How bad is the error in the volume of a transformed region, when the transformation is approximated as linear?

Question

For context, I'm trying to prove the change of variables theorem for Riemann integrals. I need to know how a $C^1$ diffeomorphism $f$ changes the volume of a region. Here's what I know already:

• The Riemann integral can be equivalently defined using partitions into arbitrary compact Jordan-measurable sets (CJM sets), instead of (hyper)rectangles. (That is, after we've defined Jordan measure using rectangles.)

• A $C^1$ diffeomorphism, applied to a CJM set, is a CJM set. (Proposition 10.5.6 in Jiří Lebl's notes)

• A linear transformation changes volume by the absolute value of its determinant.

• Given that $f$ is continuously differentiable, there is a continuous function $h$ of two variables such that $$f(x) = f(y) + f'(y)\, (x - y) + h(x, y)\, \lVert x - y\rVert,\qquad h(y, y) = 0,$$ for all $x$ and $y$. (Here $f'(y)$ is a matrix, and all the other things are vectors, except of course $\lVert x-y\rVert$.)

Denote the volume of a region $R$ as $V(R) = \int\, [x \in R]\, dx$, using Iverson brackets: $[\text{true}]=1,\; [\text{false}]=0$.

What I want is a bound on the difference $V(f(R)) - |\det f'(y)|\, V(R)$, where $y$ is some point in $R$, and $R$ is some CJM subset of a fixed CJM set. The region $R$ could be a small rectangle, to simplify things. Specifically, for any $\varepsilon>0$, I want some $\delta>0$ such that, for all $R$ with diameter smaller than $\delta$, and for all $y \in R$, that volume difference is smaller than $\varepsilon\, V(R)$.

Answer 1

$$\newcommand{\assign}{:=} \newcommand{\nosymbol}{} \newcommand{\of}{:} \newcommand{\tmop}[1]{{\operatorname{#1}}}$$

In 3.1.49 of http://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/analmv.pdf, the author claims that given $\varepsilon > 0$, if $\delta > 0$ is small enough and $\tmop{maxsize} (\mathcal{P}) \leq \delta$, then we have \begin{eqnarray*} \eta_{\alpha} + (1 + \varepsilon) H_{\alpha} & \supset & G (R_{\alpha}) \text{ for all } R_{\alpha} \in \mathcal{P}, \end{eqnarray*} that is, \begin{eqnarray*} G (\xi_{\alpha}) + (1 + \varepsilon) D G (\xi_{\alpha}) \tilde{R}_{\alpha} & \supset & G (R_{\alpha}) \text{ for all } R_{\alpha} \in \mathcal{P}. \end{eqnarray*} We can show this by brute force. By properties of the Riemann integral, we may assume each $R_{\alpha}$ is a cube $R_{\alpha} = \{ x \in \mathbb{R}^n \of | x - \xi_{\alpha} |_{\infty} \leq \delta_{\alpha} \}$. Fix $R_{\alpha} \in \mathcal{P}$. Let $\xi_{\alpha} + y \in R_{\alpha}$ (i.e. $y \in \tilde{R}_{\alpha}$). Then we want to find $v$ such that \begin{eqnarray*} G (\xi_{\alpha} + y) & = & G (\xi_{\alpha}) + D G (\xi_{\alpha}) v. \end{eqnarray*} By the fundamental theorem of calculus, \begin{eqnarray*} G (\xi_{\alpha} + y) & = & G (\xi_{\alpha}) + D G (\xi_{\alpha}) y + \Phi (\xi_{\alpha}, y) y,\\ \Phi (\xi, y) & = & \int_0^1 (D G (\xi + ty) - D G (\xi)) d t. \end{eqnarray*} Thus we want \begin{eqnarray*} G (\xi_{\alpha}) + D G (\xi_{\alpha}) y + \Phi (\xi_{\alpha}, y) y & = & G (\xi_{\alpha}) + D G (\xi_{\alpha}) v. \end{eqnarray*} Performing algebra yields \begin{eqnarray*} v & = & y + D G (\xi_{\alpha})^{- 1} \Phi (\xi_{\alpha}, y) y. \end{eqnarray*} Let us use $\tmop{maxsize} (\mathcal{P}) \assign \max_{R_{\alpha} \in \mathcal{P}} \delta_{\alpha}$. Suppose $\tmop{maxsize} (\mathcal{P}) = \delta$. Then \begin{eqnarray*} {| v |_{\infty}}_{\nosymbol} & \leq & (1 + \| D G (\xi_{\alpha})^{- 1} \Phi (\xi_{\alpha}, y) \|) | y |_{\infty}\\ & \leq & (1 + \sup_{\xi \in R, | x | \leq \delta} \| D G (\xi)^{- 1} \Phi (\xi, x) \|) | y |_{\infty}\\ & \assign & (1 + q (\delta)) | y |_{\infty} \end{eqnarray*} Thus $v \in (1 + q (\delta)) \tilde{R}_{\alpha}$. Since $y$ was arbitrary, $$ G (R_{\alpha}) \subset G (\xi_{\alpha}) + D G (\xi_{\alpha}) (1 + q (\delta)) \tilde{R}_{\alpha} = G (\xi_{\alpha}) + (1 + q (\delta)) D G (\xi_{\alpha}) \tilde{R}_{\alpha} . $$ We now need to show that $q (\delta) = \sup_{\xi \in R, \| x \| \leq \delta} \| D G (\xi)^{- 1} \Phi (\xi, x) \| \rightarrow 0$ as $\delta \rightarrow 0$. Note that since $R \subset \mathcal{O}$ can be as small as we like, we can ensure there exists $r > 0$ such that $\Phi \of R \times \overline{B_r (0)} \rightarrow M (n \times n, \mathbb{R})$ is well defined (we just need $\{ x \of d (x, R) \leq r \} \subset \mathcal{O}$). Then note that $\Phi$ is continuous, and $\Phi (\xi, 0) = 0$. Since $\Phi$ is continuous on a compact set, $\Phi$ is uniformly continuous, so $q (\delta) = \sup_{x \in R, \| x \| \leq \delta} (\Phi (\xi, x) - \Phi (\xi, 0)) \rightarrow 0$ as $\delta \rightarrow 0$.

Answer 2

First we'll prove an upper bound for cubes: $V(f(R)) \leq (1 + \varepsilon)\, |\det f'(y)|\, V(R)$. We'll use the max norm, $\lVert v\rVert=\max\{|v_k|:1\leq k\leq n\}$ (where $n$ is the dimension), so $R=\{x:\lVert x-y\rVert\leq r\}$ is a cube with edge length $2r$. (The diameter is $2r$ according to this norm, though it would be $2r\sqrt n$ according to the Euclidean norm.)

According to the fourth bullet in the OP,

$$f'(y)^{-1} (f(x) - f(y))\; =\; x - y + f'(y)^{-1} h(x,y) \lVert x - y\rVert,$$

which implies, by the triangle inequality,

$$\lVert f'(y)^{-1} (f(x) - f(y))\rVert\; \leq\; \lVert x - y\rVert\, \big(1 + \lVert f'(y)^{-1} h(x,y)\rVert\big).$$

Since both $f'^{-1}$ and $h$ are continuous on a compact set and $h(y,y)=0$, given $0<\varepsilon'\leq1$ there exists $\delta>0$ such that (for all $x$ and $y$, uniformly) if $\lVert x-y\rVert\leq\delta$ then $\lVert f'(y)^{-1} h(x,y)\rVert\leq\varepsilon'$. Now, if the cube $R$ has diameter $2r\leq2\delta$, and $y$ is its centre and $x$ is any other point in it (which is to say that $\lVert x-y\rVert\leq r\leq\delta$), then

$$\lVert f'(y)^{-1} (f(x) - f(y))\rVert\; \leq\; \lVert x - y\rVert\, \big(1 + \varepsilon'\big)$$ $$\leq\; (1 + \varepsilon') r$$ $$f'(y)^{-1} (f(x) - f(y))\; \in\; (1 + \varepsilon') (R - y)$$ $$f(x) - f(y)\; \in\; (1 + \varepsilon')\, f'(y) (R - y).$$

Since $f(x) - f(y)$ is an arbitrary point in $f(R) - f(y)$, this shows that

$$f(R) - f(y)\; \subset\; (1 + \varepsilon')\, f'(y) (R - y).$$

As translations don't change volume, and we know how linear transformations change volume, we get

$$V(f(R))\; \leq\; (1 + \varepsilon')^n\, |\det f'(y)|\, V(R)$$ $$=\; \left(1 + \varepsilon' \sum_{k=0}^{n-1} (1 + \varepsilon')^k\right) |\det f'(y)|\, V(R)$$ $$\leq\; \left(1 + \varepsilon' \sum_{k=0}^{n-1} 2^k\right) |\det f'(y)|\, V(R)$$ $$\leq\; (1 + \varepsilon'\, 2^n)\, |\det f'(y)|\, V(R).$$

Thus, given any $\varepsilon>0$, we can define $\varepsilon'=\min\{1,\, \varepsilon/2^n\}$ and go through the above process to get $\delta$. For any cube $R$ with any centre $y$ and diameter less than $2\delta$, we have $V(f(R)) \leq (1 + \varepsilon)\, |\det f'(y)|\, V(R)$, as claimed.

---

Next we'll extend that upper bound to CJM sets. Denote such a set as $S$, with $z \in S$. (The letters $R$ and $y$ remain for cubes and their centres.) We want to show $V(f(S)) \leq (1 + \varepsilon)\, |\det f'(z)|\, V(S)$ for small enough $S$.

Find upper and lower bounds $0<L\leq|\det f'(x)|\leq M$ for all $x$.

Given any $\varepsilon'>0$, get $\delta'>0$ from the previous section, and get $\delta''>0$ from the uniform continuity of $|\det f'(x)|$, and set $\delta=\min\{\delta',\delta''\}$. Suppose $S$ has diameter less than $\delta$, so it's contained in a cube $R$ with centre $z$ and diameter less than $2\delta$. As $S$ is Jordan-measurable, for any $\varepsilon''>0$ there is a cubic partition of $R$ such that

$$\bigcup_{i\in I} R_i\; \subset\; S\; \subset\; \bigcup_{i\in I\cup J} R_i\; \subset\; R,\qquad \sum_{i\in J} V(R_i) \leq \varepsilon''.$$

The index set $I$ is for the interior of $S$, and $J$ for the boundary of $S$. Noting that the cubes are essentially disjoint, i.e. they intersect only on their boundaries which have measure zero, and diffeomorphisms preserve this property, we also have

$$\bigcup_{i\in I} f(R_i)\; \subset\; f(S)\; \subset\; \bigcup_{i\in I\cup J} f(R_i),$$ $$\bigcup_{i\in I} f'(z)(R_i)\; \subset\; f'(z)(S)\; \subset\; \bigcup_{i\in I\cup J} f'(z)(R_i),$$ $$\sum_{i\in I} V(R_i)\; \leq\; V(S)\; \leq\; \sum_{i\in I\cup J} V(R_i),$$ $$\sum_{i\in I} V(f(R_i))\; \leq\; V(f(S))\; \leq\; \sum_{i\in I\cup J} V(f(R_i)).$$

These cubes of course (being smaller than $R$) have diameter less than $2\delta$. Hence,

$$V(f(S))\; \leq\; \sum_{i\in I\cup J} V(f(R_i))$$ $$\leq\; \sum_{i\in I\cup J} (1 + \varepsilon')\, |\det f'(y_i)|\, V(R_i)$$ $$\leq\; \sum_{i\in I\cup J} (1 + \varepsilon')\, \big(|\det f'(z)| + \varepsilon'\big)\, V(R_i)$$ $$=\; (1 + \varepsilon')\, \big(|\det f'(z)| + \varepsilon'\big)\, \left(\sum_{i\in I} V(R_i) + \sum_{i\in J} V(R_i)\right)$$ $$\leq\; (1 + \varepsilon')\, \big(|\det f'(z)| + \varepsilon'\big)\, \big(V(S) + \varepsilon''\big),$$

and since $\varepsilon''>0$ was arbitrary,

$$V(f(S))\; \leq\; (1 + \varepsilon')\, \big(|\det f'(z)| + \varepsilon'\big)\, V(S)$$ $$=\; \big(|\det f'(z)| + \varepsilon'|\det f'(z)| + \varepsilon' + \varepsilon'^2\big)\, V(S)$$ $$\leq\; \big(|\det f'(z)| + \varepsilon'(M + 1 + \varepsilon')\big)\, V(S).$$

Now, given any $\varepsilon>0$, we can define $\varepsilon'=\min\{1,\, \varepsilon L/(M+2)\}$ and go through the above process to get $\delta$. For any CJM set $S$ with diameter less than $\delta$, and any $z \in S$, we have

$$V(f(S))\; \leq\; \big(|\det f'(z)| + \varepsilon L\big)\, V(S)$$ $$\leq\; (1 + \varepsilon)\, |\det f'(z)|\, V(S)$$

as claimed. Note that we could just as well use the bound $V(f(S)) \leq |\det f'(z)|\, V(S) + \varepsilon\, V(S)$ as in the OP, due to the determinant bounds $L$ and $M$.

---

Lastly we'll get a lower bound on the volume, by considering the inverse diffeomorphism $g=f^{-1}$.

Given $\varepsilon>0$, get $\delta'>0$ from the previous section applied to $g$ instead of $f$. Plug this $\delta'$ into the uniform continuity of $f$ to get $\delta>0$. Suppose $R \ni y$ is a CJM set with diameter less than $\delta$, so $S=f(R) \ni z=f(y)$ is a CJM set with diameter less than $\delta'$. Then

$$V(g(S))\; \leq\; (1 + \varepsilon)\, |\det g'(z)|\, V(S),$$

that is,

$$V(R)\; \leq\; (1 + \varepsilon)\, |\det f'(y)^{-1}|\, V(f(R))$$ $$\leq\; \frac1{1 - \varepsilon}\, |\det f'(y)^{-1}|\, V(f(R)).$$

(We can assume $\varepsilon<1$.) This gives us the desired result

$$V(f(R))\; \geq\; (1 - \varepsilon)\, |\det f'(y)|\, V(R).$$

Again, we could just as well use the form in the OP, $V(f(R)) \geq |\det f'(y)|\, V(R) - \varepsilon\, V(R)$, due to the determinant bounds.