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

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An explicit invariant set for a rational rotation

Example

Assume countable choice. If $α = p / q$ with integers $p$ , $q \geq 1$ in lowest terms, then $E = ⋃_{j = 0}^{q - 1} [j / q, j / q + 1 / (2 q))$ is strictly invariant under $R_{α}$ and has Lebesgue measure $1 / 2$ . Thus this rational rotation is not ergodic.

Facts & Assumptions

[F1]

Circle rotations preserve Lebesgue probability and have measurable inverses. Circle rotations preserve Lebesgue measure.

[F2]

In a probability system ergodicity requires strict invariant sets to have measure zero or one. Ergodicity relative to an invariant measure.

Verification

Given: Assume countable choice. If $α = p / q$ with integers $p$ , $q \geq 1$ in lowest terms, then $E = ⋃_{j = 0}^{q - 1} [j / q, j / q + 1 / (2 q))$ is strictly invariant under $R_{α}$ and has Lebesgue measure $1 / 2$ . Thus this rational rotation is not ergodic.

1.1

Write $E_{j} = [j / q, j / q + 1 / (2 q))$ . For $x = j / q + t$ with $0 \leq t < 1 / (2 q)$ , one has $R_{p / q} x = ((j + p) m o d q) / q + t$ . The residue map $j \mapsto j + p m o d q$ is a permutation, with inverse subtraction of $p$ . Thus rotation maps the collection of $E_{j}$ onto itself and, being bijective, satisfies $R_{p / q}^{- 1} E = E$ . The half-open convention makes the formula exact at every included left endpoint and excludes every right endpoint.

2.1

The $q$ intervals $E_{j}$ are pairwise disjoint and each has length $1 / (2 q)$ . Hence $λ (E) = \sum_{j = 0}^{q - 1} 1 / (2 q) = 1 / 2$ . It is Borel, and its complement also has measure $1 / 2$ . By [F1] the ambient system preserves the probability, so [F2] proves nonergodicity. For $q = 1$ , this is simply the identity rotation and $E = [0, 1 / 2)$ . Countable choice is used only for the Lebesgue probability supplied by [F1].

1.1F1F2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

E–W Proposition 2.16 (standard reference, not scraped)