False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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FALSE: every atomless Borel measure on $R$ is absolutely continuous with respect to Lebesgue measure

Statement

False claim. Every atomless Borel measure on $R$ has a density with respect to Lebesgue measure. The Cantor measure already separates atomlessness from absolute continuity.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Cantor measure $μ_{c}$ .

[L1]

Assuming Countable Choice, the Cantor measure is an atomless probability measure singular with respect to Lebesgue measure. (The Cantor measure is a singular atomless probability measure concentrated on the Cantor set)

Refutation

technique · direct

1.1

By [L1], the Cantor measure $μ_{c}$ is atomless.

2.1

The same fact [L1] says that $μ_{c}$ is singular and has total mass $1$ . [step 1.1, L1] If it were absolutely continuous with respect to Lebesgue measure, its concentration on a Lebesgue-null set would force its total mass to be $0$ . Thus it is not absolutely continuous and cannot arise from a density. ∎

Depends on

The Cantor measure is a singular atomless probability measure concentrated on the Cantor set

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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 (standard reference, not scraped)

FALSE: every atomless Borel measure on RR is absolutely continuous with respect to Lebesgue measure