TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Assuming countable choice, a premeasure extends through its induced outer measure

Statement

Assume the Axiom of Countable Choice. If $μ_{0}$ is a premeasure on an algebra $A_{0}$ of subsets of $X$ and $μ^{*}$ is its induced outer set function, then $A_{0} \subseteq M_{μ^{*}}$ and $μ^{*} ∣_{A_{0}} = μ_{0}$ . The restriction of $μ^{*}$ to $σ (A_{0})$ is therefore a measure extending $μ_{0}$ .

Equivalently: assuming countable choice, the induced outer measure restricts to a complete measure on its Carathéodory sigma-algebra and to a measure on $σ (A_{0})$ extending $μ_{0}$ .

Facts & Assumptions

Given: Countable choice, a premeasure $μ_{0}$ on $A_{0}$ , and its induced outer set function $μ^{*}$ .

[L1]

Assuming countable choice, the outer set function induced by a premeasure is an outer measure. (Assuming countable choice, the outer set function induced by a premeasure is an outer measure)

[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)

[L3]

For every $A \in A_{0}$ , the outer measure induced by a premeasure satisfies $μ^{*} (A) = μ_{0} (A)$ . (The induced outer measure agrees with the premeasure on the source algebra)

[L4]

Assuming countable choice, every member of the source algebra is Carathéodory measurable for the induced outer measure. (Assuming countable choice, every source-algebra set is measurable for the induced outer measure)

[L5]

For every $E \subseteq P (X)$ , $σ_{X} (E)$ 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

1.1L1L2

By [L1], $μ^{*}$ is an outer measure, and [L2] makes its restriction to $M_{μ^{*}}$ a complete measure.

1.2L3L4

By [L4], $A_{0} \subseteq M_{μ^{*}}$ , and [L3] identifies the restricted values there with $μ_{0}$ .

2.1step 1.1step 1.2L5algebra∎

Since $M_{μ^{*}}$ is a sigma-algebra containing $A_{0}$ , [L5] gives $σ (A_{0}) \subseteq M_{μ^{*}}$ ; restricting the measure from step 1.1 to this generated sigma-algebra and using step 1.2 gives the claimed extension, with no sigma-finiteness hypothesis.

Assuming countable choice, a premeasure extends through its induced outer measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources