C1 change of variables: a finite-dimensional Jordan/Riemann proof

Learning goals and precise scope

Prove a finite-dimensional $C^1$ change-of-variables theorem using the preceding chapter's finite Jordan covers and uniform two-sided cube distortion. Then extend the continuous case to bounded Riemann integrands by finite Darboux step brackets. The theorem below specifies the proper Jordan/Riemann setting and its coverage conditions.

Throughout, $n$ is a positive finite integer. A function is Riemann integrable over a bounded Jordan set if its zero extension to an enclosing box is Riemann integrable there; different enclosing boxes give the same integral. Null boundaries may be assigned arbitrary bounded values. The preceding chapter supplies the Jordan boundary and null-image lemmas and the uniform small-cube estimate.

1. The theorem and every hypothesis

Theorem: proper $C^1$ Jordan/Riemann substitution

Let $U,V\subseteq\mathbb R^n$ be open and let $T:U\to V$ be a $C^1$ diffeomorphism: it is a bijection, $T$ and its inverse are continuously differentiable, and in particular $\det DT(x)\ne0$ for every $x\in U$. Let $K\subset U$ be compact and Jordan measurable. Let $f:T(K)\to\mathbb R$ be bounded and Riemann integrable over $T(K)$. Then $T(K)$ is Jordan measurable, the function $x\mapsto f(T(x))|\det DT(x)|$ is Riemann integrable over $K$, and

$$\int_{T(K)}f(y)\,dy =\int_K f(T(x))J(x)\,dx, \qquad J(x)=|\det DT(x)|.$$

The empty or Jordan-null case is allowed. The pointwise determinant magnitude covers both orientations and disconnected domains. A bounded Jordan region $D$ with $\overline D\subset U$ is covered as well: replace it by its compact closure and assign bounded boundary values, since the boundary and its image are null.

Compact containment is substantive. It supplies uniform derivative and inverse-matrix bounds and keeps all small boundary cells inside the map's open domain. Scope. This is a proper bounded Riemann theorem. Unbounded or noncompact limits need convergence estimates; many-to-one maps need coverage accounting. General Lebesgue and lower-dimensional surface-area formulas are separate results.

2. Prepare one uniform grid, including the boundary layer

If $K$ is empty, both integrals are zero, so assume it is nonempty. As proved in the foundations chapter, $E=T(K)$ is compact and Jordan measurable: its boundary is the null image of $\partial K$.

Choose a compact neighborhood $K_*$ of $K$ inside $U$. For all sufficiently small regular grid cubes meeting $K$, the previous chapter supplies common constants $L,C$ and a function $\beta(h)\to0$ such that, for a cube $Q$ of side $h$ and center $c$,

$$\|T(x)-T(z)\|_\infty\le L\|x-z\|_\infty\quad(x,z\in Q),$$ $$|v(T(Q))-J(c)h^n|\le\beta(h)h^n,\qquad v(T(Q))\le Ch^n.$$

The derivative bound is taken along segments inside each cube.

Let $\mathcal I_h$ be the finite collection of grid cubes wholly contained in $K$. Let $\mathcal B_h$ be the cubes meeting $K$ but not wholly contained in it. Every cube in the second class meets $\partial K$, and

$$b_h:=\sum_{Q\in\mathcal B_h}v(Q)\longrightarrow0$$

by the Jordan boundary criterion. Cubes may share faces, but faces are null, as are their images under $T$. Global injectivity therefore makes distinct interior-cell images overlap only on null images of faces. Finite additivity applies to their volumes and integrals.

The part of $E$ not covered by the interior-cell images lies in the images of boundary cubes. Its volume is at most $Cb_h$, because $v(T(Q))\le Cv(Q)$ and the cover is finite. It is itself Jordan measurable: subtract a finite Jordan union from the Jordan set $E$. This gives the target boundary-layer error bound.

3. Prove the theorem first for continuous f

Suppose $f$ is continuous on compact $E$. It is bounded by $M$ and uniformly continuous; write $\omega_f(\eta)\to0$ for a common modulus on $E$. For every interior cube $Q$, the target integral over $T(Q)$ differs from $f(T(c))v(T(Q))$ by at most $\omega_f(Lh)v(T(Q))$. The uncovered target boundary layer contributes an error at most $MCb_h$.

Since the interior-cell images have disjoint interiors up to null sets, their total volume is at most $v(E)$. Thus

$$\left|\int_Ef-\sum_{Q\in\mathcal I_h}f(T(c))v(T(Q))\right| \le\omega_f(Lh)v(E)+MCb_h\longrightarrow0.$$

Now replace each image volume by $J(c)h^n$. For a fixed box $B$ containing $K$, the total additional error is at most

$$M\beta(h)\sum_{Q\in\mathcal I_h}h^n\le M\beta(h)v(B)\longrightarrow0.$$

The resulting sum is the interior-cell tagged sum for $f(T(x))J(x)$ on $K$. That function is continuous on compact $K$, and the preceding Jordan/Darboux lemma proves these sums tend to its integral: uniform continuity controls interior cells and $b_h\to0$ controls omitted boundary cells. Therefore the two integrals are equal for continuous $f$.

The common relative cell bound makes the total volume-replacement error vanish, while the separate boundary estimate controls the omitted cells.

4. Extract the volume identity for Jordan pieces

Taking $f\equiv1$ in the proved continuous case gives

$$v(T(S))=\int_SJ(x)\,dx$$

for every compact Jordan set $S\subset U$. It also holds for bounded Jordan $S$ whose closure lies in $U$: replace it by $\overline S$. The added subset lies in $\partial S$, whose image is null, so neither volume nor integral changes.

We need this identity for pieces obtained by cutting the target with a finite grid. Let $C$ be one grid box and $S=K\cap T^{-1}(C)$. Its boundary is contained in

$$\partial K\ \cup\ T^{-1}(\partial C\cap E).$$

The first set is null. The second is null because the target grid face portion is a bounded subset of a compact null set inside $V$, and the inverse map is $C^1$ on a neighborhood of compact $E$. The finite-cover null-image lemma applies to that inverse. Consequently $S$ is Jordan measurable and has compact closure in $U$.

Choose grid cells with a consistent half-open boundary assignment if an exactly disjoint partition is desired. The resulting pieces still have boundaries in the same finite null collection. Their image is precisely $E\cap C$, with the corresponding assignment. Thus the volume identity proves, by finite linearity, the substitution formula for every finite step function on the target grid.

The continuous theorem's constant-function consequence has thus established substitution for finite target step functions.

5. Bounded Riemann integrands through finite step brackets

Take a target box containing $E$ and let $f_0$ be the zero extension of $f$. Its Riemann integrability means that for every $\varepsilon>0$ there is a finite grid whose lower and upper step functions $\ell,u$ satisfy

$$\ell\le f_0\le u,\qquad \int(u-\ell)<\varepsilon.$$

On a cell, use the infimum and supremum of $f_0$ over its closure, assigning face values consistently. The Darboux criterion gives exactly the displayed gap. Restricting the nonnegative difference to $E$ cannot increase its integral.

The source functions $p(x)=\ell(T(x))J(x)$ and $q(x)=u(T(x))J(x)$ on $K$, extended by zero outside, are Riemann integrable. Each is a finite sum of continuous $J$ times indicators of the Jordan preimage pieces just established. Such a bounded product is integrable by the Darboux product estimate, or directly by uniform continuity off those null piece boundaries. Since $J\ge0$,

$$p(x)\le f(T(x))J(x)\le q(x),$$

and the step-function identity gives

$$\int_K(q-p)=\int_E(u-\ell)<\varepsilon.$$

A bounded function with integrable lower and upper brackets of arbitrarily small integral gap is Riemann integrable. Indeed its lower integral is at least the lower bracket's integral and its upper integral is at most the upper bracket's integral, so their difference is below every positive epsilon.

This proves the transformed integrand is integrable. Its integral and the original $\int_Ef$ both lie between the same numbers $\int_E\ell=\int_Kp$ and $\int_Eu=\int_Kq$, whose gap is below epsilon. Their difference is therefore zero. The theorem, including the bounded Riemann case, is proved.

The finite Darboux brackets and the Jordan volume identity complete the bounded Riemann extension.

6. Worked nonlinear map, including a discontinuous integrand

Let $T(u,v)=(u,v+u^2)$ on $\mathbb R^2$. Its inverse is $(x,y)\mapsto(x,y-x^2)$ and its determinant is one. The unit square maps to

$$E=\{0\le x\le1,\quad x^2\le y\le x^2+1\}.$$

For continuous $f(x,y)=y$, the theorem gives

$$\iint_Ey\,dA=\int_0^1\int_0^1(v+u^2)\,dv\,du=\frac56.$$

The map bends the top and bottom boundaries but preserves area, since the integral of one remains one.

For the bounded indicator $f(x,y)=1$ when $y\ge x^2+1/2$ and zero below, the discontinuity inside $E$ lies on one smooth graph. Its bounded zero extension is Riemann integrable: cover that graph and $\partial E$ by finite thin cells of arbitrarily small total area; away from them it is constant. The transformed function is the step $1_{\{v\ge1/2\}}$, so its integral is $1/2$. The example applies the bounded Riemann extension to a jump across a smooth graph.

7. Polar coordinates as a limiting corollary

On $U=(0,\infty)\times(-\pi,\pi)$, polar $T(r,\theta)=(r\cos\theta,r\sin\theta)$ is a diffeomorphism onto the plane with the nonpositive horizontal ray removed, and $J=r$. An explicit smooth inverse on that slit plane is $r=\sqrt{x^2+y^2}$ and $\theta=2\arctan[y/(\sqrt{x^2+y^2}+x)]$: its denominator is strictly positive there, and the half-angle identities verify both compositions. For $0<\varepsilon<R$ and $0<\delta<\pi$, the compact rectangle $[\varepsilon,R]\times[-\pi+\delta,\pi-\delta]$ satisfies the theorem exactly.

For a bounded continuous integrand on a closed disk, pass from these truncated sectors to the disk only after controlling the omissions. A missing angular wedge of width $2\delta$ in an annulus can be covered using another nonsingular polar branch; its area is $\delta(R^2-\varepsilon^2)$ by the theorem with integrand one. The disk of radius epsilon lies in a square of area $4\varepsilon^2$. Multiplying these bounds by a bound on the integrand makes both omitted contributions tend to zero. The circle and radial seams have null area by finite smooth-curve covers. This proves the usual disk polar formula for such bounded integrands and, with integrand one, gives disk area $\pi R^2$.

Truncation keeps the map regular and single-valued; the displayed bounds recover the origin and angular seam in the limit.

8. Nonzero Jacobian does not imply once-only coverage

For $F(x,y)=(x^2-y^2,2xy)$ on the punctured plane, $\det DF=4(x^2+y^2)>0$. It is locally invertible everywhere, but $F(x,y)=F(-x,-y)$. The annulus $1\le r\le2$ maps twice onto the annulus $1\le\rho\le4$.

The image area is $15\pi$, whereas

$$\iint_{1\le r\le2}|\det DF|\,dA =\int_0^{2\pi}\int_1^2 4r^3\,dr\,d\theta=30\pi.$$

The factor of two records the overlapping coverage. Global injectivity was the condition making distinct image cells disjoint up to null faces. Restricting to injective branches and recording their overlap separates the two traversals from the ordinary image area.

Graded theorem and application tasks

1. Hypothesis audit. Explain why compact containment $K\subset U$ and global injectivity are logically different from a nonzero derivative determinant. Give a boundary-domain example where the proper theorem does not apply without an improper limit.
2. Proof mechanism. Identify exactly where the uniform small-cube relative error and the Jordan boundary-layer estimate enter the continuous proof. Why is neither one dispensable from that argument?
3. Nonlinear shear. Verify the inverse and determinant for $(u,v)\mapsto(u,v+u^2)$ and reproduce both its $y$ integral and the indicated upper-half step integral.
4. Coverage. Verify the determinant and the two-to-one annulus description for the complex-square map. Explain which step of the theorem's proof would overcount if injectivity were silently removed.
5. Bounded Riemann extension. Explain why the finite step brackets have Jordan-integrable pullbacks and why multiplying inequalities by $J$ preserves their direction. State what the proof does not establish for an arbitrary bounded nonintegrable function.
6. Three dimensions. Use $T(u,v,w)=(u,v,w+u^2+v^2)$ on the unit cube to find the image volume and $\iiint_{T(K)}z\,dV$, verifying all map and domain conditions.

Full solutions

1. A nonzero determinant is local. It neither bounds the derivative and inverse matrices near a missing domain boundary nor prevents distinct distant inputs from sharing an image. For example $T(x)=1/x$ on $(0,\infty)$ is a diffeomorphism, but $(0,1]$ is not compactly contained in that domain and its image is unbounded. A substitution there is an improper-integral question requiring truncation and a convergence argument, not a direct use of the proper compact theorem. The complex-square annulus gives the independent global-injectivity failure despite a nonzero determinant everywhere.
2. Uniform relative error controls the replacement of every curved-cell volume by $J(c)h^n$; summing gives at most $M\beta(h)$ times a fixed containing-box volume, tending to zero. The boundary-layer bound controls the target points not covered by wholly interior source cells, by $MCb_h$. Interior linear approximation alone misses that layer, while boundary control alone does not identify the interior volume density. Both estimates are explicit parts of the equality proof.
3. Solving $x=u,y=v+u^2$ gives $u=x,v=y-x^2$, a global smooth inverse. The determinant is one. Over the unit square the transformed $y$ is $v+u^2$, integrating to $1/2+1/3=5/6$. The indicator of $y\ge x^2+1/2$ becomes the indicator of $v\ge1/2$, whose square integral is $1/2$. Boundaries and the separating graph are Jordan null, so bounded assignments on them do not change either value.
4. The derivative matrix is $\begin{pmatrix}2x&-2y\\2y&2x\end{pmatrix}$, determinant $4(x^2+y^2)$. In polar notation it squares radius and doubles angle, producing exactly two antipodal preimages of each image point in the stated annuli. Source determinant integration yields $30\pi$ while image area is $15\pi$. The finite image-cell additivity step would fail: images from distinct source regions overlap in positive area, not merely on null faces.
5. Each preimage grid piece has boundary contained in $\partial K$ plus an inverse image of finitely many target faces. Both are Jordan null by the finite-cover lemma for $T^{-1}$, so each piece is Jordan. Continuous $J$ times its indicator is Riemann integrable. The determinant magnitude is nonnegative, so lower and upper inequalities stay in order. The argument assumes target Riemann integrability to obtain arbitrarily close Darboux brackets; arbitrary bounded functions may have no such brackets, and their pullbacks are not asserted integrable by this theorem.
6. The inverse is $(x,y,z)\mapsto(x,y,z-x^2-y^2)$, smooth on all of $\mathbb R^3$. The triangular derivative has determinant one; the cube is compact Jordan inside that open domain. Its image is the curved slab $0\le x,y\le1$, $x^2+y^2\le z\le x^2+y^2+1$, of volume one. The transformed integrand is $w+u^2+v^2$, whose cube integral is $1/2+1/3+1/3=7/6$. This illustrates the theorem's finite-dimensional form beyond planar area.

Summary

Substitution combines Jordan boundary preservation with uniform two-sided cell volumes. The interior sums converge because their relative errors are uniform, and the boundary layer has vanishing volume. Finite Darboux brackets then extend the continuous formula to bounded Riemann integrands. The nonlinear shear, polar truncation and double-cover examples show how to verify the hypotheses in calculations.