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TheoremStatement: AI-adaptedProof: AI-adaptedprecheck 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 E⊆X.

[F1]

A measurable hull of E is a Carathéodory measurable set H⊇E 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 E⊆X the infimum of ∑kμ0(Ak) over all countable algebra covers E⊆⋃kAk. (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
1.1F2givenchoosecases

If μ∗(E)=+∞, put H=X. Otherwise, for each n∈N, 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.

2.1step 1.1L1algebra

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.

3.1step 1.1step 2.1F1algebra∎

One has E⊆H. 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.

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