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

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False: an ergodic invariant sigma-algebra has only two sets

Statement

Assuming countable choice, it is false that an ergodic system has only the empty set and the whole space as strictly invariant measurable sets. For Lebesgue doubling on $[0, 1)$ , the dyadic rationals form a nonempty proper strictly invariant null set.

Facts & Assumptions

[F1]

Doubling is ergodic for Lebesgue probability. Doubling is ergodic for Lebesgue measure.

[F2]

Strict invariance means exact equality with the inverse image. Strict and mod-null invariant sigma-algebras.

[F3]

Countable real sets are measurable and Lebesgue null under countable choice. Every at most countable subset of $R^{n}$ is Lebesgue null; in particular $λ_{1} (Q) = 0$ .

[F4]

Countable unions of finite sets are countable under countable choice. Countable unions of at most countable sets, assuming ${A C}_{ω}$ .

Refutation

Given: Assuming countable choice, it is false that an ergodic system has only the empty set and the whole space as strictly invariant measurable sets. For Lebesgue doubling on $[0, 1)$ , the dyadic rationals form a nonempty proper strictly invariant null set.

1.1

Put $Q_{2} = ⋃_{n \geq 0} {k / 2^{n} : 0 \leq k < 2^{n}, k integer}$ . Each level is finite, so [F4] and [F3] give measurability and $λ (Q_{2}) = 0$ ; it is also Borel as a countable union of finite closed subsets of the circle. It contains $0$ and is proper: $1 / 3 \notin Q_{2}$ , since $1 / 3 = k / 2^{n}$ would give $2^{n} = 3 k$ , while induction gives the residue of $2^{n}$ modulo 3 as $1$ for even $n$ and $2$ for odd $n$ .

F3F4

2.1

If $x = k / 2^{n}$ is dyadic, then $D (x) = {2 x}$ is dyadic, including $n = 0$ when $x = 0$ . Conversely, if $D (x) = k / 2^{n}$ , write $2 x = k / 2^{n} + ℓ$ with $ℓ \in {0, 1}$ . Then $x = (k + ℓ 2^{n}) / 2^{n + 1}$ is dyadic and belongs to $[0, 1)$ . Hence $D^{- 1} Q_{2} = Q_{2}$ exactly, as required by [F2]. By [F1] the system is ergodic, but its invariant sigma-algebra contains this nonempty proper set. Ergodicity only constrains its measure to zero or one. Countable choice is inherited from [F1], [F3] and [F4].

1.1F1F2F3F4∎

Depends on

Used by

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Sources

E–W Proposition 2.14; doubling example specialization (standard reference, not scraped)