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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The induced outer measure agrees with the premeasure on the source algebra

Statement

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

Facts & Assumptions

Given: A premeasure $μ_{0}$ on an algebra $A_{0}$ and its induced outer measure $μ^{*}$ .

[F1]

A premeasure on an algebra $A_{0}$ vanishes at the empty set and is countably additive whenever a disjoint sequence in $A_{0}$ has its union in $A_{0}$ . (Premeasures on algebras of sets)

[L1]

Every countable algebra cover of an algebra set disjointifies inside that set into algebra members subordinate to the original cover. (A countable algebra cover disjointifies inside the covered algebra set)

[F2]

The set function induced by $μ_{0}$ assigns $E \subseteq X$ the infimum of $\sum_{k} μ_{0} (A_{k})$ over all countable algebra covers $E \subseteq ⋃_{k} A_{k}$ . (The outer set function induced by a premeasure)

Proof

technique · direct

1.1F2construct

The sequence $(A, \emptyset, \emptyset, \dots)$ covers $A$ at cost $μ_{0} (A)$ , so $μ^{*} (A) \leq μ_{0} (A)$ .

1.2F1L1algebra

For any algebra cover $(A_{k})$ of $A$ , [L1] gives disjoint $B_{k} \subseteq A_{k}$ with union $A$ ; [F1] gives $μ_{0} (A) = \sum_{k} μ_{0} (B_{k})$ , and finite additivity applied to $A_{k} = B_{k} ⊔ (A_{k} ∖ B_{k})$ gives $μ_{0} (B_{k}) \leq μ_{0} (A_{k})$ , hence $μ_{0} (A) \leq \sum_{k} μ_{0} (A_{k})$ .

2.1step 1.1step 1.2F2algebra∎

Since step 1.2 bounds every covering cost below by $μ_{0} (A)$ , [F2] gives $μ_{0} (A) \leq μ^{*} (A)$ ; combining this with step 1.1 proves equality, including infinite values without subtraction.

Remarks

No choice principle is used here. Steps 1.1, 1.2 and 2.1 read the defining infimum of The outer set function induced by a premeasure, disjointify one given cover, and apply countable additivity of the premeasure; none of them selects a cover for each index. The name outer measure is the one that Assuming countable choice, the outer set function induced by a premeasure is an outer measure earns for the induced set function under countable choice, and the identity proved here holds for the set function whether or not that hypothesis is in force.

Depends on

Dependency tree · 2 levels

Sources

G. Folland, Real Analysis, 2nd ed., Proposition 1.13(a)