PropositionStatement: 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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Closed sets are Carathéodory measurable for metric outer measures

Statement

Every closed subset of a metric space is Carathéodory measurable for every metric outer measure.

Facts & Assumptions

Given: A metric outer measure $μ^{*}$ on $(X, d)$ , a closed set $F \subseteq X$ , and a test set $A \subseteq X$ .

[L1]

For a closed set $F$ and a finite-outer-measure test set $A$ , the positive-distance layers inside $A ∖ F$ increase to $A ∖ F$ in outer measure. (Boundary layers of finite metric outer measure exhaust the complement of a closed set in outer measure)

[L2]

For every outer measure $μ^{*}$ and all $A, E \subseteq X$ , $μ^{*} (A) \leq μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ . (In the Carathéodory identity, the subadditive inequality is automatic)

[F1]

An outer measure on a metric space is a metric outer measure when $μ^{*} (A \cup B) = μ^{*} (A) + μ^{*} (B)$ for all nonempty $A, B$ with $d (A, B) > 0$ . (Metric outer measures)

Proof

technique · direct

1.1

If $μ^{*} (A) < + \infty$ , let $B_{n}$ be the layers of [L1]. The sets $A \cap F$ and $B_{n}$ are disjoint and positively separated whenever both are nonempty, while normalization handles an empty member; hence [F1] and monotonicity give $μ^{*} (A) \geq μ^{*} ((A \cap F) \cup B_{n}) = μ^{*} (A \cap F) + μ^{*} (B_{n})$ .

F1algebra

2.1

Taking the supremum over $n$ in step 1.1 and using [L1] gives $μ^{*} (A) \geq μ^{*} (A \cap F) + μ^{*} (A ∖ F)$ .

step 1.1L1

3.1

If $μ^{*} (A) = + \infty$ , the inequality in step 2.1 is automatic because the right side is at most $+ \infty$ ; in every case [L2] supplies the reverse inequality, so the Carathéodory identity holds for every test set $A$ .

step 2.1L2cases∎

Depends on

Used by

Every Borel set is Carathéodory measurable for a metric outer measure Theorem

Dependency tree · two levels

10 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., proof of Proposition 11.16 (standard reference, not scraped)