Multivariable Advanced Calculus

162 MEASURES AND MEASURABLE FUNCTIONSSince ε is arbitrary, this proves the lemma in case dist (x 1 , S) ≥ dist (x 2 , S) . The casewhere dist (x 2 , S) ≥ dist (x 1 , S) is entirely similar. This proves the lemma. The next lemma says that regularity comes free for finite measures defined on theBorel sets. Actually, it only almost says this. The following theorem will say it. Thislemma deals with closed in place of compact.Lemma 7.4.5 Let µ be a finite measure defined on B (Y ) where Y is a closed subsetof X, a finite dimensional normed vector space. Then for every F ∈ B (Y ) ,µ (F ) = sup {µ (K) : K ⊆ F, K is closed }µ (F ) = inf {µ (V ) : V ⊇ F, V is open}Proof: For convenience, I will call a measure which satisfies the above two conditions“almost regular”. It would be regular if closed were replaced with compact. First noteevery open set is the countable union of closed sets and every closed set is the countableintersection of open sets. Here is why. Let V be an open set and letK k ≡ { x ∈ V : dist ( x, V C) ≥ 1/k } .Then clearly the union of the K k equals V and each is closed because x→ dist (x, S) isalways a continuous function whenever S is any nonempty set. Next, for K closed letV k ≡ {x ∈ Y : dist (x, K) < 1/k} .Clearly the intersection of the V k equals K because if x /∈ K, then since K is closed,B (x, r) has empty intersection with K and so for k large enough that 1/k < r, V kexcludes x. Thus the only points in the intersection of the V k are those in K and inaddition each point of K is in this intersection.Therefore from what was just shown, letting V denote an open set and K a closedset, it follows from Theorem 7.3.2 thatµ (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 almost regular on open and closed sets. LetF ≡ {F ∈ B (Y ) such that µ is almost regular on F } .Then F contains the open sets. I want to show F is a σ algebra and then it will followF = B (Y ).First I will show F is closed with respect to complements. Let F ∈ F. Then sinceµ is finite and F is inner regular, there exists K ⊆ F such thatBut K C \ F C = F \ K and soµ (F \ K) = µ (F ) − µ (K) < ε.µ ( K C \ F C) = µ ( K C) − µ ( F C) < εshowing that µ is outer regular on F C . I have just approximated the measure of F C withthe measure of K C , an open set containing F C . A similar argument works to show F C