False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: the Radon-Nikodym theorem holds without sigma-finiteness

Statement

False claim: If positive measures satisfy $ν ≪ μ$ , then there is always a nonnegative measurable function $f$ with $ν (E) = \int_{E} f d μ$ for every measurable set $E$ , even when $μ$ is not sigma-finite.

Facts & Assumptions

Given: Counting measure $c$ and Lebesgue measure $λ$ on $[0, 1]$ .

[L1]

Counting measure is a measure with $c ({x}) = 1$ for each point. (Counting measure on an arbitrary set, Counting measure is a measure)

[L2]

A nonnegative density representation would satisfy $λ (E) = \int_{E} f d c$ on every measurable set (The measure with density $f$ relative to $μ$ ).

[L3]

The interval $[0, 1]$ is uncountable (Every nondegenerate interval of $R$ is uncountable).

[L4]

Lebesgue measure is the Lebesgue--Stieltjes measure of the identity, and that measure assigns $(a, b]$ the increment $b - a$ (Lebesgue measure is the Lebesgue-Stieltjes measure of the identity function, Assuming countable choice, a nondecreasing right-continuous function defines a Borel measure on $R$ ).

Refutation

technique · direct

1.1

Lebesgue measure is absolutely continuous with respect to counting measure on $[0, 1]$ , because every counting-null measurable set is empty. Moreover, a set has finite counting measure only when it is finite. If $c$ were sigma-finite, $[0, 1]$ would be a countable union of finite sets and hence countable, contradicting [L3]. Thus the dominating measure is not sigma-finite.

L1L3givenalgebra

2.1

If a density $f$ as in [L2] existed, then applying it to singletons and using [L1] would give $f (x) = 0$ for every $x \in [0, 1]$ . But [L4] gives $λ ((0, 1]) = 1$ , whereas the assumed representation gives $λ ((0, 1]) = \int_{(0, 1]} f d c = 0$ . Hence the theorem fails without sigma-finiteness of the dominating measure.

step 1.1L1L2L4contradiction: total mass onedischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

37 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

Sheldon Axler, Measure, Integration & Real Analysis, Exercise 13 (standard reference, not scraped)
John K. Hunter, Measure Theory, Example 6.28 (standard reference, not scraped)