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

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Lebesgue--Stieltjes regularity agrees with the LCH Radon convention on R

Statement

Assume the Axiom of Countable Choice. Every Lebesgue--Stieltjes measure $μ_{F}$ on $R$ is Radon in the LCH convention, and its interval convention remains $μ_{F} ((a, b]) = F (b) - F (a)$ for increasing right-continuous $F$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, an increasing right-continuous $F$ , and its Lebesgue--Stieltjes measure $μ_{F}$ .

[L1]

Under the stated choice hypothesis, $μ_{F}$ is a Borel measure finite on compact sets and regular on $R$ . (Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on $R$ , Lebesgue-Stieltjes measures on $R$ are outer regular and inner regular by compact sets)

Proof

technique · direct

1.1

By [L1], $μ_{F}$ is finite on compact sets, outer regular on Borel sets, and inner regular by compact sets on Borel sets, hence in particular on open sets. These imply every clause of the LCH Radon definition.

2.1

No measure is replaced in this comparison, so the defining half-open interval formula and the Borel sigma-algebra from the construction remain unchanged.

givenL1∎

Depends on

Used by

A Lebesgue--Stieltjes functional is represented by its Stieltjes measure Example

Dependency tree · two levels

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Sources

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