514 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

Now this is an obvious contradiction because by 3,

µ (V )≥ µ (∪ni=1Ki) =

n

∑i=1

µ (Ki)≥ nε

2

for each n, contradicting µ (V )< ∞.Next consider 1. By outer regularity, there exists an open set W ⊇ K such that µ (W )<

µ (K)+ 1. By 2, there exists compact K1 ⊆W \K such that µ ((W \K)\K1) < ε. Thenconsider V ≡W \K1. This is an open set containing K and from what was just shown,

µ ((W \K1)\K) = µ ((W \K)\K1)< ε.

Now consider the last assertion.Define S1 = {E ∈P (Ω) : E ∩K ∈S } for all compact K.First it will be shown the compact sets are in S . From this it will follow the closed sets

are in S1. Then you show S1 = S . Thus S1 = S is a σ algebra and so it contains theBorel sets since it contains the closed sets. Finally you show the inner regularity assertion.

Claim 1: Compact sets are in S .Proof of claim: Let V be an open set with µ (V )< ∞. I will show that for C compact,

µ (V )≥ µ(V \C)+µ(V ∩C).

If µ(V ) = ∞ the above is obvious. The various sets are illustrated in the following diagram.

VH CK

By 2, there exists a compact set K ⊆V \C such that µ ((V \C)\K)< ε and a compactset H ⊆V such that µ (V \H)< ε . Thus µ (V )≤ µ (V \H)+µ (H)< ε +µ (H). Then

µ (V )≤ µ (H)+ ε ≤ µ (H ∩C)+µ (H \C)+ ε

≤ µ (V ∩C)+µ (V \C)+ ε ≤ µ (H ∩C)+µ (K)+3ε

By 3,= µ (H ∩C)+µ (K)+3ε = µ ((H ∩C)∪K)+3ε ≤ µ (V )+3ε.

Since ε is arbitrary, this shows that

µ(V ) = µ(V \C)+µ(V ∩C). (19.4)

Of course 19.4 is exactly what needs to be shown for arbitrary S in place of V . It sufficesto consider only S having µ (S)< ∞. If S⊆Ω, with µ(S)< ∞, let V ⊇ S, µ(S)+ε > µ(V ).Then from what was just shown, if C is compact,

ε +µ(S)> µ(V ) = µ(V \C)+µ(V ∩C)≥ µ(S\C)+µ(S∩C).

Since ε is arbitrary, this shows the compact sets are in S . This proves the claim.