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

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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

1.1

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 $\partial g (K) \subseteq g (\partial K) .$

L1L2

1.2

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.

L3L4given

2.1

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.

L3step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

43 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

A. Leibman, Multidimensional Real Analysis, Theorem 5.5.7 (standard reference, not scraped)

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