TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $n\ge1$ , let $U\subseteq\mathbb R^n$ be open, let $g:U\to\mathbb R^n$ be injective and $C^1$ , and suppose $Dg(x)$ 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 $\mathbb{R}^m\to\mathbb{R}^m$ sends null sets to null sets), and a bounded set is Jordan measurable exactly when its boundary is null (A bounded set in $\mathbb{R}^m$ 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 $\|Df(z)v\|_2\le M\|v\|_2$ implies $\|f(y)-f(x)\|_2\le M\|y-x\|_2$ ).

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 $Dg$ 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 $\mathbb R^n$ , 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 · next 3 levels

Direct dependencies and their dependencies through the next three levels: 157 results over 25 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

An injective C1C^1C1 map with invertible derivative sends compact Jordan sets to compact Jordan sets