False statementConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Every Borel measure on an LCH space is Radon

Statement

The assertion “every Borel measure on an LCH space is Radon” is false.

Facts & Assumptions

Given: $X = [0, ω_{1}]$ and $\overset{ˉ}{m}$ is the Dieudonne Borel probability measure.

Refutation

technique · counterexample

1.1

The ordinal space $X$ is compact Hausdorff, hence LCH, and $\overset{ˉ}{m}$ is a finite Borel measure. The set $Y = [0, ω_{1})$ is open and has $\overset{ˉ}{m} (Y) = 1$ .

given

2.1

Every compact $K \subseteq Y$ is bounded below some $α < ω_{1}$ , so its complement in $Y$ contains a club and $\overset{ˉ}{m} (K) = 0$ . Therefore $\overset{ˉ}{m} (Y) = 1 \neq 0 = \sup_{K ⋐ Y} \overset{ˉ}{m} (K),$ contradicting the open-set inner-regularity clause of a Radon measure.

given∎

Depends on

Used by

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Dependency tree · two levels

29 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)