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 determines its distribution function uniquely

Statement

False claim. A Lebesgue-Stieltjes measure determines a unique distribution function. The valid statement on this page is only uniqueness modulo additive constants, with normalization at $0$ choosing one representative.

Facts & Assumptions

Given: Countable choice, a nondecreasing right-continuous function $F : R \to R$ , and the shifted function $G : = F + 1$ .

[L1]

Assuming countable choice, two nondecreasing right-continuous functions define the same Lebesgue-Stieltjes measure exactly when their difference is constant. (Assuming countable choice, finite-on-compacts Borel measures on $R$ correspond to nondecreasing right-continuous functions modulo constants)

Refutation

technique · direct

1.1

The function $G$ is nondecreasing and right-continuous whenever $F$ is, and [given] $G - F$ is the constant function $1$ .

given

2.1

Therefore [L1] gives $μ_{G} = μ_{F}$ . Unless $F$ already equals $F + 1$ , [step 1.1, L1] which no real-valued function does, the two distribution functions are distinct. So the measure does not determine a unique representative without a normalization convention.

step 1.1L1∎

Depends on

Assuming countable choice, finite-on-compacts Borel measures on $\mathbb{R}$ correspond to nondecreasing right-continuous functions modulo constants

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 1.16 (standard reference, not scraped)