Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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Reciprocal return frequency from Birkhoff

Example

Assume the Axiom of Choice. In an ergodic probability system, let E be measurable with μ(E)>0. If τm(x) is the time of the mth visit of x to E, counting time zero as a possible visit, then

τm(x)m⟶1μ(E)

for almost every x. This is the orbit-frequency form of Kac's reciprocal law.

Facts & Assumptions

Given: Full choice, an ergodic probability system (X,A,μ,T), and E∈A with μ(E)>0.

[F1]

Birkhoff's ergodic specialization gives An1E(x)→μ(E) almost everywhere (Birkhoff ergodic theorem for ergodic finite-measure systems).

[F2]

The first-return convention uses positive return times and its Kac formula is ∫ErE dμ=1, equivalently mean induced return time 1/μ(E) (First-return times and induced transformations, Kac return-time formula without invertibility).

[F3]

Full choice is the assumption recorded in The Axiom of Choice.

Verification

technique · invert the visit-frequency limit
1.1F1

Put NE(n,x)=∑k=0n−11E(Tkx). By [F1], on a conull set NE(n,x)/n→μ(E)>0. Consequently NE(n,x)→∞, so every positive visit number occurs.

2.1step 1.1construct

For such an x define canonically τm(x)=min⁡{r≥0:NE(r+1,x)=m},m≥1. Then τm(x)→∞ and NE(τm(x)+1,x)=m.

3.1step 1.1step 2.1algebra

Evaluating the limit in step 1.1 along n=τm(x)+1 gives mτm(x)+1⟶μ(E). Positivity permits reciprocals, so (τm(x)+1)/m→1/μ(E); subtracting 1/m proves τm(x)/m→1/μ(E).

4.1F1F2F3step 3.1∎

The result is orbitwise and complements [F2], which integrates the first positive return time over E. Full choice is used only through the Birkhoff specialization [F1]; the visit times in step 2.1 are least integers, not chosen from an arbitrary family.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources