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

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Ergodic averages need not equal the space mean without ergodicity

Statement

Assume the Axiom of Countable Choice. False claim: in every measure-preserving probability system, $A_{n} f$ converges almost everywhere to the constant $\int f d μ$ .

Facts & Assumptions

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

[F1]

Every circle rotation preserves Lebesgue probability (Circle rotations preserve Lebesgue measure).

[F2]

The averages are $A_{n} f = n^{- 1} \sum_{k < n} f \circ T^{k}$ (Ergodic partial sums, time averages, and the invariant L2 subspace).

[F3]

Half-open intervals have Lebesgue measure equal to their length (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

Addition of $1 / 2$ modulo one interchanges the two component intervals of $E$ . Therefore $T^{- 1} E = E$ and $f \circ T = f$ .

givenconstructalgebra

2.1

It follows from [F2] that $A_{n} f = f$ at every point for every $n \geq 1$ . Yet $f$ is nonconstant, while $\int f d λ = 1 / 4 + 1 / 4 = 1 / 2$ by [F3].

F2F3step 1.1

3.1

Thus this measure-preserving probability system has a bounded observable whose averages do not approach its constant space mean. Ergodicity is the missing hypothesis. Countable choice is used only through [F1] and [F3].

F1F3F4step 2.1discharge-construct∎

Depends on

Used by

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Sources

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