LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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.

A C^1 diffeomorphism maps Lebesgue null sets to Lebesgue null sets

Statement

Let $U, V \subseteq R^{n}$ be open and let $T : U \to V$ be a $C^{1}$ diffeomorphism. If $N \subseteq U$ is Lebesgue null, then $T (N)$ is Lebesgue null.

Facts & Assumptions

Given: Open sets $U, V \subseteq R^{n}$ , a $C^{1}$ diffeomorphism $T : U \to V$ , and a null set $N \subseteq U$ .

[L1]

Lipschitz self-maps of Euclidean space send null sets to null sets. (A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets)

[L2]

Lebesgue measure is countably subadditive. (Finite and countable subadditivity of measures)

[A1]

There are closed cubes $Q_{j} \subseteq U$ with $⋃_{j} Q_{j} = U$ such that for each $j$ there is a Lipschitz map $S_{j} : R^{n} \to R^{n}$ agreeing with $T$ on $Q_{j}$ . This is the standard cube-and-clamp construction: choose an open cube whose closure still lies in $U$ , use continuity of $D T$ there to get a derivative bound and hence a Lipschitz bound on that cube, then compose $T$ with the coordinatewise clamp onto the cube to obtain a global Lipschitz extension.

Proof

technique · direct

1.1

Write $N = ⋃_{j \geq 1} (N \cap Q_{j})$ using [A1]. Each set $N \cap Q_{j}$ is null, and the global Lipschitz extension $S_{j}$ from [A1] agrees with $T$ on $Q_{j}$ . Therefore [L1] gives $λ_{n} (T (N \cap Q_{j})) = λ_{n} (S_{j} (N \cap Q_{j})) = 0$ for every $j$ .

A1L1

2.1

Since $T (N) = ⋃_{j \geq 1} T (N \cap Q_{j})$ , [L2] implies $λ_{n} (T (N)) \leq \sum_{j \geq 1} λ_{n} (T (N \cap K_{j})) = 0.$ Hence $T (N)$ is Lebesgue null.

L2step 1.1∎

Depends on

Used by

Dependency tree · two levels

19 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

Gerald B. Folland, Real Analysis, 2nd ed., Theorem 2.47 (standard reference, not scraped)