Alphabeta Math
LemmaStatement: 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.

Mod-null invariant sets have strict representatives

Statement

If E∈I′ in a measure-preserving system, then F=lim sup⁡n→∞T−nE=⋂m≥0⋃n≥mT−nE belongs to I and satisfies μ(E△F)=0. This does not require completeness or a choice axiom.

Facts & Assumptions

[F1]

The invariant families consist of the stated measurable sets Both invariant families are sigma-algebras.

[F2]

Nonnegative iterates preserve measure; only the choice-free iteration clause is used Compositions, iterates and completions preserve invariance.

[F3]

Finite and countable unions of measurable null sets are null Finite and countable subadditivity of measures.

Proof

Given: The objects and hypotheses in the statement.

1.1F1F2F3

Write D=E△T−1E, which is measurable and null. For n≥1, E△T−nE⊆⋃k=0n−1T−kD: whenever membership at times zero and n differs, it differs at a consecutive pair. Iterates preserve measure, so every set on the right is null; finite subadditivity makes the left null. For n=0 the difference is empty.

2.1F3step 1.1

The displayed countable intersection and unions make F measurable. Outside the measurable null set N=⋃n≥0(E△T−nE) all these indicators equal 1E, so membership in their limsup equals membership in E. Thus E△F⊆N and its measure is zero.

3.1F1step 2.1∎

Pulling back the displayed formula gives T−1F=⋂m≥0⋃n≥mT−(n+1)E=F: removing the initial term does not change membership infinitely often. Hence F is strictly invariant and is the required representative.

Depends on

Used by

Dependency tree · two levels

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Sources