19.6. MEASURES ON HAUSDORFF SPACES 515

As discussed above, this verifies the closed sets are in S1 because if H is closed and Cis compact, then compact H∩C ∈S . If S1 is a σ algebra, this will show that S1 containsthe Borel sets. Thus I first show S1 is a σ algebra.

To see that S1 is closed with respect to taking complements, let E ∈ S1 and K acompact set.

K = (EC ∩K)∪ (E ∩K).

Then from the fact, just established, that the compact sets are in S , EC∩K = K \(E∩K)∈S. S1 is closed under countable unions because if K is a compact set and En ∈S1,

K∩∪∞n=1En = ∪∞

n=1K∩En ∈S

because it is a countable union of sets of S . Thus S1 is a σ algebra.Therefore, if E ∈S and K is a compact set, just shown to be in S , it follows K∩E ∈S

because S is a σ algebra which contains the compact sets and so S1 ⊇S . It remains toverify S1 ⊆S . Recall that

S1 ≡ {E : E ∩K ∈S for all K compact}

Let E ∈ S1 and let V be an open set with µ(V ) < ∞ and choose K ⊆ V such thatµ(V \K)< ε . Then since E ∈S1, it follows E ∩K,EC ∩K ∈S and so

µ (V ) ≤ µ (V \E)+µ (V ∩E)≤

The two sets are disjoint and in S︷ ︸︸ ︷µ (K \E)+µ (K∩E) +2ε

= µ (K)+2ε ≤ µ (V )+3ε

Since ε is arbitrary, this shows µ (V ) = µ (V \E)+µ (V ∩E) which would show E ∈S ifV were an arbitrary set.

Now let S⊆Ω be such an arbitrary set. If µ(S) = ∞, then µ(S) = µ(S∩E)+µ(S\E).If µ(S)< ∞, let

V ⊇ S, µ(S)+ ε ≥ µ(V ).

Thenµ(S)+ ε ≥ µ(V ) = µ(V \E)+µ(V ∩E)≥ µ(S\E)+µ(S∩E).

Since ε is arbitrary, this shows that E ∈ S and so S1 = S . Thus S ⊇ Borel sets asclaimed.

From 2 µ is inner regular on all open sets. It remains to show that

µ(F) = sup{µ(K) : K ⊆ F} (19.5)

for all F ∈S with µ(F)< ∞. It might help to refer to the following crude picture to keepthings straight. It also might not help. V is between the dotted lines.

V C∩K

µ < ε V

U \FF