TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-01

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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.

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A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets

Statement

If $T : R^{m} \to R^{m}$ is Lipschitz and $E$ is null, then $T [E]$ is null.

Facts & Assumptions

Given: A Lipschitz constant $L \geq 0$ and null $E$ .

[L1]

Lipschitz means $∥ T (x) - T (y) ∥_{2} \leq L ∥ x - y ∥_{2}$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[L2]

Norm comparisons on $R^{m}$ bound Euclidean diameter of a side- $ℓ$ cube by a fixed dimension multiple of $ℓ$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ ).

Proof

technique · cases

1.1assume-case zeroL1given

If $L = 0$ , $T [E]$ is empty or a singleton, covered by cubes of arbitrarily small side.

1.2assume-case positiveL1L2given

Suppose $L > 0$ . The image of a side- $ℓ$ cube lies in a cube of side $C_{m} L ℓ$ , where $C_{m}$ is the fixed norm-comparison factor. Its volume is $(C_{m} L)^{m} ℓ^{m}$ .

2.1step 1.2given

Given an output budget $ε$ , cover $E$ by cubes with total volume below $ε / (C_{m} L)^{m}$ . Replacing each by its image-containing cube gives a cover of $T [E]$ with total volume below $ε$ .

3.1step 1.1step 2.1cases-exhaustive∎

Both cases prove nullity. Equal domain and codomain dimensions are used in the volume scaling.

A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A Lipschitz map Rm→RmRm→Rm sends null sets to null sets

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets