CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-24

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

Statement

Assume the Axiom of Countable Choice. The outer set function induced by a premeasure is an outer measure.

Facts & Assumptions

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

[L1]

Assuming countable choice, the infimum of countable covering costs defines an outer measure. (Assuming countable choice, countable covering costs define an outer measure)

[F1]

An algebra of subsets of $X$ contains $\emptyset$ , is closed under complements relative to $X$ and binary unions, and therefore also contains $X$ . (Algebras of subsets)

Proof

technique · direct

1.1F1given

The algebra law in [F1] gives $\emptyset, X \in A_{0}$ , and the premeasure normalization gives $μ_{0} (\emptyset) = 0$ . Thus $C = A_{0}$ and $p = μ_{0}$ satisfy every hypothesis of [L1].

2.1step 1.1L1∎

Applying [L1] to the data in step 1.1 shows that the induced outer set function $μ^{*}$ is an outer measure.

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

G. Folland, Real Analysis, 2nd ed., Proposition 1.10 and formula 1.12 (standard reference, not scraped)