TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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The RMK representing measure is inner regular on open sets

Statement

For every open $U \subseteq X$ , $μ (U) = \sup {μ (K) : K \subseteq U, K compact} .$ Together with outer regularity and finiteness on compact sets, the constructed Borel measure is Radon in the convention of Radon measure on an LCH space.

Facts & Assumptions

Given: The constructed measure satisfies $μ (U) = ρ (U)$ on opens.

[L1]

The compact-set formula holds and compact sets have finite measure. (Compact-set formula and local finiteness of the RMK measure)

Proof

technique · direct

1.1

If $f ≺ U$ and $K = supp f$ , then [L1] $0 \leq f \leq 1_{K}$ . For every $h \in C_{c} (X)$ with $1_{K} \leq h$ , monotonicity gives $Λ (f) \leq Λ (h)$ ; taking the infimum in [L1] yields $Λ (f) \leq μ (K)$ . Hence $ρ (U) \leq \sup_{K ⋐ U} μ (K)$ .

2.1

The reverse inequality is monotonicity of the measure. Since $μ (U) = ρ (U)$ , the displayed equality follows. Outer regularity is built into $μ^{*}$ , and [L1] gives compact finiteness, so all Radon clauses hold.

step 1.1L1∎

Depends on

Used by

Positive functionals on C_c(X) are integration against a Radon measure Theorem

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Donald L. Cohn, Measure Theory, 2nd ed., Chapter 7 (standard reference, not scraped)