False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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FALSE: a Lebesgue-Stieltjes measure always gives every singleton measure $0$

Statement

False claim. Every Lebesgue-Stieltjes measure $μ_{F}$ satisfies $μ_{F} ({a}) = 0$ for every $a \in R$ . In fact singleton masses are exactly the jumps of $F$ by Interval formulas and atoms for a Lebesgue-Stieltjes measure.

Facts & Assumptions

Given: The Axiom of Countable Choice and the step function

$F (x) : = {\begin{cases} 0, & x < 0, \\ 1, & x \geq 0. \end{cases}$

[L1]

For a Lebesgue-Stieltjes measure, one has $μ_{F} ({a}) = F (a) - F (a^{-})$ . (Interval formulas and atoms for a Lebesgue-Stieltjes measure)

[L2]

Assuming Countable Choice, every nondecreasing right-continuous real function defines a Lebesgue-Stieltjes measure. (Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on $R$ )

Refutation

technique · direct

1.1

The function $F$ is nondecreasing and right-continuous, so [L2] defines its [given, L2] Lebesgue-Stieltjes measure $μ_{F}$ . At the point $0$ , one has $F (0) = 1$ and $F (0^{-}) = 0$ .

2.1

Applying [L1] at $a = 0$ gives

step 1.1L1

$μ_{F} ({0}) = F (0) - F (0^{-}) = 1.$

So the singleton ${0}$ has positive measure, and the claim is false. [step 1.1, L1] ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

15 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

John K. Hunter, Measure Theory, Example 2.36 (standard reference, not scraped)

FALSE: a Lebesgue-Stieltjes measure always gives every singleton measure 00