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TheoremStatement: 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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Assuming countable choice, a premeasure-induced outer measure is regular with generated measurable hulls

Statement

Assume the Axiom of Countable Choice. An outer measure induced by a premeasure is regular, and every set has a measurable hull in σ(A0).

Facts & Assumptions

Given: Countable choice, a premeasure μ0 on A0, its induced outer measure μ, and a subset EX.

[F1]

A measurable hull of E is a Carathéodory measurable set HE with μ(H)=μ(E); the outer measure is regular when every subset has a measurable hull. (Measurable hulls and regular outer measures)

[F2]

The set function induced by μ0 assigns EX the infimum of kμ0(Ak) over all countable algebra covers EkAk. (The outer set function induced by a premeasure)

[L1]

Assuming countable choice, the induced outer measure restricts to a complete measure on its Carathéodory sigma-algebra and to a measure on σ(A0) extending μ0. (Assuming countable choice, a premeasure extends through its induced outer measure)

Proof

technique · direct
1.1

If μ(E)=+, put H=X. Otherwise, for each nN, countable choice and [F2] give an algebra cover (Ank)k of E with cost below μ(E)+1/(n+1); put Un=kAnk and H=nUn.

F2givenchoosecases
2.1

In the finite case each Un and their intersection H lie in σ(A0); in the infinite case H=X lies there. Thus [L1] makes H Carathéodory measurable in either case.

step 1.1L1algebra
3.1

One has EH. In the finite case, monotonicity and each covering bound give μ(E)μ(H)μ(Un)<μ(E)+1/(n+1) for every n, hence μ(H)=μ(E); in the infinite case both values are +. Therefore [F1] makes H a generated measurable hull, and μ is regular.

step 1.1step 2.1F1algebra

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