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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-11
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On a small cube, a C1 diffeomorphism distorts Jordan content by factors arbitrarily close to its linearized absolute determinant

Statement

Let n≥1, let g:U→Rn be C1 on an open set, let a∈U, and suppose A=Dg(a) is invertible. For every 0<ε<1 there is a closed cube Q centred at a, of positive radius and contained in U, such that every Jordan set E⊆Q has Jordan image and ∣det⁡Dg(a)∣(1−ε)ncont⁡(E)≤cont⁡(g(E))≤∣det⁡Dg(a)∣(1+ε)ncont⁡(E). The cube may be chosen inside any prescribed neighborhood of a.

Facts & Assumptions

Given: The C1 map, the point a, invertible A=Dg(a), and 0<ε<1.

[L1]

A linear endomorphism maps Jordan sets to Jordan sets and scales content by its absolute determinant (A linear endomorphism of Rn sends bounded Jordan sets to bounded Jordan sets and scales their content by the absolute determinant).

[L2]

A map whose derivative is uniformly close enough to the identity sandwiches each sufficiently small cube between concentric contracted and expanded cubes (A C1 map uniformly close to the identity derivative sandwiches a cube between contracted and expanded cubes).

[L3]

Jordan inner and outer content approximate Jordan sets by finite rectangular figures (Jordan inner and outer content and Jordan measurable bounded sets in Rm).

[L4]
[L5]

Jordan content is finitely additive across Jordan pieces whose overlaps have content zero (Jordan content is finitely additive when the overlap has content zero).

Proof

technique · reduction
1.1

Normalize at a by the affine map H(x)=a+A−1(g(x)−g(a)). Choose 0<q<ε and a slightly larger cube inside U on which ∥DH−I∥2→2≤q/n. The mean-value bound in [L4] makes H−I a q-contraction in the sup norm, so H is injective and bi-Lipschitz there. The derivative bound also makes every DH invertible; the inverse function theorem in [L4] therefore makes H a homeomorphism on a neighborhood of the smaller positive-radius cube Q. Here H(a)=a and DH(a)=I; continuity of Dg permits the stated choice inside any prescribed neighborhood.

L4given
2.1

If E⊆Q is Jordan, the homeomorphism in step 1.1 gives ∂H(E)=H(∂E). Compose H on the larger cube with coordinatewise clamping onto that cube to obtain a global Lipschitz map. Since ∂E is null, [L4] makes H(∂E) null and hence makes H(E) Jordan.

L4step 1.1
3.1

Refine inner and outer figures from [L3] into finite unions of sufficiently small, interior-disjoint cubes P⊆E⊆R with arbitrarily small content gap. After translating at each cube centre, [L2] sandwiches its H-image between cubes with factors (1−q)n and (1+q)n. Step 2.1 makes those images Jordan, injectivity makes their interiors disjoint, and [L5] adds their contents. Letting the figure gap vanish gives the stronger bounds with q; since q<ε, these imply the displayed bounds for H(E). Finally g(E)=g(a)+A(H(E)−a), so [L1] multiplies every content by ∣det⁡A∣=∣det⁡Dg(a)∣.

L1L2L3L5step 1.1step 2.1∎

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