LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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A Lipschitz self-map of $R^{n}$ carries Lebesgue null sets to Lebesgue null sets

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $T : R^{n} \to R^{n}$ be Lipschitz for the Euclidean metric (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it) and let $E \subseteq R^{n}$ satisfy $λ_{n}^{*} (E) = 0$ . Then $T [E]$ is Lebesgue measurable and

$λ_{n} (T [E]) = 0.$

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, a Lipschitz map $T$ of $R^{n}$ into itself, and a set $E$ with $λ_{n}^{*} (E) = 0$ .

[L1]

Assuming countable choice, $λ_{n}^{*} (S) = 0$ if and only if $S$ is null in the covering sense of closed-cube covers (A subset of $R^{m}$ has Lebesgue outer measure zero if and only if it is null in the sense of countable closed-cube covers, Measure zero and content zero in $R^{m}$ by countable and finite cube covers).

[L2]

Assuming countable choice, every $S \subseteq R^{n}$ with $λ_{n}^{*} (S) = 0$ is Lebesgue measurable with $λ_{n} (S) = 0$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[F1]

If $T : R^{m} \to R^{m}$ is Lipschitz and $E$ is null, then $T [E]$ is null (A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets).

[F2]

$f$ is Lipschitz with constant $L \geq 0$ if $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'})$ for all $x, x^{'}$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

Proof

technique · direct

1.1

By the agreement theorem, $λ_{n}^{*} (E) = 0$ says exactly that $E$ is null in the covering sense of closed-cube covers.

2.1

The published theorem on Lipschitz images therefore applies and gives that $T [E]$ is null in that same covering sense, so the agreement theorem read the other way gives $λ_{n}^{*} (T [E]) = 0$ ; completeness then makes $T [E]$ Lebesgue measurable with $λ_{n} (T [E]) = 0$ .

step 1.1L1L2F1F2∎

Depends on

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Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Section 1.2 (standard reference, not scraped)

A Lipschitz self-map of RnRn carries Lebesgue null sets to Lebesgue null sets