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

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Birkhoff averages need not converge at every point

Statement

Assume the Axiom of Countable Choice. False claim: for every $L^{1}$ observable in a measure-preserving system, $A_{n} f$ converges at every point.

Facts & Assumptions

Given: Countable choice, the doubling map $D_{2}$ on Lebesgue probability $[0, 1)$ , and $f = 1_{[0, 1 / 2)}$ .

[F1]

Under canonical binary coding, $f (D_{2}^{k} x)$ equals $1$ exactly when digit $k + 1$ is zero (Base-b digit cylinders are orbit cylinders).

[F2]

Birkhoff asserts convergence almost everywhere, not at every point (Birkhoff pointwise ergodic theorem).

[F3]

Under countable choice, $D_{2}$ preserves Lebesgue probability (Integer-base circle maps preserve Lebesgue measure, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Refutation

technique · constructive

1.1construct

Let $x$ have the binary digit string consisting successively of a zero block of length $1$ , a one block of length $2$ , a zero block of length $4$ , a one block of length $8$ , and so on, the block of index $r \geq 0$ having length $2^{r}$ . This infinite string is neither terminating nor eventually one, so it is the canonical expansion of a point $x \notin E_{2}$ .

2.1step 1.1algebra

At the end of block $r$ the number of read digits is $N_{r} = 1 + 2 + \dots + 2^{r} = 2^{r + 1} - 1$ . At the end of zero block $r = 2 m$ , the number of zero digits is $Z_{2 m} = 1 + 4 + \dots + 4^{m} = (4^{m + 1} - 1) / 3,$ so $Z_{2 m} / N_{2 m} \to 2 / 3$ .

3.1step 2.1algebra

At the end of the following one block, the zero count is unchanged, while $N_{2 m + 1} = 2^{2 m + 2} - 1$ ; hence $Z_{2 m} / N_{2 m + 1} ⟶ 1 / 3.$

4.1F1F2F3step 2.1step 3.1discharge-construct∎

By [F1], these two zero-frequency subsequences are precisely subsequences of $A_{n} f (x)$ . Their distinct limits show that $A_{n} f (x)$ diverges. By [F3] this is a measure-preserving probability system, and the bounded indicator $f$ is integrable, so the example refutes pointwise-everywhere convergence while remaining consistent with the almost-everywhere assertion in [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alessio Del Vigna, The Birkhoff Ergodic Theorem (standard reference, not scraped)