Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 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.

Positive sets sweep out ergodic probability systems

Statement

In a measure-preserving probability system the following are equivalent: ergodicity; for every measurable A with μ(A)>0, μ(n1TnA)=1; and for every measurable A,B of positive measure there is n1 with μ(BTnA)>0.

Facts & Assumptions

[F1]

On probability systems ergodicity is equivalent to null/conull modulo-null invariant sets Equivalent invariant-set and invariant-function criteria for ergodicity.

[F2]

Every nonnegative iterate preserves measure Compositions, iterates and completions preserve invariance.

[F3]

Nested measurable sets of equal finite measure have null difference Measure of a set difference when the smaller set has finite measure.

[F4]

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

Assume ergodicity and put U=n1TnA. Then T1UU and μ(T1U)=μ(U)1. The finite-measure difference formula gives μ(UT1U)=0. Since μ(U)μ(T1A)=μ(A)>0, the modulo-null invariant-set criterion yields μ(U)=1.

F1F2F3given
2.1

If the sweep-out property holds and μ(B)>0, then μ(BU)=μ(B)>0. Were every BTnA null, their countable union BU would be null. Thus at least one intersection has positive measure.

step 1.1givenF4
3.1

If the positive-intersection property holds and T1A=A, then TnA=A for all n. Taking B=XA gives BTnA= for every n. The property excludes both A and its complement having positive measure, proving ergodicity.

F1step 2.1given

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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