ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

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 $τ_{m} (x)$ is the time of the $m$ th visit of $x$ to $E$ , counting time zero as a possible visit, then

$\frac{τ_{m} (x)}{m} ⟶ \frac{1}{μ (E)}$

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 $E \in A$ with $μ (E) > 0$ .

[F1]

Birkhoff's ergodic specialization gives $A_{n} 1_{E} (x) \to μ (E)$ almost everywhere (Birkhoff ergodic theorem for ergodic finite-measure systems).

[F2]

The first-return convention uses positive return times and its Kac formula is $\int_{E} r_{E} d μ = 1$ , 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

1.1

Put $N_{E} (n, x) = \sum_{k = 0}^{n - 1} 1_{E} (T^{k} x)$ . By [F1], on a conull set $N_{E} (n, x) / n \to μ (E) > 0$ . Consequently $N_{E} (n, x) \to \infty$ , so every positive visit number occurs.

2.1

For such an $x$ define canonically $τ_{m} (x) = \min {r \geq 0 : N_{E} (r + 1, x) = m}, m \geq 1.$ Then $τ_{m} (x) \to \infty$ and $N_{E} (τ_{m} (x) + 1, x) = m$ .

step 1.1construct

3.1

Evaluating the limit in step 1.1 along $n = τ_{m} (x) + 1$ gives $\frac{m}{τ_{m} (x) + 1} ⟶ μ (E) .$ Positivity permits reciprocals, so $(τ_{m} (x) + 1) / m \to 1 / μ (E)$ ; subtracting $1 / m$ proves $τ_{m} (x) / m \to 1 / μ (E)$ .

step 1.1step 2.1algebra

4.1

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.

F1F2F3step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

20 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

Charles Walkden, Ergodic Theory lecture notes (standard reference, not scraped)