Radon MeasuresRecall that Riesz representation theorem asserts that for a positive linear functional Λon the space Cc(X) where X is a locally compact Hausdorff space, there exists an Borelouter measure λ satisfyingΛf =Zf dλ,for all f ∈ Cc(X). We used to call λ the Riesz measure associated to Λ. Furthermore, λenjoys the following regularity properties:(a) λ(E) = inf λ(G) : E ⊂ G, G open for every E ⊂ Rn.12 CHAPTER 6. DIFFERENTIATION OF MEASURES AND FUNCTIONS(b) λ(E) = sup λ(K) : K ⊂ E, K compact for every λ-measurable E ⊂ Rn.Now, an outer measure µ in Rnis called a Radon measure if(a) It is Borel regular, that is, for every A ⊂ Rn, there is a Borel set B, A ⊂ B, suchthat µ(A) = µ(B), and(b) it is finite on compact sets.Given a Borel measure µ, we defineΛf =Zf dµ, f ∈ Cc(X).When µ is finite on compact sets, Λ is a well-defined positive functional on Cc(X). ByRiesz representation theorem there is an outer Borel measure λ satisfyingZf dλ =Zf dµ, ∀f ∈ Cc(Rn).For every compact set K sitting inside an open set G, there exists a continuous function fcompactly supported in G, equals to 1 in K and bounded between 0 and 1. Plugging suchfunctions in the relation above and passing limit by Lebsegue’s dominated convergencetheorem, we see that λ and µ coincide on open sets. Consequently, they are the same onall Borel sets. Now, let A be an arbitrary set in Rn. By Borel regularity, there is a Borelset B containing A such that µ(B) = µ(A). Therefore,µ(A) = µ(B)= λ(B)≥ λ(A) .On the other hand, by the regularity property of the Riesz measure, for every A andε > 0, there is an open set G containing A such that λ(A) + ε ≥ λ(G). Therefore,λ(A) + ε ≥ λ(G)= µ(G)≥ µ(A) ,which implies λ(A) ≥ µ(A). Summing up, we have proved that every Radon measure isa Riesz measure in Rn

measures taking values in a finite-dimensional Banach space V
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