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

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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.
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Both invariant families are sigma-algebras

Statement

For any measure-preserving system, $I$ and $I^{'}$ are sigma-algebras on $X$ , and $I \subseteq I^{'}$ .

Facts & Assumptions

[F1]

The two families use exact equality and null symmetric difference, respectively Strict and mod-null invariant sigma-algebras.

[F2]

A countable union of measurable null sets is null Finite and countable subadditivity of measures.

Proof

Given: The objects and hypotheses in the statement.

1.1

$T^{- 1} X = X$ and $T^{- 1} \emptyset = \emptyset$ . Also $T^{- 1} (X ∖ E) = X ∖ T^{- 1} E$ and $T^{- 1} (⋃_{j} E_{j}) = ⋃_{j} T^{- 1} E_{j}$ . Thus exact invariance is preserved under complements and countable unions, proving that $I$ is a sigma-algebra.

2.1

The symmetric difference of the two complements is $T^{- 1} E △ E$ . Further, $T^{- 1} (⋃_{j} E_{j}) △ ⋃_{j} E_{j} \subseteq ⋃_{j} (T^{- 1} E_{j} △ E_{j})$ . If all component differences are null, countable subadditivity makes the union null. Therefore $I^{'}$ is a sigma-algebra as well. Exact invariance gives empty symmetric difference, proving $I \subseteq I^{'}$ .

F1F2step 1.1∎

Depends on

Used by

Mod-null invariant sets have strict representatives Lemma

Cited to discharge well-definedness by Strict and mod-null invariant sigma-algebras.

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

Sarig Proposition 1.1 proof; E–W Proposition 2.14 (standard reference, not scraped)