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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

False: Poincare recurrence needs no finite total measure

Statement

Poincaré recurrence is false if the finite-total-measure hypothesis is omitted. Assuming countable choice, $T (x) = x + 1$ on $(R, L, λ)$ preserves measure, but no point of the positive-measure set $E = [0, 1)$ ever returns to $E$ at a positive time.

Facts & Assumptions

[F1]

Under countable choice, Lebesgue measure on the real line is a measure and has infinite total mass. Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume.

[F2]

The half-open interval $[0, 1)$ is measurable of measure one. 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.

[F3]

Translation preserves Lebesgue measurability and measure. Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation.

[F4]

A measurable self-map preserves measure exactly when each measurable inverse image has the original measure. Measure-preserving transformations and systems.

Refutation

Given: Poincaré recurrence is false if the finite-total-measure hypothesis is omitted. Assuming countable choice, $T (x) = x + 1$ on $(R, L, λ)$ preserves measure, but no point of the positive-measure set $E = [0, 1)$ ever returns to $E$ at a positive time.

1.1

By [F1] this is a measure space with $λ (R) = \infty$ . For every Lebesgue measurable $A$ , one has $T^{- 1} A = A - 1$ , which is measurable and has measure $λ (A)$ by [F3]. Thus $T$ is measurable and measure preserving by [F4], even though the total measure is infinite. The set $E$ has measure one by [F2]. Countable choice is used to obtain the Lebesgue measure in [F1] and the interval value in [F2].

F1F2F3F4

2.1

Induction gives $T^{n} x = x + n$ for every $n \geq 0$ . If $x \in [0, 1)$ and $n \geq 1$ , then $x + n \geq 1$ , so $T^{n} x \notin [0, 1)$ . The exceptional set for recurrence is therefore all of $E$ , of measure one, rather than a null subset. The time-zero visit does not satisfy the positive-return conclusion.

1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

38 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

E–W Example 2.12 pp.21–22 (standard reference, not scraped)