False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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FALSE: every subset of a measure-null set is measurable

Statement

False claim. In every measure space, every subset of a measurable null set is measurable. Equivalently, every measure space is complete in the sense of Complete measure spaces.

Facts & Assumptions

Given: The two-point set $X = {0, 1}$ and the family $A = {\emptyset, X}$ .

[L1]

A measure has value $0$ at the empty set and is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[L2]

Completeness requires every subset of every measurable null set to belong to the sigma-algebra (Complete measure spaces).

Refutation

technique · direct

1.1

The family $A$ is a sigma-algebra on $X$ , and the set function $μ (\emptyset) = μ (X) = 0$ is a measure: both sides of countable additivity are $0$ for every disjoint measurable sequence.

givenL1

1.2

The set $X$ is measurable and $μ (X) = 0$ , but ${0} \subseteq X$ and ${0} \notin A$ .

given

2.1

By step 1.2 the measure space of step 1.1 is not complete, so the claimed universal measurability of subsets of null sets is false.

step 1.1step 1.2L2∎

Depends on

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

G. Folland, Real Analysis, 2nd ed., §1.3 (standard reference, not scraped)