ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Kac mean return to a half-circle under irrational rotation

Example

Assume countable choice. Let $α$ be irrational, let $R_{α}$ act on the Lebesgue circle, and put $E = [0, 1 / 2]$ . Its first positive return time satisfies $\int_{E} r_{E} d λ = 1$ . On the recurrent core $E_{\infty}$ with normalized restricted probability $λ_{E}$ , the mean return time is $\int_{E_{\infty}} r_{E} d λ_{E} = 2$ .

Facts & Assumptions

[F1]

For a positive-measure set in an ergodic probability system, Kac gives unnormalized mean one and normalized mean reciprocal to the set measure. Kac return-time formula without invertibility.

[F2]

Irrational rotations are ergodic for Lebesgue probability. Circle rotation is ergodic for Lebesgue measure exactly at irrational angles.

[F3]

Finite sets, including endpoints, are Lebesgue null. Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ .

Verification

Given: Assume countable choice. Let $α$ be irrational, let $R_{α}$ act on the Lebesgue circle, and put $E = [0, 1 / 2]$ . Its first positive return time satisfies $\int_{E} r_{E} d λ = 1$ . On the recurrent core $E_{\infty}$ with normalized restricted probability $λ_{E}$ , the mean return time is $\int_{E_{\infty}} r_{E} d λ_{E} = 2$ .

1.1

The set $E$ is Borel and is the disjoint union of $[0, 1 / 2)$ and ${1 / 2}$ . The first has length $1 / 2$ and the second is null by [F3], so $λ (E) = 1 / 2 > 0$ . By [F2] irrationality gives ergodicity of the Lebesgue probability system. All hypotheses of [F1] hold; applying its first conclusion yields $\int_{E} r_{E} d λ = 1$ . This uses the return time to the closed set $E$ as stated, so no unproved comparison between return times for different endpoint conventions is involved.

F1F2F3

2.1

The recurrent core has restricted mass $λ (E_{\infty}) = λ (E) = 1 / 2$ in [F1], and normalization divides restricted measure by $1 / 2$ . The second conclusion of [F1] therefore gives $\int_{E_{\infty}} r_{E} d λ_{E} = 1 / (1 / 2) = 2$ . This is an average, not an assertion that all return times equal two. The unnormalized mean is one, and multiplying it by the normalization factor two gives the same answer. Countable choice is propagated from the Lebesgue ergodicity and endpoint-null suppliers [F2] and [F3].

1.1F1F2F3∎

Depends on

Used by

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Sources

Sarig Theorem 1.7, specialization (standard reference, not scraped)