PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-10

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Positive sets sweep out ergodic probability systems

Statement

In a measure-preserving probability system the following are equivalent: ergodicity; for every measurable $A$ with $μ (A) > 0$ , $μ (⋃_{n \geq 1} T^{- n} A) = 1$ ; and for every measurable $A, B$ of positive measure there is $n \geq 1$ with $μ (B \cap T^{- n} A) > 0$ .

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.

1.1F1F2F3given

Assume ergodicity and put $U = ⋃_{n \geq 1} T^{- n} A$ . Then $T^{- 1} U \subseteq U$ and $μ (T^{- 1} U) = μ (U) \leq 1$ . The finite-measure difference formula gives $μ (U ∖ T^{- 1} U) = 0$ . Since $μ (U) \geq μ (T^{- 1} A) = μ (A) > 0$ , the modulo-null invariant-set criterion yields $μ (U) = 1$ .

2.1step 1.1givenF4

If the sweep-out property holds and $μ (B) > 0$ , then $μ (B \cap U) = μ (B) > 0$ . Were every $B \cap T^{- n} A$ null, their countable union $B \cap U$ would be null. Thus at least one intersection has positive measure.

3.1F1step 2.1given∎

If the positive-intersection property holds and $T^{- 1} A = A$ , then $T^{- n} A = A$ for all n. Taking $B = X ∖ A$ gives $B \cap T^{- n} A = \emptyset$ for every n. The property excludes both A and its complement having positive measure, proving ergodicity.

Depends on

Used by

Inducing an ergodic system gives an ergodic system Proposition
Kac integral formula for excursions Theorem
Kac return-time formula without invertibility Theorem

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

E–W Proposition 2.14 pp.24–25 (standard reference, not scraped)