Proposition 10.7.1.
Suppose \(R \subset \R^n\) is a rectangle and \(A \colon \R^n \to \R^n\) is linear. Then \(A(R)\) is Jordan measurable and \(V\bigl(A(R)\bigr) = \babs{\det (A)} \, V(R)\text{.}\)
Section 10.7 Change of variables
Note: 1 lecture
In one variable, we have the familiar change of variables, the \(u\)-substitution,
\begin{equation}
\int_a^b f\bigl(g(x)\bigr) g'(x)\, dx =
\int_{g(a)}^{g(b)} f(u) \, du .
\end{equation}
The analogue in higher dimensions is quite a bit more complicated. The first complication is orientation. If we use the definition of integral from this chapter, then we do not have the notion of \(\int_a^b\) versus \(\int_b^a\text{.}\) We are simply integrating over an interval \([a,b]\text{.}\) With this notation, and assuming \(g\) is one-to-one, the change of variables becomes
\begin{equation}
\int_{[a,b]} f\bigl(g(x)\bigr) \, \babs{g'(x)}\, dx =
\int_{g([a,b])} f(u) \, du .
\end{equation}
In this section we will obtain the several-variable analogue of this form.
Let us remark on the role of \(\babs{g'(x)}\) in the formula. The integral measures volumes in general, so in one dimension it measures length. Notice that \(\babs{g'(x)}\) scales the \(dx\) and so it scales the lengths. If our \(g\) is linear, that is, \(g(x)=Lx\text{,}\) then \(g'(x) = L\) and the length of the interval \(g([a,b])\) is simply \(\sabs{L}(b-a)\text{.}\) That is because \(g([a,b])\) is either \([La,Lb]\) or \([Lb,La]\text{.}\) This property holds in higher dimension with \(\sabs{L}\) replaced by the absolute value of the determinant.
Proof.
It is enough to prove for elementary matrices. The proof is left as an exercise.
Let us prove that the absolute value of the Jacobian determinant \(\babs{J_g(x)} = \babs{\det \bigl(g'(x)\bigr)}\) is the replacement of \(\babs{g'(x)}\) for multiple dimensions in the change of variables formula. The following theorem holds in more generality, but this statement is sufficient for many uses.
Theorem 10.7.2.
Suppose \(U \subset \R^n\) is open, \(S \subset U\) is a compact Jordan measurable set, and \(g \colon U \to \R^n\) is a one-to-one continuously differentiable mapping, such that \(J_g\) is never zero on \(S\text{.}\) Suppose \(f \colon g(S) \to \R\) is Riemann integrable. Then \(f \circ g\) is Riemann integrable on \(S\) and
\begin{equation}
\int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx
=
\int_{g(S)} f(u) \, du .
\end{equation}
The set \(g(S)\) is Jordan measurable by Proposition 10.5.9, so the left-hand side does make sense. That the right-hand side makes sense follows by Corollary 10.4.4 (actually Exercise 10.5.7).
Proof.
The set \(S\) can be covered by finitely many closed rectangles \(P_1,P_2,\ldots,P_k\text{,}\) whose interiors do not overlap such that each \(P_j \subset U\) (Exercise 10.7.2). Proving the theorem for \(P_j \cap S\) instead of \(S\) is enough as we can simply add up the integrals. Define \(f(y) \coloneqq 0\) for all \(y \notin g(S)\text{.}\) The new \(f\) is Riemann integrable as \(g(S)\) is Jordan measurable. We can now replace the integrals over \(S\) with integrals over the whole rectangle. We therefore assume without loss of generality that \(S\) is a closed rectangle.
The matrix \(g'(x)\) is invertible for every \(x \in S\text{,}\) and it is continuous. Therefore, \({\bigl(g'(x)\bigr)}^{-1}\) and consequently \(\bnorm{{\bigl(g'(x)\bigr)}^{-1}}\) is continuous and never zero. As \(S\) is compact, then there exists an \(M > 1\) so that \(\bnorm{{\bigl(g'(x)\bigr)}^{-1}} \leq M\) for all \(x \in S\text{.}\)
Let \(\epsilon > 0\) be given. For every \(x \in S\text{,}\) let
\begin{equation}
W_x \coloneqq
\left\{ y \in U : \bnorm{g'(x)-g'(y)} < \frac{\epsilon}{2M} \right\} .
\end{equation}
By Exercise 10.7.3, \(W_x\) is open. As \(x \in W_x\) for every \(x\text{,}\) we have an open cover. By the Lebesgue covering lemma (Lemma 7.4.10), there exists a \(\delta > 0\) such that for every \(y \in S\text{,}\) there is an \(x\) such that \(B(y,\delta) \subset W_x\text{.}\) In other words, if \(Q\) is a rectangle of maximum side length less than \(\frac{\delta}{\sqrt{n}}\) and \(y \in Q\text{,}\) then \(Q \subset
B(y,\delta) \subset W_x\text{.}\) By the triangle inequality, \(\bnorm{g'(\xi)-g'(\eta)} < \nicefrac{\epsilon}{M}\) for all \(\xi, \eta \in Q\text{.}\)
Let \(\varphi(x) \coloneqq f\bigl(g(x)\bigr) \babs{J_g(x)}\text{.}\) There exists a partition \(P\) of \(S\) such that \(\epsilon + \int_S \varphi \geq U(P,\varphi)\text{.}\) We can assume \(\delta\) is sufficiently small relative to the side of each subrectangle of the partition \(P\) so that we can cut each such subrectangle into further subrectangles each of whose sides \(s\) satisfies \(\frac{\delta}{2\sqrt{n}} \leq s \leq \frac{\delta}{\sqrt{n}}\) (Exercise 10.7.4). Denote these subrectangles by \(R_1,R_2,\ldots,R_N\text{.}\) For each \(j=1,2,\dots,N\text{,}\) find \(x_j \in R_j\) so that \(\sabs{J_g(x_j)} \leq \sabs{J_g(x)}\) for all \(x \in R_j\text{,}\) which is possible as \(\sabs{J_g(x)}\) is continuous and \(R_j\) is compact.
Consider some \(R_j\text{.}\) First suppose \(x_j=0\text{,}\) \(g(0) = 0\text{,}\) and \(g'(0) = I\text{.}\) We claim that \(g(R_j)\) is contained in a rectangle of volume at most \(V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n\text{.}\) Let us prove this claim. For any given \(y \in R_j\text{,}\) apply the fundamental theorem of calculus to the function \(t \mapsto g(ty)\) to find \(g(y) = \int_0^1 g'(ty)y \,dt\text{.}\) As the side of \(R_j\) is at most \(\frac{\delta}{\sqrt{n}}\text{,}\) we have \(\snorm{y} \leq \delta\text{.}\) We note that \(\bnorm{g'(x)-I} < \epsilon\) as \(M > 1\) and so
\begin{equation}
\begin{aligned}
\bnorm{g(y)-y}
=
\norm{\int_0^1 \bigl(g'(ty) y - y\bigr) \,dt}
&
\leq
\int_0^1 \bnorm{g'(ty) y - y} \,dt
\\
&
\leq
\snorm{y} \int_0^1 \bnorm{g'(ty) - I} \,dt
\leq
\delta \epsilon .
\end{aligned}
\end{equation}
Therefore, \(g(R_j) \subset \widetilde{R}_j\text{,}\) where \(\widetilde{R}_j\) is a rectangle obtained from \(R_j\) by extending by \(\delta \epsilon\) on all sides. See Figure 10.17.
If the sides of \(R_j\) are \(s_1,s_2,\ldots,s_n\text{,}\) then \(V(R_j) = s_1 s_2 \cdots s_n\text{.}\) Recall \(\delta \leq 2\sqrt{n} \, s_j\text{.}\) Thus,
\begin{equation}
\begin{split}
V(\widetilde{R}_j) & =
(s_1+2\delta \epsilon )
(s_2+2\delta \epsilon )
\cdots
(s_n+2\delta \epsilon )
\\
& \leq
\bigl(s_1+4 \sqrt{n}\,s_1 \epsilon \bigr)
\bigl(s_2+4 \sqrt{n}\,s_2 \epsilon \bigr)
\cdots
\bigl(s_n+4 \sqrt{n}\,s_n \epsilon \bigr)
\\
& =
s_1 \bigl(1+4 \sqrt{n}\, \epsilon \bigr)
\,
s_2 \bigl(1+4 \sqrt{n}\, \epsilon \bigr)
\cdots
s_n \bigl(1+4 \sqrt{n}\, \epsilon \bigr)
=
V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n .
\end{split}
\end{equation}
The claim is proved: \(g(R_j) \subset \widetilde{R}_j\) and
\begin{equation}
V\bigl(g(R_j)\bigr) \leq V(\widetilde{R}_j) \leq V(R_j) \,
{\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n .
\end{equation}
Next, suppose \(A \coloneqq g'(0)\) is not necessarily the identity. Write \(g = A \circ \widetilde{g}\) where \(\widetilde{g}'(0) = I\text{.}\) We have that \(\snorm{A^{-1}} \leq M\) and so
\begin{equation}
\bnorm{\widetilde{g}'(x)-I} =
\bnorm{A^{-1}g'(x)-A^{-1}A} \leq
\bnorm{A^{-1}}\,\bnorm{g'(x)-A} < M \frac{\epsilon}{M} = \epsilon .
\end{equation}
In other words, the claim above applies to \(\widetilde{g}\) as what we needed in its proof was precisely that \(\bnorm{\widetilde{g}'(x)-I} < \epsilon\text{.}\) Therefore, we have that \(\widetilde{g}(R_j)\) is contained in a rectangle \(\widetilde{R}_j\) with \(V(\widetilde{R}_j) \leq V(R_j) \, {\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n\text{.}\)
By Proposition 10.7.1, \(V\bigl(A(\widetilde{R}_j)\bigr) = \babs{\det(A)} \, V(\widetilde{R}_j)\text{,}\) and hence
\begin{equation}
\begin{split}
V\bigl(g(R_j)\bigr) & =
V\bigl(A\bigl(\widetilde{g}(R_j)\bigr)\bigr)
\\
& \leq
V\bigl(A(\widetilde{R}_j)\bigr)
\\
& \leq
\babs{\det(A)} \, V(R_j) \,
{\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n \\
& =
\babs{J_g(0)} \, V(R_j) \,
{\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n .
\end{split}
\end{equation}
Translation does not change volume, and therefore for every \(R_j\text{,}\) and \(x_j \in R_j\text{,}\) including when \(x_j \neq 0\) and \(g(x_j)
\neq 0\text{,}\) we find
\begin{equation}
V\bigl(g(R_j)\bigr) \leq
\babs{J_g(x_j)} \, V(R_j) \,
{\bigl(1+4\sqrt{n} \, \epsilon\bigr)}^n .
\end{equation}
Write \(f\) as \(f = f_+ - f_-\) for two nonnegative Riemann integrable functions \(f_+\) and \(f_-\text{:}\)
\begin{equation}
f_+(u) \coloneqq \max \bigl\{ f(u) , 0 \bigr\}, \qquad
f_-(u) \coloneqq \max \bigl\{ -f(u) , 0 \bigr\} .
\end{equation}
So, if we prove the theorem for a nonnegative \(f\text{,}\) we obtain the theorem for arbitrary \(f\text{.}\) Therefore, suppose without loss of generality that \(f(u) \geq 0\) for all \(u \in g(S)\text{.}\)
As the rectangles \(R_1,R_2,\ldots,R_N\) give a refinement of the partition \(P\text{,}\)
\begin{equation}
\begin{split}
\epsilon + \int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx
& \geq
\sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \biggr) \, V(R_j)
\\
& \geq
\sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \biggr) \, \babs{J_g(x_j)} \, V(R_j)
\\
& \geq
\sum_{j=1}^N \biggl(\sup_{u \in g(R_j)} f(u) \biggr) \,
V\bigl(g(R_j)\bigr)
\frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n}
\\
& \geq
\sum_{j=1}^N \left(\int_{g(R_j)}f(u) \,du \right)
\frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n}
\\
& =
\frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n}
\int_{g(S)} f(u) \,du .
\end{split}
\end{equation}
The last equality follows because the overlaps of the rectangles are their boundaries, which are of measure zero, and hence the image of their boundaries is also measure zero. Let \(\epsilon\) go to zero to find
\begin{equation}
\int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx \geq \int_{g(S)} f(u) \,du .
\end{equation}
Recall that \(g^{-1}\) exists and \(g^{-1}\bigl(g(S)\bigr) = S\text{.}\) Also, \(1 = J_{g\circ g^{-1}} = J_g\bigl(g^{-1}(u)\bigr) \,J_{g^{-1}}(u)\) for \(u \in g(S)\text{.}\) So
\begin{equation}
\begin{split}
\int_{g(S)} f(u) \, du
& =
\int_{g(S)} f\Bigl(g\bigl(g^{-1}(u)\bigr)\Bigr) \,
\Babs{J_g\bigl(g^{-1}(u)\bigr)} \, \babs{J_{g^{-1}}(u)} \, du
\\
& \geq
\int_{g^{-1}(g(S))} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx
=
\int_{S} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx .
\qedhere
\end{split}
\end{equation}
Exercises Exercises
10.7.1.
Prove Proposition 10.7.1.
10.7.2.
Suppose \(U \subset \R^n\) is open and \(S \subset U\) is a compact Jordan measurable set. Show that there exist finitely many closed rectangles \(P_1,P_2, \ldots, P_k\) such that \(P_j \subset U\text{,}\) \(S \subset P_1 \cup P_2 \cup \cdots \cup P_k\text{,}\) and the interiors are mutually disjoint, that is, \(P_j^\circ \cap P^\circ_\ell = \emptyset\) whenever \(j \neq \ell\text{.}\)
10.7.3.
Suppose \(U \subset \R^n\) is open, \(x \in U\text{,}\) and \(g \colon U \to \R^n\) is a continuously differentiable mapping. For every \(c > 0\text{,}\) show that
\begin{equation}
\bigl\{ y \in U : \bnorm{g'(x)-g'(y)} < c \bigr\}
\end{equation}
is an open set.
10.7.4.
Suppose \(R \subset \R^n\) is a closed rectangle. Show that if \(\delta > 0\) is sufficiently small relative to the sides of \(R\text{,}\) then \(R\) can be partitioned into subrectangles where each side \(s\) of every subrectangle satisfies \(\frac{\delta}{2} \leq s \leq \delta\text{.}\)
10.7.5.
Prove the following version of the theorem: Suppose \(f \colon \R^n \to \R\) is a Riemann integrable compactly supported function. Suppose \(K \subset \R^n\) is the support of \(f\text{,}\) \(S\) is a compact set, and \(g \colon \R^n \to \R^n\) is a function that when restricted to a neighborhood \(U\) of \(S\) is one-to-one and continuously differentiable, \(g(S) = K\) and \(J_g\) is never zero on \(S\) (in the formula assume \(J_g(x) = 0\) if \(g\) is not differentiable at \(x\text{,}\) that is, when \(x \notin
U\)). Then
\begin{equation}
\int_{\R^n} f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx
=
\int_{\R^{n}} f(u) \, du .
\end{equation}
10.7.6.
Prove the following version of the theorem: Suppose \(S \subset \R^n\) is an open bounded Jordan measurable set, \(g \colon S \to \R^n\) is a one-to-one continuously differentiable mapping such that \(J_g\) is never zero on \(S\text{,}\) and such that \(g(S)\) is bounded and Jordan measurable (it is also open). Suppose \(f \colon g(S) \to \R\) is Riemann integrable. Then \(f \circ g\) is Riemann integrable on \(S\) and
\begin{equation}
\int_S f\bigl(g(x)\bigr) \, \babs{J_g(x)} \, dx
=
\int_{g(S)} f(u) \, du .
\end{equation}
Hint: Write \(S\) as an increasing union of compact Jordan measurable sets, then apply the theorem of the section to those. Then prove that you can take the limit.
