False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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A Birkhoff limit need not be constant

Statement

Assume the Axiom of Countable Choice. False claim: the almost-everywhere Birkhoff limit is constant without an ergodicity hypothesis.

Facts & Assumptions

Given: Countable choice, Lebesgue probability on the circle, $T = R_{1 / 2}$ , and $f = 1_{[0, 1 / 4) \cup [1 / 2, 3 / 4)}$ .

[F1]

The rotation $R_{1 / 2}$ preserves Lebesgue probability (Circle rotations preserve Lebesgue measure).

[F2]

The time averages use the iterates $T^{k}$ (Ergodic partial sums, time averages, and the invariant L2 subspace).

[F3]

Half-open intervals have their displayed positive Lebesgue lengths (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]

Countable choice is the standing assumption required by the circle-measure and interval-measure suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Refutation

technique · constructive invariant witness

1.1

The half-rotation interchanges $[0, 1 / 4)$ and $[1 / 2, 3 / 4)$ , so the union is invariant and $f \circ T = f$ .

givenconstructalgebra

2.1

Every summand in $A_{n} f$ is therefore $f$ , so $A_{n} f = f$ everywhere for every $n \geq 1$ .

F2step 1.1

3.1

By [F3], the function $f$ is $1$ on a set of measure $1 / 2$ and $0$ on its complement, hence is not almost everywhere constant. Its Birkhoff limit in step 2.1 is consequently nonconstant, refuting the claim. Countable choice is used only through [F1] and [F3].

F1F3F4step 2.1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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