PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Compositions, iterates and completions preserve invariance

Statement

Compositions and nonnegative iterates of measure-preserving self-maps of $(X, A, μ)$ preserve measure. Assuming countable choice, $T$ also defines a measurable measure-preserving self-map of the completion $(X, \overline{A}, \overline{μ})$ . Countable choice is needed here only for the cited construction of the completion measure.

Facts & Assumptions

[F1]

Under countable choice the completion construction is a complete measure extending the original measure Assuming countable choice, every measure space has a unique complete extension to its completion.

[F2]

Countable choice is assumed for this completion construction The Axiom of Countable Choice ( ${A C}_{ω}$ ).

[F3]

Every completion-measurable set is an original measurable set modified within an original measurable null set The completion domain and proposed completed set function of a measure space.

Proof

Given: The objects and hypotheses in the statement.

1.1

For preserving $S, T$ and measurable $E$ , $(S \circ T)^{- 1} E = T^{- 1} (S^{- 1} E)$ is measurable and has measure $μ (S^{- 1} E) = μ (E)$ . The identity preserves measure; applying this composition calculation successively gives preservation for every $T^{n}$ , $n \geq 0$ .

given

2.1

Assume countable choice. The completion theorem supplies the complete measure extending $μ$ . For $E \in \overline{A}$ choose $A, N \in A$ with $μ (N) = 0$ and $E △ A \subseteq N$ . Then $T^{- 1} E △ T^{- 1} A \subseteq T^{- 1} N$ , where $T^{- 1} N$ is measurable and null. Thus $T^{- 1} E$ is completion measurable and $\overline{μ} (T^{- 1} E) = μ (T^{- 1} A) = μ (A) = \overline{μ} (E)$ .

F1F2givenF3∎

Depends on

Used by

Mod-null invariant sets have strict representatives Lemma
Positive sets sweep out ergodic probability systems Proposition

Dependency tree · two levels

17 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

Einsiedler–Ward Exercise 2.1.3, p.19; Sarig Proposition 1.4 preservation proof, p.8 (standard reference, not scraped)