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

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Under sigma-finiteness, every Carathéodory measurable set differs from a generated measurable hull by a null set

Statement

Assume countable choice. If $μ_{0}$ is sigma-finite, $μ^{*}$ is its induced outer measure, and $E$ is Carathéodory measurable, then there is $H \in σ (A_{0})$ with $E \subseteq H$ and $μ^{*} (H ∖ E) = 0$ .

Facts & Assumptions

Given: Countable choice, a sigma-finite premeasure $μ_{0}$ , a covering sequence $(A_{n})$ of finite premeasure, and a Carathéodory measurable set $E$ .

[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]

Assuming countable choice, an outer measure induced by a premeasure is regular, and every set has a measurable hull in $σ (A_{0})$ . (Assuming countable choice, a premeasure-induced outer measure is regular with generated measurable hulls)

[L2]

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 $μ_{0}$ . (Assuming countable choice, a premeasure extends through its induced outer measure)

Proof

technique · direct

1.1F1L1choose

Put $P_{n} = ⋃_{k \leq n} A_{k}$ , so [F1] gives $P_{n} \in A_{0}$ , $P_{n} ↑ X$ , and $μ_{0} (P_{n}) < + \infty$ ; using countable choice and [L1], select $H_{n} \in σ (A_{0})$ with $E \cap P_{n} \subseteq H_{n}$ and $μ^{*} (H_{n}) = μ^{*} (E \cap P_{n})$ .

2.1step 1.1L2algebra

Both $H_{n}$ and $E \cap P_{n}$ are Carathéodory measurable by [L2], and their common measure is finite, so additivity on $H_{n} = (E \cap P_{n}) ⊔ (H_{n} ∖ (E \cap P_{n}))$ gives $μ^{*} (H_{n} ∖ (E \cap P_{n})) = 0$ .

3.1step 2.1algebra∎

The set $H = ⋃_{n} H_{n}$ belongs to $σ (A_{0})$ and contains $E$ because $P_{n} ↑ X$ ; moreover $H ∖ E$ is contained in the union of the null excesses from step 2.1, so countable subadditivity gives $μ^{*} (H ∖ E) = 0$ .

Depends on

Used by

Assuming countable choice, the Carathéodory domain is the completion of the sigma-finite extension Theorem

Dependency tree · two levels

23 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., Exercise 18(c) in Section 1.4 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory, Exercise 1.7.9(i) (standard reference, not scraped)