Change of variables foundations: Jordan covers and two-sided cell distortion

Learning goals and prerequisites

Prove that a differentiable coordinate change preserves negligible Jordan boundaries and obtain a uniform two-sided volume estimate for each small cube. The inner-coverage argument supplies the lower bound alongside the outer linearization estimate. The next chapter uses these estimates to prove the integral theorem in every finite dimension.

We use finite-dimensional Euclidean topology, the fundamental theorem along a line segment, determinants, compact uniform continuity and the ordinary Darboux integral on boxes. Volume and integration are developed through finite Jordan covers and Darboux sums.

For estimates use $\|x\|_\infty=\max_j|x_j|$ and its induced matrix norm. A closed cube of side $h$ centered at $c$ has $\|x-c\|_\infty\le h/2$. Its ordinary volume is $h^n$. Changing the norm does not change continuity or differentiability in finite dimension, but this norm makes the cube constants transparent.

1. Jordan volume and a finite-cover boundary test

A bounded set is Jordan null if, for every $\varepsilon>0$, it can be covered by finitely many axis-parallel boxes with total volume below epsilon. One may require the boxes to be cubes of arbitrarily small sides: subdivide each of a finite box cover finely, allowing arbitrarily small total extra volume at its faces. Jordan volume $v(E)$ is the Riemann integral of the indicator of $E$ in a containing box, when that indicator is integrable.

Lemma: bounded Jordan sets have null boundaries, and conversely

A bounded set $E$ is Jordan measurable exactly when $\partial E$ is Jordan null. Moreover, on a sufficiently fine regular cube grid, the total volume of cubes meeting $\partial E$ tends to zero.

Proof from Darboux sums. A cell whose interior meets both $E$ and its complement has indicator oscillation one, and cells not meeting the boundary have oscillation zero. If the boundary is null, compactness of its closure permits a finite open box cover of arbitrarily small total volume. Enlarge a null cover slightly to make it open while keeping its total volume small. A sufficiently fine grid's boundary cells lie in that open cover: otherwise cells of shrinking diameter would supply points arbitrarily close to the compact boundary but outside its neighborhood, a contradiction. Their total volume is then small, proving the Darboux criterion.

Conversely, choose a grid with indicator upper-lower gap below epsilon. Every boundary point not on a grid face lies in a cell meeting both sets, hence one of those oscillating cells. The finitely many grid faces themselves are null: cover each bounded face by a slab of arbitrarily small thickness. The oscillating cells together with thin face slabs cover the whole boundary with arbitrarily small total volume. This proves nullity. The preceding open-cover argument then gives the regular-grid boundary-cell assertion. The argument also establishes the regular-grid boundary-cell estimate used below.

Finite unions, intersections and differences of bounded Jordan sets are Jordan: their boundaries are contained in the union of the original boundaries. Jordan volume and Riemann integration are finitely additive on pieces whose overlaps are null. To see this, refine their box approximations together; outside a cover of the finitely many null boundaries the relevant indicators add exactly, and the remaining error is bounded by the total covering volume. The same argument proves that changing a bounded integrand on a Jordan-null set does not change its integral.

If a function is continuous on a compact Jordan set $K$, its zero extension to a containing box is integrable. In cubes inside $K$, uniform continuity controls its oscillation. Cubes meeting the boundary have arbitrarily small total volume and a common bound on the function controls their contribution. Outside $K$, the extension is zero. This also shows that tagged sums over the grid cubes wholly inside $K$ converge to its integral; the omitted boundary cubes have vanishing total volume.

2. Affine box volume

Lemma: parallelepiped volume

For an invertible real matrix $A$ and an axis-parallel box $Q$,

$$v(AQ+b)=|\det A|\,v(Q).$$

The image of a box is a parallelepiped.

Elementary geometric proof. Gaussian elimination expresses $A$ as a product of coordinate permutations, nonzero coordinate scalings and shears that add a multiple of one coordinate to another. Permutations preserve box-cover volumes. A coordinate scaling multiplies their volumes by the absolute scale, first exactly for boxes and then for finite box approximations. A shear preserves the volume of a parallelepiped: a slice parallel to the changed coordinate remains an interval of the same length, merely translated by a quantity depending on the other coordinates.

The needed slicing principle for polytopes follows directly from Darboux sums. A bounded polytope has boundary in finitely many hyperplanes, each null. For a bounded hyperplane piece, solve for one coordinate with nonzero coefficient and subdivide the remaining coordinates into cubes of side $h$; its height variation is $O(h)$ on each column, so covering volume is $O(h)$ overall. Slice lengths of a polytope are piecewise affine on finitely many pieces, hence Riemann integrable. Integrating the cellwise lower and upper indicator sums bounds their slice-length integral between the same sums; the boundary-null criterion makes their gap vanish. This proves volume equals the integral of the slice lengths for these polytopes. Therefore a shear's unchanged slice lengths imply unchanged volume.

Apply these elementary operations successively to $Q$. Every intermediate image is a parallelepiped, so the slicing justification continues to apply. The volume factors are one for permutations and shears, and the absolute diagonal factors for scalings. Their product is $|\det A|$ by determinant algebra. Translation preserves the finite box approximations. This proves the lemma.

The absolute value is compulsory for ordinary volume. A coordinate reflection changes determinant sign but cannot turn a nonnegative volume into a negative one.

3. A local Lipschitz map sends a compact null set to a null set

Lemma: finite-cover null preservation

Let $N$ be a compact Jordan-null subset of an open set $U$, and let $F:U\to\mathbb R^n$ be $C^1$. Then $F(N)$ is Jordan null.

Proof. Choose a compact neighborhood of $N$ inside $U$, and bound $\|DF\|_\infty$ there by some $L>0$. Cover $N$ by finitely many sufficiently small cubes in that neighborhood, with total volume as small as desired. On each cube, segments stay in the neighborhood. The fundamental theorem along a segment gives $\|F(x)-F(y)\|_\infty\le L\|x-y\|_\infty$. A cube of side $h$ centered at $c$ consequently has its image in the cube centered at $F(c)$ of side $Lh$. Those image-enclosing cubes have total volume at most $L^n$ times the original cover total. Make that total below $\varepsilon/L^n$ to obtain the required finite image cover. If desired enlarge $L$ to at least one. The cube-by-cube estimate keeps every segment in the open domain.

The same conclusion holds for any bounded subset of such a compact null set, because its image lies in the null image cover. It also applies to the inverse of a $C^1$ diffeomorphism on compact subsets of its target.

Let now $T:U\to V$ be a $C^1$ diffeomorphism between open subsets of $\mathbb R^n$. If $K\subset U$ is compact and Jordan measurable, then $T(K)$ is compact and Jordan measurable. Indeed a homeomorphism of open neighborhoods maps the boundary to the boundary:

$$\partial T(K)=T(\partial K).$$

For justification, $T$ maps the interior of $K$ to the interior of $T(K)$ in both directions because it and its inverse are open; compactness makes both sets closed. Subtracting interiors proves the boundary identity. Apply null preservation to $\partial K$ and then the Jordan boundary criterion.

This is also valid for bounded Jordan subsets whose closure lies in $U$, by applying the same local boundary argument. It will justify pieces cut by a finite grid in the target.

4. Uniform cube distortion: prove both inclusions

Fix compact $K\subset U$. Choose a compact neighborhood $K_*$ of $K$ lying inside $U$. On it, both $DT$ and its inverse matrices are bounded, because their entries are continuous and every determinant is nonzero. Write $\kappa$ for a bound on $\|DT(x)^{-1}\|_\infty$ and $J_*$ for a bound on $J(x)=|\det DT(x)|$.

Uniform continuity gives a modulus $\omega(h)\to0$ bounding $\|DT(x)-DT(y)\|_\infty$ when $x,y\in K_*$ and $\|x-y\|_\infty\le h$. Consider any sufficiently small closed grid cube $Q$ of side $h$ meeting $K$; choose the size so that all of $Q$ lies in $K_*$. Let $c$ be its center and $A=DT(c)$. The segment fundamental theorem gives

$$T(x)=T(c)+A(x-c)+R(x),\qquad \|R(x)\|_\infty\le\omega(h)h/2\quad(x\in Q).$$

Set $\delta(h)=\kappa\omega(h)/2$ and normalize the image by

$$H(x)=c+A^{-1}[T(x)-T(c)].$$

Then $H(c)=c$ and $\|H(x)-x\|_\infty\le\delta(h)h$ throughout $Q$. Let $Q^+$ and $Q^-$ be concentric cubes of side $h(1+2\delta)$ and $h(1-2\delta)$, taking $h$ small enough that $\delta<1/4$.

Outer inclusion. The coordinate error bound immediately gives $H(Q)\subseteq Q^+$.

Inner inclusion. A boundary point $x\in\partial Q$ has some coordinate at distance $h/2$ from the center. Its image under $H$ therefore has that coordinate at distance at least $h/2-\delta h$. Hence $H(\partial Q)$ misses the open cube with the same interior as $Q^-$. Since $H$ is a homeomorphism on an open neighborhood, $\partial H(Q)=H(\partial Q)$, and $c$ is an interior point of $H(Q)$.

If a point of the open inner cube were outside the closed set $H(Q)$, the segment joining it to $c$, wholly in that inner cube, would have to hit $\partial H(Q)$ on leaving the set. That contradicts the just-proved boundary exclusion. Therefore the open inner cube lies in $H(Q)$, and closedness gives $Q^-\subseteq H(Q)$. This is the elementary first-exit property of a segment from an interior point of a closed set.

Undo the affine normalization. The linear-volume lemma and set inclusion give the two-sided estimate

$$J(c)(1-2\delta(h))^n h^n \le v(T(Q))\le J(c)(1+2\delta(h))^n h^n.$$

Thus there is a single $\beta(h)\to0$, independent of the selected cube near $K$, with

$$|v(T(Q))-J(c)v(Q)|\le\beta(h)v(Q).$$

For example take $\beta(h)=J_*\max((1+2\delta)^n-1,\ 1-(1-2\delta)^n)$. Also $v(T(Q))\le C v(Q)$ with fixed $C=J_*(3/2)^n$ for these small cubes. The affine-volume lemma converts the two enclosures into the stated volume bounds.

The uniform relative error allows the cell errors to be summed over a refining grid; the same constants also control boundary cubes.

Worked estimate

In dimension two, if $\delta\le1/100$ and the center Jacobian is $J(c)$, the image area lies between $0.98^2J(c)h^2$ and $1.02^2J(c)h^2$. The relative bounds remain useful as the number of grid cells grows on the order of $h^{-2}$. Uniform derivative continuity forces delta to zero as $h$ decreases.

Graded proof tasks

1. Null cover. Reproduce the Lipschitz null-image proof and explain why the derivative bound need only hold on small cover cubes, not on every segment joining arbitrary points of $U$.
2. Boundary. Prove $\partial T(K)=T(\partial K)$ under the stated compact/homeomorphism hypotheses. Which facts would need rechecking if the set reached the boundary of $U$?
3. Inner coverage. Explain why the outer image enclosure alone cannot establish the stated volume asymptotic. Reproduce the boundary-and-segment argument for the inner enclosure.
4. Uniformity. Identify where compactness supplies common derivative, inverse-matrix and continuity bounds. Explain why a separate little-o statement for every unrelated cell center is not the summed estimate needed later.

Full solutions

1. A finite cube cover of total volume $\eta$ has image enclosed by cubes of total volume at most $L^n\eta$, so choose $\eta<\varepsilon/L^n$. Each small cube and its internal segments stay in a fixed compact neighborhood inside $U$, where the derivative bound is uniform. A distant connecting segment might leave a nonconvex domain; the proof never needs it.
2. Compactness makes $K$ and $T(K)$ closed in the ambient spaces. A homeomorphism between open neighborhoods carries interior to interior in both directions. Thus $T(K\setminus\operatorname{int}K)=T(K)\setminus\operatorname{int}T(K)$, giving the boundary identity. At a domain boundary there may be no open neighborhood for that argument, no uniform derivative bound and no defined image of missing boundary points. The theorem deliberately keeps compact closure inside $U$.
3. An upper enclosure only bounds image volume above and does not show the map covers most of the affine cell. Here the normalized image boundary lies outside the open inner cube, while its center is inside the image. A segment to any allegedly missing inner point would leave the closed image and hence meet its boundary within the inner cube, impossible. Taking closure includes the inner cube's faces. This provides the necessary lower bound before computing the asymptotic.
4. Continuity of $DT$ and its inverse matrices on $K_*$ gives finite common bounds; uniform continuity of $DT$ gives one modulus for all small cubes near $K$. Consequently every volume error is at most the same $\beta(h)$ times cell volume, whose sum stays bounded. Pointwise little-o constants or scales depending independently on cell centers would not by themselves provide a common vanishing bound for the whole grid.

Summary and terms

A Jordan-null set has arbitrarily small finite box covers. A cell-distortion estimate bounds image volume above and below by nearby determinant-scaled volume. Uniformity lets the relative cell errors be summed safely. A diffeomorphism supplies neighborhood homeomorphism and invertible derivatives, not just a formal determinant. These finite-cover and finite-cell facts establish the geometric foundation of the next chapter's Riemann theorem.