LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (gpt-6.1-sol)

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.

C1 material regions have compatible Jordan volume integrals

Statement

Let $V \subset R^{3}$ be a bounded open set whose boundary is locally, after a rigid coordinate change, the graph of a $C^{1}$ function, with $V$ lying locally on one side of that graph. Let $U \subset R^{3}$ be open with $\overline{V} \subset U$ . Let $ϕ : U \to ϕ (U)$ be a bijection onto an open image with $ϕ$ and $ϕ^{- 1}$ both $C^{1}$ , and let $f : ϕ (U) \to R$ be continuous. A bounded set is Jordan measurable when its indicator is Riemann integrable by the zero-extension convention; a set has content zero when it has finite cube covers of arbitrarily small total volume. Then $V$ and $ϕ (V)$ are bounded open sets with locally $C^{1}$ graph boundaries, their compact closures are Jordan measurable, $\partial V$ and $ϕ (\partial V)$ have content zero, and

$\int_{ϕ (V)} f (x) d x = \int_{ϕ (\overline{V})} f (x) d x .$

Thus a balance integral over an open $C^{1}$ material region agrees with the compact-Jordan integral on its closure after a $C^{1}$ diffeomorphic motion.

Statement references

The graph of a continuous function on a closed nondegenerate rectangle in $R^{m}$ has content zero in $R^{m + 1}$

A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero

The Euclidean inverse function theorem

Jiří Lebl, Basic Analysis II, §10.5 Jordan measurable sets — Proposition10.5.1 (boundary criterion), Definition10.5.4 and Proposition10.5.5 (zero-extension Jordan integral and continuous functions), Proposition10.5.6 (linearity/order/a.e. equality), Proposition10.5.9 (compact Jordan images under injective C1 maps); ingredients adapted in the local proof, not an attribution of the complete item verbatim.

Jiří Lebl, Basic Analysis II, §8.5 Inverse and implicit function theorems — Theorem8.5.1: local C1 inverse from invertible derivative, used for the local graph-coordinate argument. The complete C1-region compatibility statement is proved locally from the declared published suppliers.

Facts & Assumptions

Given: The open set $V$ , its local $C^{1}$ graph-boundary property, the open neighborhood $U$ , the map $ϕ$ , and the continuous function $f$ from the statement.

[F1]

The graph of a continuous function on a closed nondegenerate rectangle in $R^{m}$ has content zero in $R^{m + 1}$ : the graph of a continuous function on a closed nondegenerate rectangle in $R^{2}$ has content zero in $R^{3}$ .

[F2]

Measure zero and content zero in $R^{m}$ by countable and finite cube covers: content zero means finite cube covers of arbitrarily small total volume; it passes to subsets and finite unions, and padding such covers with degenerate cubes makes the set null.

[F3]

A bounded set in $R^{m}$ is Jordan measurable iff its boundary is null, equivalently of content zero: a bounded subset of Euclidean space is Jordan measurable if and only if its boundary has content zero.

[F4]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line: closed bounded subsets of $R^{3}$ are compact.

[F5]

A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism: the continuous image of a compact space is compact, and a continuous real-valued function on a compact space is bounded.

[F6]

Lebesgue's criterion in $R^{m}$ : a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null: a bounded function on a rectangle is Riemann integrable when its discontinuity set is null.

[F7]

A bounded set is Jordan measurable iff its indicator is Riemann integrable, and the integral is its Jordan content: the indicator of a Jordan measurable set is Riemann integrable and its integral equals its Jordan content.

[F8]

Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ : the Riemann integral is linear, monotone, and bounded in absolute value by the integral of the absolute value.

[F9]

The Riemann integral of a bounded function over a bounded Jordan measurable set: a bounded function is integrated over a bounded Jordan set by extending it by zero to a containing rectangle.

[F10]

The Euclidean inverse function theorem: a $C^{1}$ map with invertible derivative at a point has a $C^{1}$ local inverse near that point.

[F11]

The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ : derivatives of composable totally differentiable Euclidean maps multiply in order.

Proof

Given: [F1]–[F11], the stated local graph condition, and the $C^{1}$ diffeomorphism $ϕ$ .

1.1F1F4algebra

Put $S = \partial V$ . It is closed and bounded, hence compact by [F4]. At each $p \in S$ , the local graph condition gives rigid coordinates and an open planar set on which $S$ is the graph of a $C^{1}$ function. Choose a small closed nondegenerate rectangle inside that planar set so its graph lies within the boundary chart and covers a neighborhood of $p$ in $S$ . Compactness gives finitely many such graph patches covering $S$ . Each patch has content zero by [F1]. A rigid image of a cube of side $ℓ$ lies in an axis-parallel cube of side $\sqrt{3} ℓ$ , so rigid coordinate changes enlarge the total covering volume by at most $3 \sqrt{3}$ . Thus each patch has content zero in the original coordinates.

1.2F10F11given

Parameterize a local graph patch at $p \in S$ by $γ (y) = (y, h (y))$ on an open subset of $R^{2}$ ; its derivative has rank two because its first two coordinates are $y$ . Differentiate $ϕ^{- 1} \circ ϕ = i d$ and use [F11] to see that $D ϕ$ is invertible. Thus $D (ϕ \circ γ)$ has rank two. Choose rigid target coordinates so the first two coordinates of this derivative are invertible. The inverse function theorem [F10] gives a local $C^{1}$ inverse for those two coordinates, so the third coordinate expresses $ϕ (S)$ locally as a $C^{1}$ graph. The homeomorphism $ϕ$ maps the local one-sided region $V$ to one side of that image graph. Hence $ϕ (V)$ has locally $C^{1}$ graph boundary.

2.1step 1.1F2

By [F2], the finite union of the source graph patches in step 1.1 shows that $S$ has content zero.

2.2F1F2F3F4F5step 1.1step 1.2

The compact image $ϕ (S)$ has a finite cover by the local graph patches from step 1.2. By [F1] each patch has content zero in its rigid coordinates, and the cube-cover comparison in step 1.1 transfers that property to the original coordinates. Therefore $ϕ (S)$ has content zero. It is closed and bounded by compactness and [F4], so its boundary is a subset of that content-zero set and [F3] makes $ϕ (S)$ Jordan measurable.

3.1F2F3F4F5step 2.1step 2.2

The Jordan boundary criterion [F3] makes the bounded open set $V$ Jordan measurable by step 2.1. The closure $\overline{V}$ is closed and bounded, and $\partial \overline{V} \subseteq S$ , so [F2]–[F4] make it compact Jordan. The compact image $ϕ (\overline{V})$ is bounded by [F5], and its boundary is contained in $ϕ (S)$ ; [F2] and [F3] make it Jordan measurable. The open image $ϕ (V)$ is bounded as a subset of $ϕ (\overline{V})$ , and its boundary is $ϕ (S)$ ; [F3] makes it Jordan measurable.

4.1F5F6F7F8F9step 2.2∎

The function $f$ is bounded by some $M \geq 0$ on the compact set $ϕ (\overline{V})$ by [F5]. In a common containing rectangle, its zero extensions from $ϕ (V)$ and $ϕ (\overline{V})$ differ only on $ϕ (S)$ , which has content zero by step 2.2. Their bounded difference $g$ has discontinuities only on that content-zero set, so [F6] makes it integrable. Moreover $∣ g ∣ \leq M 1_{ϕ (S)}$ . By [F7], $\int 1_{ϕ (S)} = cont (ϕ (S)) = 0$ ; monotonicity and the absolute-value estimate [F8] imply $∣ \int g ∣ \leq \int ∣ g ∣ \leq M \int 1_{ϕ (S)} = 0$ . The zero-extension definition [F9] therefore gives equal integrals: $\int_{ϕ (V)} f (x) d x = \int_{ϕ (\overline{V})} f (x) d x .$

Source notes

These are source ingredients adapted in the local proof, rather than an attribution of this complete lemma verbatim.

C1 material regions have compatible Jordan volume integrals

Statement

Statement references

Facts & Assumptions

Proof

Source notes

Depends on

Dependency tree · 2 levels

Sources