TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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.

An injective $C^{1}$ map with invertible derivative sends compact Jordan sets to compact Jordan sets

Statement

Let $n \geq 1$ , let $U \subseteq R^{n}$ be open, let $g : U \to R^{n}$ be injective and $C^{1}$ , and suppose $D g (x)$ is invertible for every $x \in U$ . If $K \subseteq U$ is compact and Jordan measurable, then $g (K)$ is compact and Jordan measurable.

Facts & Assumptions

Given: The open set $U$ , injective $C^{1}$ map $g$ , and compact Jordan set $K \subseteq U$ .

[L1]

The Euclidean inverse function theorem makes $g$ a local $C^{1}$ diffeomorphism wherever its derivative is invertible (The Euclidean inverse function theorem).

[L2]

A continuous image of a compact space is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L3]

Lipschitz self-maps of Euclidean space preserve null sets (A Lipschitz map $R^{m} \to R^{m}$ sends null sets to null sets), and a bounded set is Jordan measurable exactly when its boundary is null (A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero).

[L4]

A uniform derivative bound on a convex open set gives a Lipschitz bound there (On a convex open set, a uniform bound $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ implies $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ ).

Proof

technique · local-to-global

1.1L1L2

Continuity and [L2] make $g (K)$ compact, hence closed and bounded. If $y = g (x) \in \partial g (K)$ , then $x$ cannot lie in the interior of $K$ : otherwise [L1], together with global injectivity on $U$ , would map a neighborhood of $x$ contained in $K$ onto a neighborhood of $y$ contained in $g (K)$ . Thus $\partial g (K) \subseteq g (\partial K) .$

1.2L3L4given

Around each point of the compact set $\partial K$ , choose a closed cube in a slightly larger convex cube inside $U$ on which $D g$ is bounded. By [L4], $g$ is Lipschitz on the smaller cube. Composing its restriction with coordinatewise clamping onto that cube produces a Lipschitz self-map of $R^{n}$ , so [L3] sends the null set $\partial K$ inside the cube to a null set. A finite subcover shows that $g (\partial K)$ is null.

2.1L3step 1.1step 1.2∎

Step 1.1 makes $\partial g (K)$ a subset of the null set from step 1.2. Since $g (K)$ is bounded, the boundary criterion in [L3] proves it is Jordan measurable.

Depends on

Dependency tree · 2 levels

Sources

A. Leibman, Multidimensional Real Analysis, Theorem 5.5.7

An injective C1C1 map with invertible derivative sends compact Jordan sets to compact Jordan sets