Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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Every Borel set is Carathéodory measurable for a metric outer measure

Statement

Every Borel subset of a metric space is Carathéodory measurable for every metric outer measure.

Facts & Assumptions

Given: A metric space (X,d), a metric outer measure μ, and its Carathéodory family Mμ.

[L1]

Every closed subset of a metric space is Carathéodory measurable for every metric outer measure. (Closed sets are Carathéodory measurable for metric outer measures)

[L2]

The Carathéodory measurable subsets form a sigma-algebra, and the restriction of the outer measure to it is a complete measure. (Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure)

[F1]

The Borel sigma-algebra B(X) of a topological space is the sigma-algebra generated by its open sets. (The Borel sigma-algebra of a topological space)

[L3]

For every EP(X), σX(E) is the unique smallest sigma-algebra on X containing E. (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal)

Proof

technique · direct
1.1

By [L2], Mμ is a sigma-algebra. It contains every closed set by [L1], and hence contains every open set by closure under complements.

L1L2algebra
2.1

By [F1], the Borel sigma-algebra is generated by the open sets, so minimality in [L3] and step 1.1 give B(X)Mμ; this also covers the empty metric space and the one-point space.

step 1.1F1L3

Depends on

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Dependency tree · two levels

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