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

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Countable disjoint unions of Carathéodory measurable sets are measurable and split every test set

Statement

Let $(E_{k})_{k \in N}$ be a pairwise disjoint sequence of Carathéodory measurable subsets of $X$ , and put $E = ⋃_{k} E_{k}$ . Then $E$ is Carathéodory measurable and

$μ^{*} (A \cap E) = \sum_{k = 0}^{\infty} μ^{*} (A \cap E_{k})$

for every $A \subseteq X$ . Equivalently: 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 $A \subseteq X$ .

Facts & Assumptions

Given: A pairwise disjoint sequence $(E_{k})$ of Carathéodory measurable sets, its union $E$ , and a test set $A \subseteq X$ .

[L1]

If $E_{0}, \dots, E_{n - 1}$ are pairwise disjoint Carathéodory measurable sets, then outer measure splits every test set over those pieces and the remaining complement. (Outer measure splits exactly over finite Carathéodory-measurable partitions)

[F1]

An outer measure on a set $X$ is a function $μ^{*} : P (X) \to [0, + \infty]$ that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

Proof

technique · direct

1.1L1algebra

Applying [L1] to the first $n$ sets and to $A$ gives $μ^{*} (A) \geq \sum_{k < n} μ^{*} (A \cap E_{k}) + μ^{*} (A ∖ E)$ , because $A ∖ E \subseteq A ∖ ⋃_{k < n} E_{k}$ and outer measure is monotone; this includes $n = 0$ .

2.1step 1.1F1algebra∎

Taking the supremum over $n$ in step 1.1 yields $μ^{*} (A) \geq \sum_{k} μ^{*} (A \cap E_{k}) + μ^{*} (A ∖ E)$ . Apply step 1.1 again with the test set $A \cap E$ : its remainder outside $E$ is empty, so taking the supremum gives $μ^{*} (A \cap E) \geq \sum_{k} μ^{*} (A \cap E_{k})$ without subtracting an infinite quantity. Countable subadditivity gives the reverse inequality. Substituting this equality into the first bound and using subadditivity on $(A \cap E) \cup (A ∖ E)$ proves the Carathéodory identity, even when the series is $+ \infty$ .

Depends on

Used by

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

Dependency tree · two levels

9 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 17 in Section 1.4 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory, proof of Theorem 1.7.3 (standard reference, not scraped)