# Basic Analysis I & II: Introduction to Real Analysis, Volumes I & II

## Section10.7Change of variables

Note: 1 lecture
In one variable, we have the familiar change of variables
\begin{equation*} \int_a^b f\bigl(g(x)\bigr) g'(x)\, dx = \int_{g(a)}^{g(b)} f(x) \, dx . \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, the change of variables becomes
\begin{equation*} \int_{[a,b]} f\bigl(g(x)\bigr) \, \sabs{g'(x)}\, dx = \int_{g([a,b])} f(x) \, dx . \end{equation*}
In this section we will obtain the several-variable analogue of this form.
Let us remark the role of $$\sabs{g'(x)}$$ in the formula. The integral measures volumes in general, so in one dimension it measures length. Notice that $$\sabs{g'(x)}$$ 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 absolute value of the Jacobian determinant $$\sabs{J_g(x)} = \babs{\det \bigl(g'(x)\bigr)}$$ is the replacement of $$\sabs{g'(x)}$$ for multiple dimensions in the change of variables formula. The following theorem holds in more generality, but this statement is sufficient for many uses.
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. Define $$f(y) \coloneqq 0$$ for all $$y \notin g(S)\text{.}$$ The new $$f$$ is still Riemann integrable since $$g(S)$$ is Jordan measurable. We can now replace the integrals over $$S$$ with integrals over the whole rectangle. We therefore assume that $$S$$ is equal to a rectangle $$R\text{.}$$
Let $$\epsilon > 0$$ be given. For every $$x \in R\text{,}$$ let
\begin{equation*} W_x \coloneqq \bigl\{ y \in U : \snorm{g'(x)-g'(y)} < \nicefrac{\epsilon}{2} \bigr\} . \end{equation*}
By Exercise 10.7.3, $$W_x$$ is open. As $$x \in W_x$$ for every $$x\text{,}$$ it is an open cover. By the Lebesgue covering lemma (Lemma 7.4.10), there exists a $$\delta > 0$$ such that for every $$y \in R\text{,}$$ there is an $$x$$ such that $$B(y,\delta) \subset W_x\text{.}$$ In other words, if $$P$$ is a rectangle of maximum side length less than $$\frac{\delta}{\sqrt{n}}$$ and $$y \in P\text{,}$$ then $$P \subset B(y,\delta) \subset W_x\text{.}$$ By triangle inequality, $$\snorm{g'(\xi)-g'(\eta)} < \epsilon$$ for all $$\xi, \eta \in P\text{.}$$
Let $$R_1,R_2,\ldots,R_N$$ be subrectangles partitioning $$R$$ such that the maximum side of every $$R_j$$ is less than $$\frac{\delta}{\sqrt{n}}\text{.}$$ We also make sure that the minimum side length is at least $$\frac{\delta}{2\sqrt{n}}\text{,}$$ which we can do if $$\delta$$ is sufficiently small relative to the sides of $$R$$ (Exercise 10.7.4).
Consider some $$R_j$$ and some fixed $$x_j \in R_j\text{.}$$ First suppose $$x_j=0\text{,}$$ $$g(0) = 0\text{,}$$ and $$g'(0) = I\text{.}$$ 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 $$\frac{\delta}{\sqrt{n}}\text{,}$$ then $$\snorm{y} \leq \delta\text{.}$$ So
\begin{equation*} \snorm{g(y)-y} = \norm{\int_0^1 \bigl(g'(ty) y - y\bigr) \,dt} \leq \int_0^1 \snorm{g'(ty) y - y} \,dt \leq \snorm{y} \int_0^1 \snorm{g'(ty) - I} \,dt \leq \delta \epsilon . \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 (s_1+4 \sqrt{n}\,s_1 \epsilon ) (s_2+4 \sqrt{n}\,s_2 \epsilon ) \cdots (s_n+4 \sqrt{n}\,s_n \epsilon ) \\ & = s_1 (1+4 \sqrt{n}\, \epsilon ) \, s_2 (1+4 \sqrt{n}\, \epsilon ) \cdots s_n (1+4 \sqrt{n}\, \epsilon ) = V(R_j) \, {(1+4\sqrt{n} \, \epsilon)}^n . \end{split} \end{equation*}
In other words,
\begin{equation*} V\bigl(g(R_j)\bigr) \leq V(\widetilde{R}_j) \leq V(R_j) \, {(1+4\sqrt{n} \, \epsilon)}^n . \end{equation*}
Next, suppose $$A \coloneqq g'(0)$$ is not necessarily the identity. Write $$g = A \circ \widetilde{g}$$ where $$\widetilde{g}'(0) = I\text{.}$$ By Proposition 10.7.1, $$V\bigl(A(R_j)\bigr) = \sabs{\det(A)} \, V(R_j)\text{,}$$ and hence
\begin{equation*} \begin{split} V\bigl(g(R_j)\bigr) & \leq \sabs{\det(A)} \, V(R_j) \, {(1+4\sqrt{n} \, \epsilon)}^n \\ & = \sabs{J_g(0)} \, V(R_j) \, {(1+4\sqrt{n} \, \epsilon)}^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 $$x_j \not= 0$$ and $$g(x_j) \not= 0\text{,}$$ we find
\begin{equation*} V\bigl(g(R_j)\bigr) \leq \sabs{J_g(x_j)} \, V(R_j) \, {(1+4\sqrt{n} \, \epsilon)}^n . \end{equation*}
Write $$f$$ as $$f = f_+ - f_-$$ for two nonnegative Riemann integrable functions $$f_+$$ and $$f_-\text{:}$$
\begin{equation*} f_+(x) \coloneqq \max \bigl\{ f(x) , 0 \bigr\}, \qquad f_-(x) \coloneqq \max \bigl\{ -f(x) , 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 that $$f(y) \geq 0$$ for all $$y \in R\text{.}$$
For a small enough $$\delta > 0\text{,}$$ we have
\begin{equation*} \begin{split} \epsilon + \int_R f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx & \geq \sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \biggr) \, V(R_j) \\ & \geq \sum_{j=1}^N \biggl(\sup_{x \in R_j} f\bigl(g(x)\bigr) \biggr) \, \sabs{J_g(x_j)} \, V(R_j) \\ & \geq \sum_{j=1}^N \biggl(\sup_{y \in g(R_j)} f(y) \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(y) \,dy \right) \frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n} \\ & = \frac{1}{{(1+4\sqrt{n} \, \epsilon)}^n} \int_{g(R)} f(y) \,dy . \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_R f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx \geq \int_{g(R)} f(y) \,dy . \end{equation*}
By adding this result for several rectangles covering an $$S$$ we obtain the result for an arbitrary bounded Jordan measurable $$S \subset U\text{,}$$ and nonnegative integrable function $$f\text{:}$$
\begin{equation*} \int_S f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx \geq \int_{g(S)} f(y) \,dy . \end{equation*}
Recall that $$g^{-1}$$ exists and $$g^{-1}\bigl(g(S)\bigr) = S\text{.}$$ Also, $$1 = J_{g\circ g^{-1}} = J_g\bigl(g^{-1}(y)\bigr) \,J_{g^{-1}}(y)$$ for $$y \in g(S)\text{.}$$ So
\begin{equation*} \begin{split} \int_{g(S)} f(y) \, dy & = \int_{g(S)} f\bigl(g\bigl(g^{-1}(y)\bigr)\bigr) \, \sabs{J_g\bigl(g^{-1}(y)\bigr)} \, \sabs{J_{g^{-1}}(y)} \, dy \\ & \geq \int_{g^{-1}(g(S))} f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx = \int_{S} f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx . \end{split} \end{equation*}
The conclusion of the theorem holds for all nonnegative $$f$$ and as we mentioned above, it thus holds for all Riemann integrable $$f\text{.}$$

### Subsection10.7.1Exercises

#### Exercise10.7.2.

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 $$j \not= \ell\text{.}$$

#### Exercise10.7.3.

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 $$\epsilon > 0\text{,}$$ show that
\begin{equation*} W_x \coloneqq \bigl\{ y \in U : \snorm{g'(x)-g'(y)} < \nicefrac{\epsilon}{2} \bigr\} \end{equation*}
is an open set.

#### Exercise10.7.4.

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 of every subrectangle is between $$\frac{\delta'}{2}$$ and $$\delta'\text{.}$$

#### Exercise10.7.5.

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 $$g$$ not differentiable at $$x\text{,}$$ that is when $$x \notin U$$). Then
\begin{equation*} \int_{\R^{n}} f(x) \, dx = \int_{\R^n} f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx . \end{equation*}

#### Exercise10.7.6.

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_{g(S)} f(x) \, dx = \int_S f\bigl(g(x)\bigr) \, \sabs{J_g(x)} \, dx . \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.
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