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

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Kac normalization needs ergodicity

Statement refuted

Assume countable choice. Removing ergodicity from Kac’s probability normalization is invalid. For the identity $T$ on $([0, 1], B ([0, 1]), λ)$ and $E = [0, 1 / 2]$ , every point of $E$ has $r_{E} = 1$ , but $\int_{E} r_{E} d λ = 1 / 2$ , not one. The normalized return mean is one, not $1 / λ (E) = 2$ .

Facts & Assumptions

[F1]

The return time is the least positive return time, with infinity when there is none. First-return times and induced transformations.

[F2]

For an ergodic probability-preserving system and a positive-measure set $E$ , Kac's theorem gives $\int_{E} r_{E} d μ = 1$ and normalized mean $1 / μ (E)$ . Kac return-time formula without invertibility.

[F3]

Under countable choice, closed intervals have their lengths as Lebesgue measure. A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included.

[F4]

A strict invariant set of intermediate probability disproves ergodicity. Ergodicity relative to an invariant measure.

Counterexample

Given: Assume countable choice. Removing ergodicity from Kac’s probability normalization is invalid. For the identity $T$ on $([0, 1], B ([0, 1]), λ)$ and $E = [0, 1 / 2]$ , every point of $E$ has $r_{E} = 1$ , but $\int_{E} r_{E} d λ = 1 / 2$ , not one. The normalized return mean is one, not $1 / λ (E) = 2$ .

1.1

By [F3] the restricted Lebesgue measure has $λ ([0, 1]) = 1$ and $λ (E) = 1 / 2$ . The identity is measurable and satisfies $T^{- 1} A = A$ for every Borel $A$ , so it preserves this probability. In particular $E$ is strictly invariant of measure $1 / 2$ , and [F4] shows that the system is not ergodic. Countable choice is inherited from the Lebesgue measure in [F3].

F3F4

2.1

For each $x \in E$ and each positive integer $n$ , $T^{n} x = x \in E$ . Thus the least positive return time in [F1] is 1 and the infinitely-returning core is all of $E$ . The integral of the constant one over $E$ is $λ (E) = 1 / 2$ . The normalized restricted probability has total mass one, so the same constant return time has mean one there. Both values differ from the respective ergodic conclusions of [F2], namely one before normalization and $1 / (1 / 2) = 2$ afterwards. All endpoints return as well, so no exceptional-point convention is involved.

1.1F1F2∎

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Sources

Sarig Theorem 1.7 hypotheses (standard reference, not scraped)