False statementConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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 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 $μ^{*} (\emptyset) = 0$ , $μ^{*} (A) = 1$ for $A \neq \emptyset$ .

[F1]

An outer measure on a set $X$ is a function $μ^{*} : P (X) \to [0, + \infty]$ that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

[F2]

A set $E \subseteq X$ is Carathéodory measurable for $μ^{*}$ when $μ^{*} (A) = μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ for every $A \subseteq X$ . (Carathéodory measurable sets)

Refutation

technique · direct

1.1

The function vanishes at $\emptyset$ 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 $0 \leq 0$ . Thus $μ^{*}$ is an outer measure.

F1algebra

2.1

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

step 1.1F2algebra∎

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

G. Folland, Real Analysis, 2nd ed., Section 1.4 (standard reference, not scraped)