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: every measure is absolutely continuous or singular with respect to Lebesgue measure

Statement

False claim: Every finite Borel measure on $R$ is either absolutely continuous with respect to Lebesgue measure or singular with respect to Lebesgue measure.

Facts & Assumptions

Given: The measure $ν : = λ ⁣ ↾_{[0, 1]} + μ_{c}$ .

[L1]

The Cantor measure $μ_{c}$ is singular with respect to Lebesgue measure, and it is concentrated on the Cantor set $C$ with $μ_{c} (C) = 1$ . (The Cantor measure is a singular atomless probability measure concentrated on the Cantor set)

[L2]

Lebesgue measure is the Lebesgue--Stieltjes measure of the identity (Lebesgue measure is the Lebesgue-Stieltjes measure of the identity function).

[L3]

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

Refutation

technique · direct

1.1

The measure $ν$ is not absolutely continuous with respect to $λ$ : the Cantor set $C$ is Lebesgue null by [L1], but $ν (C) = λ (C) + μ_{c} (C) = 1.$

L1givenalgebra

2.1

By [L2] and [L3], $λ ((0, 1]) = 1$ , so a Lebesgue-null set $N$ satisfies $λ ((0, 1] ∖ N) = 1$ . The measure $ν$ is therefore not singular with respect to $λ$ : if it were concentrated on $N$ , then $0 < λ ((0, 1] ∖ N) \leq ν ((0, 1] ∖ N),$ contradicting concentration on $N$ . Thus $ν$ is neither absolutely continuous nor singular.

step 1.1L2L3contradiction: concentration on a null setdischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

17 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 2.37 and §6.8 (standard reference, not scraped)