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

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Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure

Statement

For an outer measure $μ^{*}$ on $X$ , the Carathéodory measurable subsets form a sigma-algebra, and the restriction of the outer measure to it is a complete measure.

Facts & Assumptions

Given: An outer measure $μ^{*}$ on $X$ and its family $M_{μ^{*}}$ of Carathéodory measurable sets.

[L1]

The Carathéodory measurable subsets of $X$ form an algebra of subsets. (Carathéodory measurable sets form an algebra)

[L2]

A countable disjoint union of Carathéodory measurable sets is Carathéodory measurable, and $μ^{*} (A \cap ⋃_{k} E_{k}) = \sum_{k} μ^{*} (A \cap E_{k})$ for every $A \subseteq X$ . (Countable disjoint unions of Carathéodory measurable sets are measurable and split every test set)

[L3]

Every set of outer measure zero, and every subset of it, is Carathéodory measurable and has outer measure zero. (Every outer-null set is Carathéodory measurable)

[L4]

Let $A$ be an algebra of subsets of $X$ . If the union of every pairwise disjoint sequence in $A$ belongs to $A$ , then $A$ is a sigma-algebra on $X$ . (An algebra closed under countable disjoint unions is a sigma-algebra)

Proof

technique · direct

1.1L1L2L4

The family $M_{μ^{*}}$ is an algebra by [L1] and is closed under countable disjoint unions by [L2], so [L4] makes it a sigma-algebra.

1.2L2algebra

For a pairwise disjoint sequence $(E_{k})$ in $M_{μ^{*}}$ , use $A = ⋃_{k} E_{k}$ in [L2]; then $A \cap E_{k} = E_{k}$ , so $μ^{*} (⋃_{k} E_{k}) = \sum_{k} μ^{*} (E_{k})$ , including infinite values, and the restriction is a measure.

2.1L3algebra∎

If $N \in M_{μ^{*}}$ has restricted measure zero and $S \subseteq N$ , then [L3] makes $S$ measurable with outer measure zero; hence the restricted measure is complete.

Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources