User:TakuyaMurata/Sandbox6

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In the following, X denotes a locally compact space. Unless stated otherwise, Radon measures are complex-valued.

Proposition ''Let μ be a Radon measure. Let $$f$$ is a μ-measurable function on X, and Φ be the set of all compact set $$K \subset X$$ such that $$f|_K$$ is continuous. Then for every compact subset L of X and ε ＞ 0 we have:
 * |μ|(L - K) ＜ ε,

for some K ∈ Φ with K ⊂ L.

Proposition ''Let A ⊂ X be a μ-measurable subset. Then there exists Φ' ⊂ Φ such that
 * (i) The complement of the union of Φ' has μ-measure zero.
 * (ii) For every K ∈ Φ, supp(μ_K) = K

Proof: Zorn's lemma.

Proposition For $$f: X \to \mathbf{C}$$ measurable on a locally compact subspace Y ⊂ X, the mapping
 * $$f \to \int f d\mu$$

is a Radon measure on Y.

Proof:
 * $$\int |f| d|\mu| = \int |f|_K d|\mu| < \infty$$

Proposition ''Let μ be a positive Radon measure. Let K ⊂ X be a compact subset. Then
 * μ_K(H) = μ(H)

for every compact subset H ⊂.