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PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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Carathéodory measurable sets form an algebra

Statement

The Carathéodory measurable subsets of X form an algebra of subsets. In particular, is measurable, complements of measurable sets are measurable, and finite unions of measurable sets are measurable (Algebras of subsets).

Facts & Assumptions

Given: An outer measure μ on X and Carathéodory measurable sets E,FX.

[F1]

A set EX is Carathéodory measurable for μ when μ(A)=μ(AE)+μ(AE) for every AX. (Carathéodory measurable sets)

[L1]

For every outer measure μ and all A,EX, μ(A)μ(AE)+μ(AE). (In the Carathéodory identity, the subadditive inequality is automatic)

Proof

technique · direct
1.1

For every test set A, the split by reads μ(A)=0+μ(A), so is measurable; the identity for XE is the identity for E with its two summands exchanged. Applying [F1] first to E and then, inside each resulting piece, to F splits A into the four cells AEF, AEF, AFE, and A(EF), with their outer measures summing to μ(A).

F1algebra
2.1

By subadditivity, μ(A(EF)) is at most the sum of the first three cell values from step 1.1, while A(EF) is the fourth cell; hence μ(A)μ(A(EF))+μ(A(EF)). The reverse inequality is [L1], so EF is measurable and the measurable family is an algebra.

step 1.1F1L1algebra

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Used by

Dependency tree · two levels

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Sources