Alphabeta Math
False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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FALSE: every subset is Carathéodory measurable for every outer measure

Statement

Every subset of every set is Carathéodory measurable for every outer measure on that set.

Facts & Assumptions

Given: The two-point set X={0,1} and the set function μ∗(∅)=0, μ∗(A)=1 for A≠∅.

[F1]

An outer measure on a set X is a function μ∗:P(X)→[0,+∞] that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

[F2]

A set E⊆X is Carathéodory measurable for μ∗ when μ∗(A)=μ∗(A∩E)+μ∗(A∖E) for every A⊆X. (Carathéodory measurable sets)

Refutation

technique · direct
1.1F1algebra

The function vanishes at ∅ and is monotone; if a countable union is nonempty, at least one member is nonempty, so the union has value 1 and the sum of the member values is at least 1, while an empty union gives 0≤0. Thus μ∗ is an outer measure.

2.1step 1.1F2algebra∎

For E={0} and test set A=X, [F2] would require 1=μ∗(X)=μ∗({0})+μ∗({1})=1+1, which is false.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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