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

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Mod-null invariance need not be strict invariance

Statement refuted

The assertion that every invariant-modulo-null-sets measurable set is strictly invariant is false. Assuming countable choice, for Lebesgue doubling the set $E = {0}$ satisfies $λ (D^{- 1} E △ E) = 0$ , but $D^{- 1} E \neq E$ .

Facts & Assumptions

[F1]

Doubling preserves Lebesgue probability. Doubling preserves Lebesgue measure.

[F2]

Strict invariance is set equality, whereas mod-null invariance is null symmetric difference. Strict and mod-null invariant sigma-algebras.

[F3]

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

Counterexample

Given: The assertion that every invariant-modulo-null-sets measurable set is strictly invariant is false. Assuming countable choice, for Lebesgue doubling the set $E = {0}$ satisfies $λ (D^{- 1} E △ E) = 0$ , but $D^{- 1} E \neq E$ .

1.1

For $0 \leq x < 1$ , the equation $D (x) = 0$ means $2 x$ is an integer. The only possibilities are $2 x = 0$ or $2 x = 1$ , giving $D^{- 1} {0} = {0, 1 / 2}$ . Its symmetric difference with $E = {0}$ is exactly ${1 / 2}$ , a null measurable set by [F3]. Thus $E$ is invariant modulo null sets in the probability system of [F1], by [F2].

F1F2F3

2.1

The point $1 / 2$ belongs to $D^{- 1} E$ and not to $E$ , so the sets are not equal and $E$ is not strictly invariant by [F2]. Both sets are finite Borel sets; this is an exact failure for the given representative, despite their equality modulo null sets. Countable choice is used only through the Lebesgue measure and null-set suppliers [F1] and [F3].

1.1F1F2F3∎

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Sources

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