LemmaStatement: AI-adaptedProof: 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.

Outer measure splits exactly over finite Carathéodory-measurable partitions

Statement

Let $E_{0}, \dots, E_{n - 1}$ be pairwise disjoint Carathéodory measurable subsets of $X$ , where $n \in N$ . For every $A \subseteq X$ ,

$μ^{*} (A) = \sum_{k < n} μ^{*} (A \cap E_{k}) + μ^{*} (A ∖ ⋃_{k < n} E_{k}) .$

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.

Facts & Assumptions

Given: An outer measure $μ^{*}$ , a natural number $n$ , pairwise disjoint Carathéodory measurable sets $E_{0}, \dots, E_{n - 1}$ , and a test set $A \subseteq X$ .

[F1]

A set $E \subseteq X$ is Carathéodory measurable for $μ^{*}$ when $μ^{*} (A) = μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ for every $A \subseteq X$ . (Carathéodory measurable sets)

[L1]

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

Proof

technique · induction

1.1L1base

At $n = 0$ the finite union is empty and the finite sum is the empty sum $0$ , so the formula is $μ^{*} (A) = 0 + μ^{*} (A)$ ; moreover every partial union $U_{n} : = ⋃_{k < n} E_{k}$ is measurable by [L1].

1.2ih

Assume the displayed formula at $n$ and put $R_{n} : = A ∖ U_{n}$ .

2.1step 1.2F1algebra

Since $E_{n}$ is disjoint from $U_{n}$ , [F1] gives $μ^{*} (R_{n}) = μ^{*} (A \cap E_{n}) + μ^{*} (A ∖ U_{n + 1})$ . Substituting this equality into the induction hypothesis in step 1.2 gives the formula for $n + 1$ , including empty pieces and infinite values without cancellation.

3.1step 1.1step 2.1discharge-induction∎

Step 1.1 is the base case and step 2.1 proves the successor case, so the formula holds for every $n \in N$ .

Depends on

Used by

Countable disjoint unions of Carathéodory measurable sets are measurable and split every test set 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., proof of Theorem 1.11 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory, proof of Theorem 1.7.3 (standard reference, not scraped)