294 CHAPTER 12. THE CONSTRUCTION OF MEASURES
Corollary 12.3.13 Let (Ω,τ) be a locally compact Hausdorff space which is also σ com-pact meaning
Ω = ∪∞n=1Ωn, Ωn is compact,
and let L be a positive linear functional defined on Cc (Ω) . Then if (µ1,S1) , and (µ2,S2)are two Radon measures, together with their σ algebras which represent L then the two σ
algebras are equal and the two measures are equal.
Proof: Suppose (µ1,S1) and (µ2,S2) both work. It will be shown the two measuresare equal on every compact set. Let K be compact and let V be an open set containing K.Then let K ≺ f ≺V. Then
µ1 (K) =∫
Kdµ1 ≤
∫f dµ1 = L( f ) =
∫f dµ2 ≤ µ2 (V ) .
Therefore, taking the infimum over all V containing K implies µ1 (K)≤ µ2 (K) . Reversingthe argument shows µ1 (K) = µ2 (K) . This also implies the two measures are equal on allopen sets because they are both inner regular on open sets. It is being assumed the twomeasures are regular. Now let F ∈S1 with µ1 (F) < ∞. Then there exist sets, H,G suchthat H ⊆ F ⊆ G such that H is the countable union of compact sets and G is a countableintersection of open sets such that µ1 (G) = µ1 (H) which implies µ1 (G\H) = 0. Now G\H can be written as the countable intersection of sets of the form Vk \Kk where Vk is open,µ1 (Vk)< ∞ and Kk is compact. From what was just shown, µ2 (Vk \Kk) = µ1 (Vk \Kk) soit follows µ2 (G\H) = 0 also. Since µ2 is complete, and G and H are in S2, it followsF ∈S2 and µ2 (F) = µ1 (F) . Now for arbitrary F possibly having µ1 (F) = ∞, considerF∩Ωn. From what was just shown, this set is in S2 and µ2 (F ∩Ωn)= µ1 (F ∩Ωn). Takingthe union of these F ∩Ωn gives F ∈S2 and also µ1 (F) = µ2 (F) . This shows S1 ⊆S2.Similarly, S2 ⊆S1.
The following lemma is often useful.
Lemma 12.3.14 Let (Ω,F ,µ) be a measure space where Ω is a topological space. Sup-pose µ is a Radon measure and f is measurable with respect to F . Then there exists aBorel measurable function, g, such that g = f a.e.
Proof: Assume without loss of generality that f ≥ 0. Then let sn ↑ f pointwise. Say
sn (ω) =Pn
∑k=1
cnkXEn
k(ω)
where Enk ∈F . By the outer regularity of µ , there exists a Borel set, Fn
k ⊇ Enk such that
µ(Fn
k
)= µ
(En
k
). In fact Fn
k can be assumed to be a Gδ set. Let
tn (ω)≡Pn
∑k=1
cnkXFn
k(ω) .
Then tn is Borel measurable and tn (ω) = sn (ω) for all ω /∈ Nn where Nn ∈ F is a setof measure zero. Now let N ≡ ∪∞
n=1Nn. Then N is a set of measure zero and if ω /∈ N,then tn (ω)→ f (ω). Let N′ ⊇ N where N′ is a Borel set and µ (N′) = 0. Then tnX(N′)C
converges pointwise to a Borel measurable function, g, and g(ω) = f (ω) for all ω /∈ N′.Therefore, g = f a.e. and this proves the lemma.