9.8. MEASURES AND REGULARITY 255

Lemma 9.8.5 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}

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 , ∪mnk=1 B

(ak,

1n

)≡Cn.

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 since it is complete and totally bounded, Theorem 3.5.8.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. The last claimfollows from Lemma 9.8.4. ■

The next theorem is the main result. It says that if the measure is outer regular and µ isσ finite then there is an approximation for E ∈F in terms of Fσ and Gδ sets in which theFσ set is a countable union of compact sets. Also µ is inner and outer regular on F .

Theorem 9.8.6 Suppose (X ,F ,µ) ,F ⊇B (X) is a measure space for X a metricspace and µ is σ finite, X =∪nXn with µ (Xn)< ∞ and the Xn disjoint. Suppose also that µ

is outer regular. Then for each E ∈F , there exists F,G an Fσ and Gδ set respectively suchthat F ⊆ E ⊆ G and µ (G\F) = 0. In particular, µ is inner and outer regular on F . Incase X is a complete separable metric space (Polish space), one can have F in the abovebe the countable union of compact sets and µ is inner regular in the sense of 9.20.

Proof: Since µ is outer regular and µ (Xn) < ∞, there exists an open set Vn ⊇ E ∩Xnsuch that

µ (Vn \ (E ∩Xn)) = µ (Vn)−µ (E ∩Xn)<ε

2n .

Then let V ≡ ∪nVn so that V ⊇ E. Then E = ∪nE ∩Xn and so

µ (V \E)≤ µ (∪n (Vn \ (E ∩Xn)))≤∑n

µ (Vn \ (E ∩Xn))< ∑n

ε

2n = ε

Similarly, there exists Un open such that µ(Un \

(EC ∩Xn

))< ε

2n ,Un ⊇ EC ∩Xn so if U ≡∪nUn,µ

(U \EC

)= µ

(E \UC

)< ε. Now UC is closed and contained in E because U ⊇EC.

Hence, letting ε = 12n , there exist closed sets Cn, and open sets Vn such that Cn ⊆E ⊆Vn and

µ (Vn \Cn)<1

2n−1 . Letting G≡∩nVn,F ≡∪nCn,F ⊆ E ⊆G and µ (G\F)≤ µ (Vn \Cn)<1

2n−1 . Since n is arbitrary, µ (G\F) = 0.