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

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Assuming countable choice, regular outer measures are continuous from below on all subsets

Statement

Assume the Axiom of Countable Choice. Let $μ^{*}$ be a regular outer measure on $X$ , and let $(E_{n})_{n \in N}$ be increasing with $E = ⋃_{n} E_{n}$ . Then

$μ^{*} (E) = \sup_{n \in N} μ^{*} (E_{n}) .$

Facts & Assumptions

Given: The Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), a regular outer measure $μ^{*}$ , and an increasing sequence $(E_{n})$ with union $E$ .

[F1]

A measurable hull of $E$ is a Carathéodory measurable set $H \supseteq E$ with $μ^{*} (H) = μ^{*} (E)$ ; the outer measure is regular when every subset has a measurable hull. (Measurable hulls and regular outer measures)

[L1]

If $(E_{n})_{n \in N}$ is an increasing sequence of measurable sets for a measure $μ$ , then $μ (⋃_{n} E_{n}) = \sup_{n} μ (E_{n})$ , with no finiteness hypothesis. (Continuity from below for measures)

[L2]

For every outer measure, 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)

Proof

technique · direct

1.1F1L2givenchoose

If some $μ^{*} (E_{n}) = + \infty$ , monotonicity gives the result immediately. Otherwise countable choice and [F1] give measurable hulls $H_{n} \supseteq E_{n}$ with $μ^{*} (H_{n}) = μ^{*} (E_{n})$ ; put $G_{n} : = ⋂_{k \geq n} H_{k}$ . Sigma-algebra closure in [L2] makes each $G_{n}$ measurable, while $G_{n} \subseteq G_{n + 1}$ and $E_{n} \subseteq G_{n} \subseteq H_{n}$ , so monotonicity makes $μ^{*} (G_{n}) = μ^{*} (E_{n})$ .

2.1step 1.1L1algebra∎

Let $G = ⋃_{n} G_{n}$ . Since $E_{n} \subseteq G_{n}$ , one has $E \subseteq G$ , while [L1] for the Carathéodory restriction gives $μ^{*} (G) = \sup_{n} μ^{*} (G_{n}) = \sup_{n} μ^{*} (E_{n})$ ; hence monotonicity gives $μ^{*} (E) \leq \sup_{n} μ^{*} (E_{n})$ , and the reverse inequality follows from $E_{n} \subseteq E$ .

Depends on

Used by

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Sources

G. Folland, Real Analysis, 2nd ed., Exercises 18 and 20 in Section 1.4 (standard reference, not scraped)