19.6. MEASURES ON HAUSDORFF SPACES 513
In words, it assumes the outer measure is inner regular on open sets and outer regular on allsets. Also it assumes you can approximate the measure of an open set with a compact setand the measure of a compact set with an open set. Recall that the Borel sets are those sets inthe smallest σ algebra that contains the open sets. The big result is the Riesz representationtheorem for positive linear functionals and the following lemma is the technical part of theproof of this big theorem in addition to being interesting for its own sake. It holds in aHausdorff space, not just one which is locally compact.
Lemma 19.6.1 Let Ω be a Hausdorff space and suppose µ is an outer measure satisfy-ing µ is finite on compact sets and the following conditions,
1. µ (E) = inf{µ (V ) ,V ⊇ E,V open} for all E. (Outer regularity.)
2. For every open set V,µ (V ) = sup{µ (K) : K ⊆V,K compact} (Inner regularity onopen sets.)
3. If A,B are compact disjoint sets, then µ (A∪B) = µ (A)+µ (B).
Then the following hold.
1. If ε > 0 and if K is compact, there exists V open such that V ⊇ K and
µ (V \K)< ε
2. If ε > 0 and if V is open with µ (V )< ∞, there exists a compact subset K of V suchthat
µ (V \K)< ε
3. The µ measurable sets S defined as
S ≡ {E ⊆Ω : µ (S) = µ (S\E)+µ (S∩E) for all S}
contains the Borel sets and also µ is inner regular on every open set,
µ (V ) = sup{µ (K) : K ⊆V,K compact}
and for every E ∈S with µ(E)< ∞,
µ (E) = sup{µ (K) : K ⊆ E,K compact}
Proof: First we establish 1 and 2 and use them to establish the last assertion. Consider2. Suppose it is not true. Then there exists an open set V having µ (V ) < ∞ but for allK ⊆ V,µ (V \K) ≥ ε for some ε > 0. By inner regularity on open sets, there exists K1 ⊆V,K1 compact, such that µ (K1)≥ ε/2. Now by assumption, µ (V \K1)≥ ε and so by innerregularity on open sets again, there exists compact K2 ⊆ V \K1 such that µ (K2) ≥ ε/2.Continuing this way, there is a sequence of disjoint compact sets contained in V {Ki} suchthat µ (Ki)≥ ε/2.
V
K1 K4K2
K3