236 CHAPTER 10. REGULAR MEASURES

If U is open and contains V, then µ (U)≥ µ (V ) and so

µ (V )≤ inf{µ (U) : U ⊇V, U open} ≤ µ (V ) since V ⊆V.

Thus µ is inner and outer regular on open sets. In what follows, K will be closed and Vwill be open.

Let K be the open sets. This is a π system since it is closed with respect to finiteintersections. Let

G ≡ {E ∈B (X) : µ is inner and outer regular on E} so G ⊇K .

For E ∈ G , let V ⊇ E ⊇ K such that µ (V \K) = µ (V \E)+µ (E \K)< ε . Thus KC ⊇ EC

and so µ(KC \EC

)= µ (E \K)< ε. Thus µ is outer regular on EC because

µ(KC)= µ

(EC)+µ

(KC \EC)< µ

(EC)+ ε, KC ⊇ EC

Also, EC ⊇ VC and µ(EC \VC

)= µ (V \E) < ε so µ is inner regular on EC and so G is

closed for complements. If the sets of G {Ei} are disjoint, let Ki⊆Ei⊆Vi with µ (Vi \Ki)<ε2−i. Then for E ≡ ∪iEi,and choosing m sufficiently large,

µ (E) = ∑i

µ (Ei)≤m

∑i=1

µ (Ei)+ ε ≤m

∑i=1

µ (Ki)+2ε = µ (∪mi=1Ki)+2ε

and so µ is inner regular on E ≡ ∪iEi. It remains to show that µ is outer regular on E.Letting V ≡ ∪iVi,

µ (V \E)≤ µ (∪i (Vi \Ei))≤∑i

ε2−i = ε.

Hence µ is outer regular on E since µ (V ) = µ (E)+µ (V \E)≤ µ (E)+ ε and V ⊇ E.By Dynkin’s lemma, G = σ (K )≡B (X).2.) Suppose that µ is outer regular on sets of F ⊇B (X). Letting E ∈F , by outer

regularity, there exists an open set V ⊇ EC such that µ (V )−µ(EC)< ε . Since µ is finite,

ε > µ (V )− µ(EC)= µ

(V \EC

)= µ

(E \VC

)= µ (E)− µ

(VC)

and VC is a closed setcontained in E. Therefore, if 8.17 holds, then so does 8.16. The converse is proved in thesame way. There is K ⊆ EC with ε > µ

(EC \K

)= µ

(KC \E

)showing outer regularity

from inner regularity.3.) The last claim is obtained by letting G = ∩nVn where Vn is open, contains E,

Vn ⊇Vn+1, and µ (Vn)< µ (E)+ 1n and Kn, increasing closed sets contained in E such that

µ (E) < µ (Kn)+1n . Then let F ≡ ∪Fn and G ≡ ∩nVn. Then F ⊆ E ⊆ G and µ (G\F) ≤

µ (Vn \Kn)< 2/n. ■Next is a lemma which allows the replacement of closed with compact in the definition

of inner regular.

Lemma 10.1.3 Let µ be a finite measure on a σ algebra containing B (X) , the Borelsets of X , a separable complete metric space, Polish space. Then if C is a closed set,

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

It follows that for a finite measure on B (X) where X is a Polish space, µ is inner regularin the sense that for all F ∈B (X),µ (F) = sup{µ (K) : K ⊆ F and K is compact}