PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck 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.

Every outer-null set is Carathéodory measurable

Statement

Every set of outer measure zero, and every subset of it, is Carathéodory measurable and has outer measure zero.

Facts & Assumptions

Given: An outer measure $μ^{*}$ on $X$ , a set $N \subseteq X$ with $μ^{*} (N) = 0$ , and a subset $S \subseteq N$ .

[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)

[L1]

For every outer measure $μ^{*}$ and all $A, E \subseteq X$ , $μ^{*} (A) \leq μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ . (In the Carathéodory identity, the subadditive inequality is automatic)

Proof

technique · direct

1.1F1algebra

Monotonicity gives $0 \leq μ^{*} (S) \leq μ^{*} (N) = 0$ , so $μ^{*} (S) = 0$ ; for every test set $A$ , the same argument gives $μ^{*} (A \cap S) = 0$ .

2.1step 1.1L1algebra∎

Since $A ∖ S \subseteq A$ , monotonicity gives $μ^{*} (A) \geq μ^{*} (A ∖ S) = μ^{*} (A \cap S) + μ^{*} (A ∖ S)$ by step 1.1; [L1] supplies the reverse inequality, so $S$ is Carathéodory measurable.

Every outer-null set is Carathéodory measurable

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources