple criterionfor a set of measures to be tight.