7.1: Definition and Existence of the Multiple Integral

We now consider the Riemann integral of a real-valued function \(f\) defined on a subset of \(\R^n\), where \(n\ge2\). Much of this development will be analogous to the development in Sections~3.1–3 for \(n=1\), but there is an important difference: for \(n=1\), we considered integrals over closed intervals only, but for \(n>1\) we must consider more complicated regions of integration. To defer complications due to geometry, we first consider integrals over rectangles in \(\R^n\), which we now define.

-.4em The \[ S_1\times S_2\times\cdots\times S_n \nonumber \] of subsets \(S_1\), \(S_2\), , \(S_n\) of \(\R\) is the set of points \((x_1,x_2, \dots,x_n)\) in \(\R^n\) such that \(x_1\in S_1, x_2\in S_2, \dots,x_n\in S_n\). For example, the Cartesian product of the two closed intervals

\[ [a_1,b_1]\times [a_2,b_2]=\set{(x,y)}{a_1\le x\le b_1,\ a_2\le y\le b_2} \nonumber \] is a rectangle in \(\R^2\) with sides parallel to the \(x\)- and \(y\)-axes (Figure~).

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The Cartesian product of three closed intervals \[ [a_1,b_1]\times [a_2,b_2]\times [a_3,b_3]=\set{(x,y,z)}{a_1\le x\le b_1,\ a_2\le y\le b_2,\ a_3\le z\le b_3} \nonumber \] is a rectangular parallelepiped in \(\R^3\) with faces parallel to the coordinate axes (Figure~).

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If \(n=1\), \(2\), or \(3\), then \(V(R)\) is, respectively, the length of an interval, the area of a rectangle, or the volume of a rectangular parallelepiped. Henceforth, rectangle'' orcube’‘will always mean coordinate rectangle'' orcoordinate cube’’ unless it is stated otherwise.

If \[ R=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n] \nonumber \] and \[ P_r\colon\, a_r=a_{r0}<a_{r1}<\cdots<a_{rm_r}=b_r\quad \nonumber \] is a partition of \([a_r,b_r]\), \(1\le r\le n\), then the set of all rectangles in \(\R^n\) that can be written as \[ [a_{1,j_1-1},a_{1j_1}]\times[a_{2,j_2-1},a_{2j_2}]\times\cdots\times [a_{n,j_n-1},a_{nj_n}],\quad\ 1\le j_r\le m_r,\quad 1\le r\le n, \nonumber \] is a {} of \(R\). We denote this partition by \[\begin{equation}\label{eq:7.1.1} {\bf P}=P_1\times P_2\times\cdots\times P_n \end{equation} \nonumber \] and define its {} to be the maximum of the norms of \(P_1\), \(P_2\), , \(P_n\), as defined in Section~3.1; thus, \[ \|{\bf P}\|=\max\{\|P_1\|,\,\|P_2\|, \dots,\|P_n\|\}. \nonumber \] Put another way, \(\|{\bf P}\|\) is the largest of the edge lengths of all the subrectangles in~\({\bf P}\).

Geometrically, a rectangle in \(\R^2\) is partitioned by drawing horizontal and vertical lines through it (Figure~); in \(\R^3\), by drawing planes through it parallel to the coordinate axes. Partitioning divides a rectangle \(R\) into finitely many subrectangles that we can number in arbitrary order as \(R_1\), \(R_2\), , \(R_k\). Sometimes it is convenient to write \[ {\bf P}=\{R_1,R_2, \dots,R_k\} \nonumber \] rather than .

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If \({\bf P}=P_1\times P_2\times\cdots\times P_n\) and \({\bf P}'=P'_1 \times P'_2\times\cdots\times P'_n\) are partitions of the same rectangle, then \({\bf P}'\) is a {} of \({\bf P}\) if \(P'_i\) is a refinement of \(P_i\), \(1\le i\le n\), as defined in Section~3.1.

Suppose that \(f\) is a real-valued function defined on a rectangle \(R\) in \(\R^n\), \({\bf P}=\{R_1,R_2, \dots,R_k\}\) is a partition of \(R\), and \(\mathbf{X}_j\) is an arbitrary point in \(R_j\), \(1\le j\le k\). Then \[ \sigma=\sum_{j=1}^k f(\mathbf{X}_j)V(R_j) \nonumber \] is a {}. Since \(\mathbf{X}_j\) can be chosen arbitrarily in \(R_j\), there are infinitely many Riemann sums for a given function \(f\) over any partition \({\bf P}\) of \(R\).

The following definition is similar to Definition~.

If \(R\) is degenerate, then Definition~ implies that \(\int_R f(\mathbf{X})\,d\mathbf{X}=0\) for any function \(f\) defined on \(R\) (Exercise~). Therefore, it should be understood henceforth that whenever we speak of a rectangle in \(\R^n\) we mean a nondegenerate rectangle, unless it is stated to the contrary.

The integral \(\int_R\, f(\mathbf{X})d\mathbf{X}\) is also written as \[ \int_R f(x,y)\, d(x,y)\quad (n=2), \quad \int_R f(x,y,z) \,d(x,y,z)\quad (n=3), \nonumber \] or \[ \int_R f(x_1,x_2, \dots,x_n) \,d(x_1,x_2, \dots,x_n) \mbox{\quad($n$ arbitrary)}. \nonumber \]

Here \(d\mathbf{X}\) does not stand for the differential of \(\mathbf{X}\), as defined in Section~6.2. It merely identifies \(x_1\), \(x_2\), , \(x_n\), the components of \(\mathbf{X}\), as the variables of integration. To avoid this minor inconsistency, some authors write simply \(\int_R f\) rather than \(\int_R f(\mathbf{X})\, d\mathbf{X}\).

As in the case where \(n=1\), we will say simply integrable'' orintegral’‘when we mean Riemann integrable'' orRiemann integral.’’ If \(n\ge2\), we call the integral of Definition~ a {}; for \(n=2\) and \(n=3\) we also call them {} and {}, respectively. When we wish to distinguish between multiple integrals and the integral we studied in Chapter~ \((n=1)\), we will call the latter an {} integral.

Let \(P_1\) and \(P_2\) be partitions of \([a,b]\) and \([c,d]\); thus, \[ P_1: a=x_0<x_1<\cdots<x_r=b \mbox{\quad and \quad} P_2: c=y_0<y_1<\cdots<y_s=d. \nonumber \] A typical Riemann sum of \(f\) over \({\bf P}=P_1\times P_2\) is given by \[\begin{equation}\label{eq:7.1.2} \sigma=\sum_{i=1}^r\sum_{j=1}^s (\xi_{ij}+\eta_{ij})(x_i-x_{i-1}) (y_j-y_{j-1}), \end{equation} \nonumber \] where \[\begin{equation}\label{eq:7.1.3} x_{i-1}\le\xi_{ij}\le x_i\mbox{\quad and\quad} y_{j-1}\le\eta_{ij} \le y_j. \end{equation} \nonumber \] The midpoints of \([x_{i-1}, x_i]\) and \([y_{j-1}, y_j]\) are \[\begin{equation}\label{eq:7.1.4} \overline{x}_i=\frac{x_i+x_{i-1}}{2}\mbox{\quad and\quad} \overline{y}_j=\frac{y_j+y_{j-1}}{2}, \end{equation} \nonumber \] and implies that \[\begin{eqnarray} |\xi_{ij}-\overline{x}_i|\ar\le\frac{x_i-x_{i-1}}{2}\le \frac{\|P_1\|}{ 2}\le \frac{\|{\bf P}\|}{2}\label{eq:7.1.5}\\ \arraytext{and}\nonumber\\ |\eta_{ij}-\overline{y}_j|\ar\le\frac{y_j-y_{j-1}}{2}\le \frac{\|P_2\|}{2} \le \frac{\|{\bf P}\|}{2}. \label{eq:7.1.6} \end{eqnarray} \nonumber \]

Now we rewrite as \[\begin{equation}\label{eq:7.1.7} \begin{array}{rcl} \sigma\ar=\dst{\sum_{i=1}^r\sum_{j=1}^s (\overline{x}_i+ \overline{y}_j)(x_i-x_{i-1}) (y_j-y_{j-1})}\\[2\jot] \ar{}+\dst{\sum_{i=1}^r\sum_{j=1}^s\left[(\xi_{ij}-\overline{x}_i)+ (\eta_{ij}-\overline{y}_j)\right] (x_i-x_{i-1})(y_j-y_{j-1})}. \end{array} \end{equation} \nonumber \] To find \(\int_R f(x,y) \,d(x,y)\) from , we recall that \[\begin{equation}\label{eq:7.1.8} \sum_{i=1}^r (x_i-x_{i-1})=b-a,\quad\sum_{j=1}^s (y_j-y_{j-1})=d-c \end{equation} \nonumber \] (Example~), and \[\begin{equation}\label{eq:7.1.9} \sum_{i=1}^r (x^2_i-x^2_{i-1})=b^2-a^2,\quad\sum_{j=1}^s (y_j^2- y_{j-1}^2)=d^2-c^2 \end{equation} \nonumber \] (Example~).

Because of and the absolute value of the second sum in does not exceed \[\begin{eqnarray*} \|{\bf P}\|\sum_{j=1}^r\sum_{j=1}^s (x_i-x_{i-1})(y_j-y_{j-1})\ar= \|{\bf P}\| \left[\sum_{i=1}^r(x_i-x_{i-1})\right]\left[\sum_{j=1}^s(y_j-y_{j-1}) \right]\\ \ar=\|{\bf P}\| (b-a)(d-c) \end{eqnarray*} \nonumber \] (see ), so implies that \[\begin{equation}\label{eq:7.1.10} \left|\sigma-\sum_{i=1}^r\sum_{j=1}^s (\overline{x}_i+ \overline{y}_j)(x_i-x_{i-1})(y_j-y_{j-1})\right|\le\|{\bf P}\|(b-a)(d-c). \end{equation} \nonumber \] It now follows that \[ \begin{array}{rcl} \dst{\sum_{i=1}^r}\dst{\sum_{j=1}^s}\,\overline{x}_i(x_i-x_{i-1})(y_j-y_{j-1}) \ar=\dst{\left[\sum_{i=1}^r\overline{x}_i(x_i-x_{i-1})\right]} \dst{\left[\sum_{j=1}^s(y_j-y_{j-1})\right]}\\ \ar=(d-c)\dst{\sum_{i=1}^r}\overline{x}_i(x_i-x_{i-1})\mbox{\quad(from \eqref{eq:7.1.8})}\\ \ar=\dst\frac{d-c}{2}\sum_{i=1}^r (x^2_i-x^2_{i-1})\hspace*{2.1em}\mbox{(from \eqref{eq:7.1.4})}\\ \ar=\dst\frac{d-c}{2}(b^2-a^2)\hspace*{4.6em}\mbox{(from \eqref{eq:7.1.9})}. \end{array} \nonumber \] Similarly, \[ \sum_{i=1}^r\sum_{j=1}^s\overline{y}_j(x_i-x_{i-1}) (y_j-y_{j-1})= \frac{b-a}{2}(d^2-c^2). \nonumber \]

Therefore, can be written as \[ \left|\sigma-\frac{d-c}{2}(b^2-a^2)-\frac{b-a}{2}(d^2-c^2)\right|\le \|{\bf P}\|(b-a)(d-c). \nonumber \] Since the right side can be made as small as we wish by choosing \(\|{\bf P}\|\) sufficiently small, \[ \int_R (x+y) \,d(x,y)=\frac{1}{2}\left[(d-c)(b^2-a^2)+(b-a)(d^2-c^2)\right]. \nonumber \]

The following theorem is analogous to Theorem~.

We will show that if \(f\) is unbounded on \(R\), \({\bf P}=\{R_1,R_2, \dots,R_k\}\) is any partition of \(R\), and \(M>0\), then there are Riemann sums \(\sigma\) and \(\sigma'\) of \(f\) over \({\bf P}\) such that \[\begin{equation} \label{eq:7.1.11} |\sigma-\sigma'|\ge M. \end{equation} \nonumber \] This implies that \(f\) cannot satisfy Definition~. (Why?)

Let \[ \sigma=\sum_{j=1}^kf(\mathbf{X}_j)V(R_j) \nonumber \] be a Riemann sum of \(f\) over \({\bf P}\). There must be an integer \(i\) in \(\{1,2, \dots,k\}\) such that \[\begin{equation} \label{eq:7.1.12} |f(\mathbf{X})-f(\mathbf{X}_i)|\ge\frac{M }{ V(R_i)} \end{equation} \nonumber \] for some \(\mathbf{X}\) in \(R_i\), because if this were not so, we would have \[ |f(\mathbf{X})-f(\mathbf{X}_j)|<\frac{M}{ V(R_j)},\quad \mathbf{X}\in R_j,\quad \quad 1\le j\le k. \nonumber \] If this is so, then \[\begin{eqnarray*} |f(\mathbf{X})|\ar=|f(\mathbf{X}_j)+f(\mathbf{X})-f(\mathbf{X}_j)|\le|f(\mathbf{X}_j)|+|f(\mathbf{X})-f(\mathbf{X}_j)|\\ \ar\le |f(\mathbf{X}_j)|+\frac{M}{ V(R_j)},\quad \mathbf{X}\in R_j,\quad 1\le j\le k. \end{eqnarray*} \nonumber \] However, this implies that \[ |f(\mathbf{X})|\le\max\set{|f(\mathbf{X}_j)|+\frac{M}{ V(R_j)}}{1\le j\le k}, \quad \mathbf{X}\in R, \nonumber \] which contradicts the assumption that \(f\) is unbounded on \(R\).

Now suppose that \(\mathbf{X}\) satisfies , and consider the Riemann sum \[ \sigma'=\sum_{j=1}^nf(\mathbf{X}_j')V(R_j) \nonumber \] over the same partition \({\bf P}\), where \[ \mathbf{X}_j'=\left\{\casespace\begin{array}{ll} \mathbf{X}_j,&j \ne i,\\ \mathbf{X},&j=i.\end{array}\right. \nonumber \] Since \[ |\sigma-\sigma'|=|f(\mathbf{X})-f(\mathbf{X}_i)|V(R_i), \nonumber \] implies .

Because of Theorem~, we need consider only bounded functions in connection with Definition~. As in the case where \(n=1\), it is now convenient to define the upper and lower integrals of a bounded function over a rectangle. The following definition is analogous to Definition~.

The following theorem is analogous to Theorem~.

Exercise~.

If \[ m\le f(\mathbf{X})\le M\mbox{\quad for $\mathbf{X}$ in $R$}, \nonumber \] then \[ mV(R)\le s({\bf P})\le S({\bf P})\le MV(R); \nonumber \] therefore, \(\overline{\int_R}\, f(\mathbf{X})\,d\mathbf{X}\) and \(\underline{\int_R}\, f(\mathbf{X})\, d\mathbf{X}\) exist, are unique, and satisfy the inequalities \[ mV(R)\le\overline{\int_R}\, f(\mathbf{X})\,d\mathbf{X}\le MV(R) \nonumber \] and \[ mV(R)\le\underline{\int_R}\, f(\mathbf{X})\,d\mathbf{X}\le MV(R). \nonumber \]

The upper and lower integrals are also written as \[ \overline{\int_R}\, f(x,y) \,d(x,y)\mbox{\quad and\quad}\underline{\int_R}\, f(x,y) \,d(x,y)\quad (n=2), \nonumber \] \[ \overline{\int_R}\, f(x,y,z) \,d(x,y,z)\mbox{\quad and\quad} \underline{\int_R}\, f(x,y,z) \,d(x,y,z)\quad (n=3), \nonumber \] or \[ \overline{\int_R}\, f(x_1,x_2, \dots,x_n) \,d(x_1,x_2, \dots,x_n) \nonumber \] and \[ \underline{\int_R}\, f(x_1,x_2, \dots,x_n)\,d(x_1,x_2, \dots,x_n)\quad \mbox{\quad ($n$ arbitrary)}. \nonumber \]

Let \(P_1\) and \(P_2\) be partitions of \([a,b]\) and \([c,d]\); thus, \[ P_1: a=x_0<x_1<\cdots<x_r=b\mbox{\quad and \quad} P_2: c=y_0<y_1<\cdots<y_s=d. \nonumber \]

The maximum and minimum values of \(f\) on the rectangle \([x_{i-1}, x_i] \times [y_{j-1},y_j]\) are \(x_i+y_j\) and \(x_{i-1}+y_{j-1}\), respectively. Therefore, \[\begin{eqnarray} S({\bf P})\ar=\sum_{i=1}^r\sum_{j=1}^s (x_i+y_j)(x_i-x_{i-1}) (y_j-y_{j-1}) \label{eq:7.1.13}\\ \arraytext{and}\nonumber\\ s({\bf P})\ar=\sum_{i=1}^r\sum_{j=1}^s (x_{i-1}+y_{j-1}) (x_i-x_{i-1}) (y_j-y_{j-1}). \label{eq:7.1.14} \end{eqnarray} \nonumber \] By substituting \[ x_i+y_j=\frac{1}{2}[(x_i+x_{i-1})+(y_j+y_{j-1})+ (x_i-x_{i-1})+(y_j-y_{j-1})] \nonumber \] into , we find that \[\begin{equation}\label{eq:7.1.15} S({\bf P})=\frac{1}{2}(\Sigma_1+\Sigma_2+\Sigma_3+\Sigma_4), \end{equation} \nonumber \] where \[ \begin{array}{rclcl} \Sigma_1\ar=\dst{\sum_{i=1}^r(x_i^2-x_{i-1}^2) \sum_{j=1}^s(y_j-y_{j-1})}\ar=(b^2-a^2)(d-c),\\[2\jot] \Sigma_2\ar=\dst{\sum_{i=1}^r(x_i-x_{i-1}) \sum_{j=1}^s(y_j^2-y_{j-1}^2)}\ar=(b-a)(d^2-c^2),\\[2\jot] \Sigma_3\ar=\dst{\sum_{i=1}^r(x_i-x_{i-1})^2 \sum_{j=1}^s(y_j-y_{j-1})}\ar\le \|\mathbf{P}\|(b-a)(d-c),\\[2\jot] \Sigma_4\ar=\dst{\sum_{i=1}^r(x_i-x_{i-1}) \sum_{j=1}^s(y_j-y_{j-1})^2}\ar\le \|\mathbf{P}\|(b-a)(d-c). \end{array} \nonumber \] Substituting these four results into shows that \[ I<S({\bf P})<I+\|{\bf P}\|(b-a)(d-c), \nonumber \] where \[ I=\frac{(d-c)(b^2-a^2)+(b-a)(d^2-c^2)}{2}. \nonumber \] From this, we see that \[ \overline{\int_R}\,(x+y) \,d(x,y)=I. \nonumber \]

After substituting \[ x_{i-1}+y_{j-1}=\frac{1}{2}[(x_i+x_{i-1})+(y_j+y_{j-1})-(x_i-x_{i-1}) -(y_j-y_{j-1})] \nonumber \] into , a similar argument shows that \[ I-\|{\bf P}\|(b-a)(d-c)<s({\bf P})<I, \nonumber \]

so \[ \underline{\int_R}\, (x+y)\,d(x,y)=I. \eqno{\bbox} \nonumber \]

We now prove an analog of Lemma~.

We will prove and leave the proof of to you (Exercise~). First suppose that \(P_1'\) is obtained by adding one point to \(P_1\), and \(P_j'=P_j\) for \(2\le j\le n\). If \(P_r\) is defined by \[ P_r: a_r=a_{r0}<a_{r1}<\cdots<a_{rm_r}=b_r,\quad 1\le r\le n, \nonumber \] then a typical subrectangle of \({\bf P}\) is of the form \[ R_{j_1j_2\cdots j_n}=[a_{1,j_1-1}, a_{1j_1}]\times [a_{2,j_2-1},a_{2j_2}]\times\cdots\times [a_{n,j_n-1}, a_{nj_n}]. \nonumber \] Let \(c\) be the additional point introduced into \(P_1\) to obtain \(P_1'\), and suppose that \[ a_{1,k-1}<c<a_{1k}. \nonumber \] If \(j_1\ne k\), then \(R_{j_1j_2\cdots j_n}\) is common to \({\bf P}\) and \({\bf P}'\), so the terms associated with it in \(S({\bf P}')\) and \(S({\bf P})\) cancel in the difference \(S({\bf P})-S({\bf P}')\). To analyze the terms that do not cancel, define \[ \begin{array}{rcl} R^{(1)}_{kj_2\cdots j_n}\ar=[a_{1,k-1}, c]\times [a_{2,j_2-1}, a_{2j_2}] \times\cdots\times [a_{n,j_n-1},a_{nj_n}],\\[2\jot] R^{(2)}_{kj_2\cdots j_n}\ar=[c,a_{1k}]\times [a_{2,j_2-1}, a_{2j_2}] \times\cdots\times [a_{n,j_n-1}, a_{nj_n}], \end{array} \nonumber \] \[\begin{equation} \label{eq:7.1.18} M_{kj_2\cdots j_n}=\sup\set{f(\mathbf{X})}{\mathbf{X}\in R_{kj_2\cdots j_n}} \end{equation} \nonumber \] and \[\begin{equation} \label{eq:7.1.19} \begin{array}{rcl} M^{(i)}_{kj_2\cdots j_n}=\sup\set{f(\mathbf{X})}{\mathbf{X} \in R^{(i)}_{kj_2\cdots j_n}},\quad i=1,2. \end{array} \end{equation} \nonumber \]

Then \(S({\bf P})-S({\bf P}')\) is the sum of terms of the form \[\begin{equation}\label{eq:7.1.20} \begin{array}{c} \left[M_{kj_2\cdots j_n}(a_{1k}-a_{1,k-1})-M^{(1)}_{kj_2 \cdots j_n} (c-a_{1,k-1})-M^{(2)}_{kj_2\cdots j_n}(a_{1k}-c)\right]\\ \times (a_{2j_2}-a_{2,j_2-1})\cdots (a_{nj_n}-a_{n,j_n-1}). \end{array} \end{equation} \nonumber \] The terms within the brackets can be rewritten as \[\begin{equation}\label{eq:7.1.21} (M_{kj_2\cdots j_n}-M^{(1)}_{kj_2\cdots j_n})(c-a_{1,k-1})+(M_{kj_2\cdots j_n}-M^{(2)}_{kj_2\cdots j_n})(a_{1k}-c), \end{equation} \nonumber \]

which is nonnegative, because of and . Therefore, \[\begin{equation}\label{eq:7.1.22} S({\bf P}')\le S({\bf P}). \end{equation} \nonumber \] Moreover, the quantity in is not greater than \(2M(a_{1k}-a_{1,k-1})\), so implies that the general surviving term in \(S({\bf P})-S({\bf P}')\) is not greater than \[ 2M\|{\bf P}\|(a_{2j_2}-a_{2,j_2-1})\cdots (a_{nj_n}-a_{n,j_n-1}). \nonumber \] The sum of these terms as \(j_2\), , \(j_n\) assume all possible values \(1 \le j_i\le m_i\), \(2\le i\le n\), is \[ 2M\|{\bf P}\|(b_2-a_2)\cdots (b_n-a_n)=\frac{2M\|{\bf P}\|V(R)}{ b_1-a_1}. \nonumber \] This implies that \[ S({\bf P})\le S({\bf P}')+\frac{2M\|{\bf P}\|V(R)}{ b_1-a_1}. \nonumber \] This and imply for \(r_1=1\) and \(r_2=\cdots=r_n=0\).

Similarly, if \(r_i=1\) for some \(i\) in \(\{1, \dots,n\}\) and \(r_j=0\) if \(j\ne i\), then \[ S({\bf P})\le S({\bf P}')+\frac{2M\|{\bf P}\|V(R)}{ b_i-a_i}. \nonumber \] To obtain in the general case, repeat this argument \(r_1+r_2+\cdots+r_n\) times, as in the proof of Lemma~.

Lemma~ implies the following theorems and lemma, with proofs analogous to the proofs of their counterparts in Section~3.2.

Exercise~.

The next theorem is analogous to Theorem~3.2.3.

Exercise~.

Exercise~.

The next theorem is analogous to Theorem~3.2.5.

Exercise~.

Theorems~ and imply the following theorem, which is analogous to Theorem~.

The next theorem translates this into a test that can be conveniently applied. It is analogous to Theorem~.

Exercise~.

Theorem~ provides a useful criterion for integrability. The next theorem is an important application. It is analogous to Theorem~.

Let \(\epsilon>0\). Since \(f\) is uniformly continuous on \(R\) (Theorem~), there is a \(\delta>0\) such that \[\begin{equation} \label{eq:7.1.23} |f(\mathbf{X})-f(\mathbf{X}')|<\frac{\epsilon}{ V({\bf R})} \end{equation} \nonumber \] if \(\mathbf{X}\) and \(\mathbf{X}'\) are in \(R\) and \(|\mathbf{X}-\mathbf{X}'|<\delta\). Let \({\bf P}=\{R_1,R_2, \dots,R_k\}\) be a partition of \(R\) with \(\|P\|<\delta/\sqrt n\). Since \(f\) is continuous on \(R\), there are points \(\mathbf{X}_j\) and \(\mathbf{X}_j'\) in \(R_j\) such that \[ f(\mathbf{X}_j)=M_j=\sup_{\mathbf{X}\in R_j}f(\mathbf{X}) \mbox{\quad and \quad} f(\mathbf{X}_j')=m_j=\inf_{\mathbf{X}\in R_j}f(\mathbf{X}) \nonumber \] (Theorem~). Therefore, \[ S(\mathbf{P})-s(\mathbf{P})=\sum_{j=1}^n(f(\mathbf{X}_j)- f(\mathbf{X}_j'))V(R_j). \nonumber \] Since \(\|{\bf P}\|<\delta/\sqrt n\), \(|\mathbf{X}_j-\mathbf{X}_j'|<\delta\), and, from with \(\mathbf{X}=\mathbf{X}_j\) and \(\mathbf{X}'=\mathbf{X}_j'\), \[ S(\mathbf{P})-s(\mathbf{P})<\frac{\epsilon}{ V(R)} \sum_{j=1}^kV(R_j)=\epsilon. \nonumber \] Hence, \(f\) is integrable on \(R\), by Theorem~.

The next definition will enable us to establish the existence of \(\int_Rf(\mathbf{X})\,d\mathbf{X}\) in cases where \(f\) is bounded on the rectangle \(R\), but is not necessarily continuous for all \(\mathbf{X}\) in \(R\).

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The next lemma follows immediately from Definition~.

The following theorem will enable us to define multiple integrals over more general subsets of \(\R^n\).

Suppose that \(\epsilon>0\). Since \(E\) has zero content, there are rectangles \(T_1\), \(T_2\), , \(T_m\) such that \[\begin{equation} \label{eq:7.1.31} E\subset\bigcup_{j=1}^m T_j \end{equation} \nonumber \] and \[\begin{equation} \label{eq:7.1.32} \sum_{j=1}^m V(T_j)<\epsilon. \end{equation} \nonumber \] We may assume that \(T_1\), \(T_2\), , \(T_m\) are contained in \(R\), since, if not, their intersections with \(R\) would be contained in \(R\), and still satisfy and . We may also assume that if \(T\) is any rectangle such that \[\begin{equation}\label{eq:7.1.33} T\bigcap\left(\bigcup_{j=1}^m T_j^0\right)=\emptyset, \mbox{\quad then \quad}T\cap E=\emptyset \end{equation} \nonumber \]

since if this were not so, we could make it so by enlarging \(T_1\), \(T_2\), , \(T_m\) slightly while maintaining . Now suppose that \[ T_j=[a_{1j},b_{1j}]\times [a_{2j},b_{2j}]\times\cdots\times [a_{nj},b_{nj}],\quad 1\le j\le m, \nonumber \] let \(P_{i0}\) be the partition of \([a_i,b_i]\) (see ) with partition points \[ a_i,b_i,a_{i1},b_{i1},a_{i2},b_{i2}, \dots,a_{im},b_{im} \vspace*{1pt} \nonumber \] (these are not in increasing order), \(1\le i\le n\), and let \[ {\bf P}_0=P_{10}\times P_{20}\times\cdots\times P_{n0}. \nonumber \] -.3em Then \({\bf P}_0\) consists of rectangles whose union equals \(\cup_{j=1}^m T_j\) and other rectangles \(T'_1\), \(T'_2\), , \(T'_k\) that do not intersect \(E\). (We need to be sure that \(T'_i\cap E=\emptyset, 1\le i\le k.)\) If we let \[ B=\bigcup_{j=1}^m T_j\mbox{\quad and\quad} C=\bigcup^k_{i=1} T'_i, \nonumber \] then \(R=B\cup C\) and \(f\) is continuous on the compact set \(C\). If \({\bf P}=\{R_1,R_2, \dots,R_k\}\) is a refinement of \({\bf P}_0\), then every subrectangle \(R_j\) of \({\bf P}\) is contained entirely in \(B\) or entirely in \(C\). Therefore, we can write \[\begin{equation}\label{eq:7.1.34} S({\bf P})-s({\bf P})=\Sigma_1(M_j-m_j) V(R_j)+\Sigma_2(M_j-m_j)V(R_j), \end{equation} \nonumber \] -.3em where \(\Sigma_1\) and \(\Sigma_2\) are summations over values of \(j\) for which \(R_j\subset B\) and \(R_j\subset C\), respectively. Now suppose that \[ |f(\mathbf{X})|\le M\mbox{\quad for $\mathbf{X}$ in $R$}. \nonumber \] Then \[\begin{equation}\label{eq:7.1.35} \Sigma_1(M_j-m_j) V(R_j)\le2M\,\Sigma_1 V(R_j)=2M\sum_{j=1}^m V(T_j)< 2M\epsilon, \end{equation} \nonumber \] from . Since \(f\) is uniformly continuous on the compact set \(C\) (Theorem~), there is a \(\delta>0\) such that \(M_j-m_j<\epsilon\) if \(\|{\bf P}\|< \delta\) and \(R_j\subset C\); hence, \[ \Sigma_2(M_j-m_j)V(R_j)<\epsilon\Sigma_2\, V(R_j)\le\epsilon V(R). \nonumber \] This, , and imply that \[ S({\bf P})-s({\bf P})<[2M+V(R)]\epsilon \nonumber \] if \(\|{\bf P}\|<\delta\) and \({\bf P}\) is a refinement of \({\bf P}_0\). Therefore, Theorem~ implies that \(f\) is integrable on \(R\).

6pt

12pt

We can now define the integral of a bounded function over more general subsets of~\(\R^n\).

To see that this definition makes sense, we must show that if \(R_1\) and \(R_2\) are two rectangles containing \(S\) and \(\int_{R_1} f_S({\bf X})\, d\mathbf{X}\) exists, then so does \(\int_{R_2} f_S (\mathbf{X})\,d{\bf X}\), and the two integrals are equal. The proof of this is sketched in Exercise~.

Let \(f_S\) be as in . Since a discontinuity of \(f_S\) is either a discontinuity of \(f\) or a point of \(\partial S\), the set of discontinuities of \(f_S\) is the union of two sets of zero content and therefore is of zero content (Lemma~). Therefore, \(f_S\) is integrable on any rectangle containing \(S\) (from Theorem~), and consequently on \(S\) (Definition~).

, defined as follows, form an important class of sets of zero content in \(\R^n\).

Let \(S\), \(D\), and \(\mathbf{G}\) be as in Definition~. From Lemma~, there is a constant \(M\) such that \[\begin{equation}\label{eq:7.1.37} |\mathbf{G}(\mathbf{X})-\mathbf{G}(\mathbf{Y})|\le M|\mathbf{X}-\mathbf{Y}|\mbox{\quad if\quad}\mathbf{X},\mathbf{Y}\in D. \end{equation} \nonumber \] Since \(D\) is bounded, \(D\) is contained in a cube \[ C=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_m,b_m], \nonumber \] where \[ b_i-a_i=L,\quad 1\le i\le m. \nonumber \] Suppose that we partition \(C\) into \(N^m\) smaller cubes by partitioning each of the intervals \([a_i,b_i]\) into \(N\) equal subintervals. Let \(R_1\), \(R_2\), , \(R_k\) be the smaller cubes so produced that contain points of \(D\), and select points \(\mathbf{X}_1\), \(\mathbf{X}_2\), , \(\mathbf{X}_k\) such that \(\mathbf{X}_i\in D\cap R_i\), \(1\le i\le k\). If \(\mathbf{Y} \in D\cap R_i\), then implies that \[\begin{equation}\label{eq:7.1.38} |\mathbf{G}(\mathbf{X}_i)-\mathbf{G}(\mathbf{Y})|\le M|\mathbf{X}_i-\mathbf{Y}|. \end{equation} \nonumber \] Since \(\mathbf{X}_i\) and \(\mathbf{Y}\) are both in the cube \(R_i\) with edge length \(L/N\), \[ |\mathbf{X}_i-\mathbf{Y}|\le\frac{L\sqrt{m}}{ N}. \nonumber \] This and imply that \[ |\mathbf{G}(\mathbf{X}_i)-\mathbf{G}(\mathbf{Y})|\le\frac{ML\sqrt m}{ N}, \nonumber \] which in turn implies that \(\mathbf{G}(\mathbf{Y})\) lies in a cube \(\widetilde{R}_i\) in \(\R^n\) centered at \(\mathbf{G}(\mathbf{X}_i)\), with sides of length \(2ML\sqrt{m}/N\). Now \[ \sum_{i=1}^k V(\widetilde{R}_i)= k\left(\frac{2ML\sqrt{m}}{ N}\right)^n\le N^m\left(\frac{2ML\sqrt{m}}{ N}\right)^n=(2ML\sqrt{m})^n N^{m-n}. \nonumber \] Since \(n>m\), we can make the sum on the left arbitrarily small by taking \(N\) sufficiently large. Therefore, \(S\) has zero content.

Theorems~ and imply the following theorem.

12pt 6pt

12pt

We now list some theorems on properties of multiple integrals. The proofs are similar to those of the analogous theorems in Section~3.3.

Note: Because of Definition~, if we say that a function \(f\) is integrable on a set \(S\), then \(S\) is necessarily bounded.

Exercise~.

Exercise~.

Exercise~.

Exercise~.

Exercise~.

Exercise~.

From Definition~ with \(f\) and \(S\) replaced by \(f_S\) and \(T\), \[ (f_S)_T(\mathbf{X})=\left\{\casespace\begin{array}{ll} f_S(\mathbf{X}),&\mathbf{X}\in T,\\ 0,&\mathbf{X}\not\in T.\end{array}\right. \nonumber \] Since \(S\subset T\), \((f_S)_T=f_S\). (Verify.) Now suppose that \(R\) is a rectangle containing \(T\). Then \(R\) also contains \(S\) (Figure~), 6pt

12pt so \[ \begin{array}{rcll} \dst\int_Sf(\mathbf{X})\,d\mathbf{X}\ar=\dst\int_Rf_S(\mathbf{X})\,d\mathbf{X}& \mbox{(Definition~\ref{thmtype:7.1.17}, applied to $f$ and $S$})\\[4\jot] \ar=\dst\int_R(f_S)_T(\mathbf{X})\,d\mathbf{X}& \mbox{(since $(f_S)_T=f_S$)}\\[4\jot] \ar=\dst\int_Tf_S(\mathbf{X})\,d\mathbf{X}& \mbox{(Definition~\ref{thmtype:7.1.17}, applied to $f_S$ and $T$}), \end{array} \nonumber \] which completes the proof.

For \(i=1\), \(2\), let \[ f_{S_i}(\mathbf{X})=\left\{\casespace\begin{array}{ll} f(\mathbf{X}),&\mathbf{X}\in S_i,\\[2\jot] 0,&\mathbf{X}\not\in S_i.\end{array}\right. \nonumber \] From Lemma~ with \(S=S_i\) and \(T=S_1\cup S_2\), \(f_{S_i}\) is integrable on \(S_1\cup S_2\), and \[ \int_{S_1\cup S_2} f_{S_i}(\mathbf{X})\,d\mathbf{X} =\int_{S_i} f(\mathbf{X})\,d\mathbf{X},\quad i=1,2. \nonumber \] Theorem~ now implies that \(f_{S_1}+f_{S_2}\) is integrable on \(S_1\cup S_2\) and \[\begin{equation}\label{eq:7.1.40} \int_{S_1\cup S_2} (f_{S_1}+f_{S_2})(\mathbf{X})\,d\mathbf{X}=\int_{S_1} f(\mathbf{X})\,d\mathbf{X}+\int_{S_2} f(\mathbf{X})\, d\mathbf{X}. \end{equation} \nonumber \]

Since \(S_1\cap S_2=\emptyset\), \[ \left(f_{S_1}+f_{S_2}\right)(\mathbf{X})= f_{S_1}(\mathbf{X})+f_{S_2}(\mathbf{X}) =f(\mathbf{X}),\quad \mathbf{X}\in S_1\cup S_2. \nonumber \] Therefore, implies .

We leave it to you to prove the following extension of Theorem_{. (Exercise}

).

6pt

12pt

We will discuss this example further in the next section.

\begin{exerciselist}

Prove: If \(R\) is degenerate, then Definition~ implies that \(\int_R f(\mathbf{X})\,d\mathbf{X}=0\) if \(f\) is bounded on \(R\).

Evaluate directly from Definition~.

Suppose that \(\int_a^b f(x)\,dx\) and \(\int_c^d g(y)\,dy\) exist, and let \(R=[a,b]\times [c,d]\). Criticize the following ``proof’’ that \(\int_R f(x)g(y)\,d(x,y)\) exists and equals \[ \left(\int_a^b f(x)\,dx\right)\left(\int_c^d g(y)\,dy\right). \nonumber \] (See Exercise~ for a correct proof of this assertion.)

``Proof.’’ Let \[ P_1: a=x_0<x_1<\cdots<x_r=b\mbox{\quad and\quad} P_2:c=y_0<y_1<\cdots<y_s=d \nonumber \] be partitions of \([a,b]\) and \([c,d]\), and \({\bf P}=P_1\times P_2\). Then a typical Riemann sum of \(fg\) over \({\bf P}\) is of the form \[ \sigma=\sum_{i=1}^r\sum_{j=1}^s f(\xi_i)g(\eta_j)(x_i-x_{i-1}) (y_j-y_{j-1})=\sigma_1\sigma_2, \nonumber \] where \[ \sigma_1=\sum_{i=1}^r f(\xi_i)(x_i-x_{i-1})\mbox{\quad and\quad}\sigma_2 =\sum_{j=1}^sg(\eta_j)(y_j-y_{j-1}) \nonumber \]

are typical Riemann sums of \(f\) over \([a,b]\) and \(g\) over \([c,d]\). Since \(f\) and \(g\) are integrable on these intervals, \[ \left|\sigma_1-\int_a^b f(x)\,dx\right|\mbox{\quad and\quad} \left|\sigma_2-\int_c^d g(y)\,dy\right| \nonumber \] can be made arbitrarily small by taking \(\|P_1\|\) and \(\|P_2\|\) sufficiently small. From this, it is straightforward to show that \[ \left|\sigma-\left(\int_a^b f(x)\,dx\right)\left(\int_c^d g(y)\,dy \right)\right| \nonumber \] can be made arbitrarily small by taking \(\|{\bf P}\|\) sufficiently small. This implies the stated result.

Suppose that \(f(x,y)\ge0\) on \(R=[a,b]\times [c,d]\). Justify the interpretation of \(\int_Rf(x,y)\,d(x,y)\), if it exists, as the volume of the region in \(\R^3\) bounded by the surfaces \(z=f(x,y)\) and the planes \(z=0\), \(x=a\), \(x=b\), \(y=c\), and \(y=d\).

Prove Theorem~.

Suppose that \[ f(x,y)=\left\{\casespace\begin{array}{ll} 0& \mbox{\quad if $x$ and $y$ are rational,\quad}\\ 1&\mbox{\quad if $x$ is rational and $y$ is irrational,\quad}\\ 2&\mbox{\quad if $x$ is irrational and $y$ is rational,\quad}\\ 3&\mbox{\quad if $x$ and $y$ are irrational.\quad}\end{array}\right. \nonumber \] Find \[ \overline{\int_R}\, f(x,y)\,d(x,y)\mbox{\quad and\quad}\underline{\int_R}\, f(x,y)\,d(x,y)\mbox{\quad if\quad} R=[a,b]\times [c,d]. \nonumber \]

Prove Eqn.~ of Lemma~.

Prove Theorem~

Prove Theorem~

Prove Lemma~

Prove Theorem~

Prove Theorem~

Give an example of a denumerable set in \(\R^2\) that does not have zero content.

%:14 Prove:

Show that a degenerate rectangle has zero content.

Suppose that \(f\) is continuous on a compact set \(S\) in \(\R^n\). Show that the surface \(z=f(\mathbf{X})\), \(\mathbf{X}\in S\), has zero content in \(\R^{n+1}\).

Let \(S\) be a bounded set such that \(S\cap\partial S\) does not have zero content.

Suppose that \(f\) is integrable on a set \(S\) and \(S_0\) is a subset of \(S\) such that \(\partial S_0\) has zero content. Show that \(f\) is integrable on \(S_0\).

Prove Theorem~

Prove Theorem~.

Prove Theorem~

Prove Theorem~

Prove Theorem~

Prove Theorem~

Prove: If \(f\) is integrable on a rectangle \(R\), then \(f\) is integrable on any subrectangle of \(R\).

Suppose that \(R\) and \(\widetilde{R}\) are rectangles, \(R\subset \widetilde{R}\), \(g\) is bounded on \(\widetilde{R}\), and \(g({\bf X})=0\) if \(\mathbf{X} \not\in R\).

Suppose that \(f\) is continuously differentiable on a rectangle \(R\). Show that there is a constant \(M\) such that \[ \left|\sigma-\int_R f(\mathbf{X})\,d\mathbf{X}\right|\le M\|{\bf P}\| \nonumber \] if \(\sigma\) is any Riemann sum of \(f\) over a partition \({\bf P}\) of \(R\).

Suppose that \(\int_a^b f(x)\,dx\) and \(\int_c^d g(y)\,dy\) exist, and let \(R=[a,b]\times [c,d]\).

\end{exerciselist}

Except for very simple examples, it is impractical to evaluate multiple integrals directly from Definitions~ and . Fortunately, this can usually be accomplished by evaluating \(n\) successive ordinary integrals. To motivate the method, let us first assume that \(f\) is continuous on \(R=[a,b]\times [c,d]\). Then, for each \(y\) in \([c,d]\), \(f(x,y)\) is continuous with respect to \(x\) on \([a,b]\), so the integral \[ F(y)=\int_a^b f(x,y)\,dx \nonumber \] exists. Moreover, the uniform continuity of \(f\) on \(R\) implies that \(F\) is continuous (Exercise~) and therefore integrable on \([c,d]\). We say that \[ I_1=\int_c^d F(y)\,dy=\int_c^d\left(\int_a^b f(x,y)\,dx\right)dy \nonumber \]

is an {} of \(f\) over \(R\). We will usually write it as \[ I_1=\int_c^d dy\int_a^b f(x,y)\,dx. \nonumber \] Another iterated integral can be defined by writing \[ G(x)=\int_c^d f(x,y)\,dy,\quad a\le x\le b, \nonumber \] and defining \[ I_2=\int_a^b G(x)\,dx=\int_a^b\left(\int_c^d f(x,y)\,dy\right) dx, \nonumber \] which we usually write as \[ I_2=\int_a^b dx\int_c^d f(x,y)\,dy. \nonumber \]

In this example, \(I_1=I_2\); moreover, on setting \(a=0\), \(b=1\), \(c=1\), and \(d=2\) in Example~, we see that \[ \int_R (x+y)\,d(x,y)=2, \nonumber \] so the common value of the iterated integrals equals the multiple integral. The following theorem shows that this is not an accident.

Let \[ P_1: a=x_0<x_1<\cdots<x_r=b\mbox{\quad and \quad} P_2: c=y_0<y_1<\cdots<y_s=d \nonumber \] be partitions of \([a,b]\) and \([c,d]\), and \(\mathbf{P}=P_1\times P_2\). Suppose that \[\begin{equation}\label{eq:7.2.3} y_{j-1}\le\eta_j\le y_j,\quad 1\le j\le s, \end{equation} \nonumber \] so \[\begin{equation} \label{eq:7.2.4} \sigma=\sum_{j=1}^s F(\eta_j)(y_j-y_{j-1}) \end{equation} \nonumber \] is a typical Riemann sum of \(F\) over \(P_2\). Since \[ F(\eta_j)=\int_a^b f(x,\eta_j)\,dx=\sum_{i=1}^r\int^x_{x_{i-1}} f(x,\eta_j)\,dx, \nonumber \] implies that if \[\begin{eqnarray*} m_{ij}\ar=\inf\set{f(x,y)}{x_{i-1}\le x\le x_i,\, y_{j-1}\le y\le y_j}\\ \arraytext{and}\\ M_{ij}\ar=\sup\set{f(x,y)}{x_{i-1}\le x\le x_i,\, y_{j-1}\le y\le y_j}, \end{eqnarray*} \nonumber \] then \[ \sum_{i=1}^r m_{ij} (x_i-x_{i-1})\le F(\eta_j)\le\sum_{i=1}^r M_{ij} (x_i-x_{i-1}). \nonumber \] Multiplying this by \(y_j-y_{j-1}\) and summing from \(j=1\) to \(j=s\) yields \[\begin{eqnarray*} \sum_{j=1}^s\sum_{i=1}^r m_{ij} (x_i-x_{i-1})(y_j-y_{j-1})\ar\le \sum_{j=1}^sF(\eta_j)(y_j-y_{j-1})\\ \ar\le \sum_{j=1}^s\sum_{i=1}^r M_{ij}(x_i-x_{i-1})(y_j-y_{j-1}), \end{eqnarray*} \nonumber \]

which, from , can be rewritten as \[\begin{equation}\label{eq:7.2.5} s_f(\mathbf{P})\le\sigma\le S_f(\mathbf{P}), \end{equation} \nonumber \] where \(s_f(\mathbf{P})\) and \(S_f(\mathbf{P})\) are the lower and upper sums of \(f\) over \(\mathbf{P}\). Now let \(s_F(P_2)\) and \(S_F(P_2)\) be the lower and upper sums of \(F\) over \(P_2\); since they are respectively the infimum and supremum of the Riemann sums of \(F\) over \(P_2\) (Theorem~), implies that \[\begin{equation}\label{eq:7.2.6} s_f(\mathbf{P})\le s_F(P_2)\le S_F(P_2)\le S_f(\mathbf{P}). \end{equation} \nonumber \] Since \(f\) is integrable on \(R\), there is for each \(\epsilon>0\) a partition \(\mathbf{P}\) of \(R\) such that \(S_f(\mathbf{P})-s_f(\mathbf{P})<\epsilon\), from Theorem~. Consequently, from , there is a partition \(P_2\) of \([c,d]\) such that \(S_F(P_2)-s_F(P_2)<\epsilon\), so \(F\) is integrable on \([c,d]\), from Theorem~.

It remains to verify . From and the definition of \(\int_c^dF(y)\,dy\), there is for each \(\epsilon>0\) a \(\delta>0\) such that \[ \left|\int_c^d F(y)\,dy-\sigma\right|<\epsilon\mbox{\quad if\quad} \|P_2\|<\delta; \nonumber \] that is, \[ \sigma-\epsilon<\int_c^d F(y)\,dy<\sigma+\epsilon\mbox{\quad if \quad} \|P_2\|<\delta. \nonumber \] This and imply that \[ s_f(\mathbf{P})-\epsilon<\int_c^d F(y)\,dy<S_f(\mathbf{P})+\epsilon\mbox{\quad if\quad} \|\mathbf{P}\|<\delta, \nonumber \] and this implies that \[\begin{equation}\label{eq:7.2.7} \underline{\int_R}\, f(x,y)\,d(x,y)-\epsilon\le\int_c^d F(y)\,dy \le\overline{\int_R} f(x,y)\,d(x,y)+\epsilon \end{equation} \nonumber \] (Definition_{). Since \[ \underline{\int_R}\, f(x,y)\,d(x,y)=\overline{\int_R} f(x,y)\,d(x,y) \nonumber \] (Theorem}) and \(\epsilon\) can be made arbitrarily small, implies .

If \(f\) is continuous on \(R\), then \(f\) satisfies the hypotheses of Theorem~ (Exercise~), so is valid in this case.

If \(\int_R f(x,y)\,d(x,y)\) and \[ \int_c^d f(x,y)\,dy,\quad a\le x\le b, \nonumber \]

exist, then by interchanging \(x\) and \(y\) in Theorem~, we see that \[ \int_a^b dx\int_c^d f(x,y)\,dy=\int_R f(x,y)\,d(x,y). \nonumber \] This and yield the following corollary of Theorem~.

-2em2em

A plausible partial converse of Theorem~ would be that if \(\int_c^d dy \int_a^b f(x,y)\,dx\) exists then so does \(\int_R f(x,y)\,d(x,y)\); however, the next example shows that this need not be so. \begin{example}\rm If \(f\) is defined on \(R=[0,1]\times [0,1]\) by \[ f(x,y)=\left\{\casespace\begin{array}{ll} 2xy&\mbox{if $y$ is rational},\\ y&\mbox{if $y$ is irrational}, \end{array}\right. \nonumber \]

then \[ \int_0^1 f(x,y)\,dx=y,\quad 0\le y\le1, \nonumber \] and \[ \int_0^1 dy\int_0^1 f(x,y)\,dx=\int_0^1 y\,dy=\frac{1}{2}. \nonumber \] However, \(f\) is not integrable on \(R\) (Exercise~). \end{example}

The next theorem generalizes Theorem~ to \(\R^n\).

For convenience, denote \((x_{p+1},x_{p+2}, \dots,x_n)\) by \(\mathbf{Y}\). Denote \(\widehat R=I_1\times I_2\times\cdots\times I_p\) and \(T=I_{p+1}\times I_{p+2}\times\cdots\times I_n\). Let \(\widehat{\mathbf{P}}=\{\widehat R_1, \widehat R_2, \dots,\widehat R_k\}\) and \(\mathbf{Q}=\{T_1,T_2, \dots,T_s\}\) be partitions of \(\widehat R\) and \(T\), respectively. Then the collection of rectangles of the form \(\widehat R_i\times T_j\) (\(1\le i\le k\), \(1\le j\le s\)) is a partition \(\mathbf{P}\) of \(R\); moreover, every partition \(\mathbf{P}\) of \(R\) is of this form.

Suppose that \[\begin{equation} \label{eq:7.2.8} \mathbf{Y}_j\in T_j,\quad1\le j\le s, \end{equation} \nonumber \] so \[\begin{equation} \label{eq:7.2.9} \sigma=\sum_{j=1}^s F_p(\mathbf{Y}_j)V(T_j) \end{equation} \nonumber \] is a typical Riemann sum of \(F_p\) over \(\mathbf{Q}\). Since \[\begin{eqnarray*} F_p(\mathbf{Y_j})\ar=\int_{\widehat R} f(x_1,x_2, \dots,x_p,\mathbf{Y}_j)\,d(x_1,x_2, \dots,x_p)\\ \ar=\sum_{j=1}^k\int_{\widehat R_j} f(x_1,x_2, \dots,x_p,\mathbf{Y}_j)\,d(x_1,x_2, \dots,x_p), \end{eqnarray*} \nonumber \] implies that if \[\begin{eqnarray*} m_{ij}\ar=\inf\set{f(x_1,x_2, \dots,x_p,\mathbf{Y})} {(x_1,x_2, \dots,x_p)\in \widehat R_i,\, \mathbf{Y}\in T_j}\\ \arraytext{and}\\ M_{ij}\ar=\sup\set{f(x_1,x_2, \dots,x_p,\mathbf{Y})} {(x_1,x_2, \dots,x_p)\in \widehat R_i,\, \mathbf{Y}\in T_j}, \end{eqnarray*} \nonumber \]

then \[ \sum_{i=1}^k m_{ij}V(\widehat R_i)\le F_p(\mathbf{Y_j})\le\sum_{i=1}^k M_{ij} V(\widehat R_i). \nonumber \] Multiplying this by \(V(T_j)\) and summing from \(j=1\) to \(j=s\) yields \[ \sum_{j=1}^s\sum_{i=1}^k m_{ij} V(\widehat R_i)V(T_j)\le \sum_{j=1}^sF_p(\mathbf{Y}_j)V(T_j) \le\sum_{j=1}^s\sum_{i=1}^k M_{ij}V(\widehat R_i)V(T_j), \nonumber \] which, from , can be rewritten as \[\begin{equation} \label{eq:7.2.10} s_f(\mathbf{P})\le\sigma\le S_f(\mathbf{P}), \end{equation} \nonumber \] where \(s_f(\mathbf{P})\) and \(S_f(\mathbf{P})\) are the lower and upper sums of \(f\) over \(\mathbf{P}\). Now let \(s_{F_p}(\mathbf{Q})\) and \(S_{F_p}(\mathbf{Q})\) be the lower and upper sums of \(F_p\) over \(\mathbf{Q}\); since they are respectively the infimum and supremum of the Riemann sums of \(F_p\) over \(\mathbf{Q}\) (Theorem~), implies that \[\begin{equation} \label{eq:7.2.11} s_f(\mathbf{P})\le s_{F_p}(\mathbf{Q})\le S_{F_p}(\mathbf{Q})\le S_f(\mathbf{P}). \end{equation} \nonumber \] Since \(f\) is integrable on \(R\), there is for each \(\epsilon>0\) a partition \(\mathbf{P}\) of \(R\) such that \(S_f(\mathbf{P})-s_f(\mathbf{P})<\epsilon\), from Theorem~. Consequently, from , there is a partition \(\mathbf{Q}\) of \(T\) such that \(S_{F_p}(\mathbf{Q})-s_{F_p}(\mathbf{Q})<\epsilon\), so \(F_p\) is integrable on \(T\), from Theorem~.

It remains to verify that \[\begin{equation} \label{eq:7.2.12} \int_R f(\mathbf{X})\,d\mathbf{X}= \int_TF_p(\mathbf{Y})\,d\mathbf{Y}. \end{equation} \nonumber \] From and the definition of \(\int_TF_p(\mathbf{Y})\,d\mathbf{Y}\), there is for each \(\epsilon>0\) a \(\delta>0\) such that \[ \left|\int_TF_p(\mathbf{Y})\,d\mathbf{Y} -\sigma\right|<\epsilon\mbox{\quad if\quad} \|\mathbf{Q}\|<\delta; \nonumber \] that is, \[ \sigma-\epsilon<\int_TF_p(\mathbf{Y})\,d\mathbf{Y} <\sigma+ \epsilon\mbox{\quad if \quad}\|\mathbf{Q}\|<\delta. \nonumber \] This and imply that \[ s_f(\mathbf{P})-\epsilon< \int_TF_p(\mathbf{Y})\,d\mathbf{Y} <S_f(\mathbf{P})+\epsilon\mbox{\quad if\quad}\|\mathbf{P}\|<\delta, \nonumber \] and this implies that \[\begin{equation} \label{eq:7.2.13} \underline{\int_R}\, f(\mathbf{X})\,d\mathbf{X}- \epsilon\le \int_TF_p(\mathbf{Y})\,d\mathbf{Y} \le \overline{\int_R} f(\mathbf{X})\,d\mathbf{X}+\epsilon. \end{equation} \nonumber \] Since \(\dst\underline{\int_R}\, f(\mathbf{X})\,d\mathbf{X}= \overline{\int_R} f(\mathbf{X})\,d\mathbf{X}\) (Theorem~) and \(\epsilon\) can be made arbitrarily small, implies .

The proof is by induction. From Theorem~, the proposition is true for \(n=2\). Now assume \(n>2\) and the proposition is true with \(n\) replaced by \(n-1\). Holding \(x_n\) fixed and applying this assumption yields \[ F_n(x_n)= \int^{b_{n-1}}_{a_{n-1}} dx_{n-1}\int_{a_{n-2}}^{b_{n-2}}dx_{n-2}\cdots \int^{b_2}_{a_2} dx_2\int^{b_1}_{a_1} f(\mathbf{X})\,dx_1. \nonumber \] Now Theorem~ with \(p=n-1\) completes the induction.

The hypotheses of Theorems~ and ~ are stated so as to justify successive integrations with respect to \(x_1\), then \(x_2\), then \(x_3\), and so forth. It is legitimate to use other orders of integration if the hypotheses are adjusted accordingly. For example, suppose that

\(\{i_1,i_2, \dots,i_n\}\) is a permutation of \(\{1,2, \dots,n\}\) and \(\int_R f(\mathbf{X})\,d\mathbf{X}\) exists, along with \[\begin{equation}\label{eq:7.2.14} \int_{I_{i_1}\times I_{i_2}\times\cdots\times I_{i_j}} f(\mathbf{X})\, d(x_{i_1},x_{i_2}, \dots,x_{i_j}),\quad 1\le j\le n-1, \end{equation} \nonumber \] for each \[\begin{equation}\label{eq:7.2.15} (x_{i_{j+1}}, x_{i_{j+2}}, \dots, x_{i_n}) \mbox{\quad in\quad} I_{i_{j+1}}\times I_{i_{j+2}}\times\cdots\times I_{i_n}. \end{equation} \nonumber \] Then, by renaming the variables, we infer from Theorem~ that \[\begin{equation}\label{eq:7.2.16} \int_R f(\mathbf{X})\,d\mathbf{X}=\int^{b_{i_n}}_{a_{i_n}} dx_{i_n} \int^{b_{i_{n-1}}}_{a_{i_{n-1}}} dx_{i_{n-1}}\cdots \int^{b_{i_2}}_{a_{i_2}} dx_{i_2}\int^{b_{i_1}}_{a_{i_1}} f(\mathbf{X})\, dx_{i_1}. \end{equation} \nonumber \]

Since there are \(n!\) permutations of \(\{1,2, \dots,n\}\), there are \(n!\) ways of evaluating a multiple integral over a rectangle in \(\R^n\), provided that the integrand satisfies appropriate hypotheses. In particular, if \(f\) is continuous on \(R\) and \(\{i_1,i_2, \dots,i_n\}\) is any permutation of \(\{1,2, \dots,n\}\), then \(f\) is continuous with respect to \((x_{i_1},x_{i_2}, \dots,x_{i_j})\) on \(I_{i_1}\times I_{i_2}\times\cdots \times I_{i_j}\) for each fixed \((x_{i_{j+1}}, x_{i_{j+2}}, \dots,x_{i_n})\) satisfying . Therefore, the integrals exist for every permutation of \(\{1,2, \dots,n\}\) (Theorem~). We summarize this in the next theorem, which now follows from Theorem~.

We now consider the problem of evaluating multiple integrals over more general sets. First, suppose that \(f\) is integrable on a set of the form \[\begin{equation}\label{eq:7.2.17} S=\set{(x,y)}{u(y)\le x\le v(y),\ c\le y\le d} \end{equation} \nonumber \] (Figure~).

If \(u(y)\ge a\) and \(v(y)\le b\) for \(c\le y\le d\), and \[\begin{equation}\label{eq:7.2.18} f_S(x,y)=\left\{\casespace\begin{array}{ll} f(x,y),&(x,y)\in S,\\[2\jot] 0,&(x,y)\not\in S,\end{array}\right. \end{equation} \nonumber \] then \[ \int_S f(x,y) \,d(x,y)=\int_R f_S(x,y) \,d(x,y), \nonumber \] where \(R=[a,b]\times [c,d]\).. From Theorem~, \[ \int_R f_S(x,y) \,d(x,y)=\int_c^d dy\int_a^b f_S (x,y)\,dx \nonumber \] provided that \(\int_a^b f_S(x,y)\,dx\) exists for each \(y\) in \([c,d]\). From and , this integral can be written as \[\begin{equation}\label{eq:7.2.19} \int^{v(y)}_{u(y)} f(x,y)\,dx. \end{equation} \nonumber \] Thus, we have proved the following theorem.

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From Theorem~, the assumptions of Theorem~ are satisfied if \(f\) is continuous on \(S\) and \(u\) and \(v\) are continuously differentiable on \([c,d]\).

Interchanging \(x\) and \(y\) in Theorem~ shows that if \(f\) is integrable on \[\begin{equation}\label{eq:7.2.21} S=\set{(x,y)}{u(x)\le y\le v(x),\ a\le x\le b} \end{equation} \nonumber \] (Figure~) and \[ \int^{v(x)}_{u(x)} f(x,y)\,dy \nonumber \] exists for \(a\le x\le b\), then \[\begin{equation}\label{eq:7.2.22} \int_S f(x,y) \,d(x,y)=\int_a^b dx\int^{v(x)}_{u(x)} f(x,y)\,dy. \end{equation} \nonumber \]

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12pt Theorem~ has an analog for \(n>2\). Suppose that \(f\) is integrable on a set \(S\) of points \(\mathbf{X}=(x_1,x_2, \dots,x_n)\) satisfying the inequalities \[ u_j(x_{j+1}, \dots,x_n)\le x_j\le v_j(x_{j+1}, \dots,x_n),\quad 1\le j\le n-1, \nonumber \] and \[ a_n\le x_n\le b_n. \nonumber \] Then, under appropriate additional assumptions, it can be shown by an argument analogous to the one that led to Theorem~ that \[ \int_S f(\mathbf{X})\,d\mathbf{X}=\int^{b_n}_{a_n} dx_n\int^{v_n(x_n)}_{u_n(x_n)} dx_{n-1}\cdots\int^{v_2(x_3, \dots,x_n)}_{u_2(x_3, \dots,x_n)} dx_2 \int^{v_1(x_2, \dots,x_n)}_{u_1(x_2, \dots,x_n)} f(\mathbf{X})\,dx_1. \nonumber \] These additional assumptions are tedious to state for general \(n\). The following theorem contains a complete statement for \(n=3\).

2em

Thus far we have viewed the iterated integral as a tool for evaluating multiple integrals. In some problems the iterated integral is itself the object of interest. In this case a result

like Theorem~ can be used to evaluate the iterated integral. The procedure is as follows.

This procedure is called {} of an iterated integral.

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\begin{exerciselist}

Evaluate

Let \(I_j=[a_j, b_j]\), \(1\le j\le 3\), and suppose that \(f\) is integrable on \(R=I_1\times I_2\times I_3\). Prove:

Prove: If \(f\) is continuous on \([a,b]\times [c,d]\), then the function \[ F(y)=\int_a^b f(x,y)\,dx \nonumber \] is continuous on \([c,d]\).

Suppose that \[ f(x',y')\ge f(x,y)\mbox{\quad if\quad}\ a\le x\le x'\le b,\ c\le y\le y' \le d. \nonumber \] Show that \(f\) satisfies the hypotheses of Theorem~ on \(R=[a,b]\times [c,d]\).

Evaluate by means of iterated integrals:

Let \(A\) be the set of points of the form \((2^{-m}p, 2^{-m}q)\), where \(p\) and \(q\) are odd integers and \(m\) is a nonnegative integer. Let \[ f(x,y)=\left\{\casespace\begin{array}{ll} 1,&(x,y)\not\in A,\\[1\jot] 0,&(x,y)\in A.\end{array}\right. \nonumber \] Show that \(f\) is not integrable on any rectangle \(R=[a,b]\times [c,d]\), but \[ \int_a^b dx\int_c^d f(x,y)\,dy=\int_c^d dy\int_a^b f(x,y)\,dx= (b-a)(d-c). \eqno{\rm(A)} \nonumber \]

Let \[ f(x,y)=\left\{\casespace\begin{array}{ll} 2xy&\mbox{\quad if $y$ is rational},\\ y&\mbox{\quad if $y$ is irrational}, \end{array}\right. \nonumber \] and \(R=[0,1]\times [0,1]\) (Example~).

Let \(R=[0,1]\times [0,1]\times [0,1]\), \(\widetilde{R}=[0,1]\times [0,1]\), and \[ f(x,y,z)=\left\{\casespace\begin{array}{ll} 2xy+2xz& \mbox{\quad if $y$ and $z$ are rational},\\ y+2xz&\mbox{\quad if $y$ is irrational and $z$ is rational},\\ 2xy+z&\mbox{\quad if $y$ is rational and $z$ is irrational},\\ y+z&\mbox{\quad if $y$ and $z$ are irrational}. \end{array}\right. \nonumber \] Calculate Suppose that \(f\) is bounded on \(R=[a,b]\times[c,d]\). Prove:

Use Exercise~ to prove the following generalization of Theorem~: If \(f\) is integrable on \(R=[a,b]\times [c,d]\), then \[ \overline{\int_a^b} f(x,y)\,dy\mbox{\quad and\quad}\underline{\int_c^d} f(x,y)\,dy \nonumber \] are integrable on \([a,b]\), and \[ \int_a^b\left(\overline{\int_c^d} f(x,y)\,dy\right)\,dx=\int_a^b \left(\underline{\int_c^d} f(x,y)\,dy\right)\,dx=\int_R f(x,y)\, d(x,y). \nonumber \]

Evaluate Evaluate

Evaluate \(\int_S(x+y)\,d(x,y)\), where \(S\) is bounded by \(y=x^2\) and \(y=2x\), using iterated integrals of both possible types.

Find the area of the set bounded by the given curves.

In Example~, verify the last five representations of \(\int_S f(x,y,z)\,d(x,y,z)\) as iterated integrals.

Let \(S\) be the region in \(\R^3\) bounded by the coordinate planes and the plane \(x+2y+3z=1\). Let \(f\) be continuous on \(S\). Set up six iterated integrals that equal \(\int_S f(x,y,z)\,d(x,y,z)\).

Evaluate Find the volume of \(S\).

Let \(R=[a_1,b_2]\times [a_2,b_2]\times\cdots\times [a_n,b_n]\). Evaluate

Assuming that \(f\) is continuous, express \[ \int^1_{1/2} dy\int^{\sqrt{1-y^2}}_{-\sqrt{1-y^2}} f(x,y)\,dx \nonumber \] as an iterated integral with the order of integration reversed.

Evaluate \(\int_S (x+y)\,d(x,y)\) of Example~ by means of iterated integrals in which the first integration is with respect to \(x\).

Evaluate \(\dst\int_0^1 x\,dx\int^{\sqrt{1-x^2}}_0\frac{dy}{\sqrt{x^2+y^2}}.\)

Suppose that \(f\) is continuous on \([a,\infty)\), \[ y^{(n)}(x)=f(x),\quad t\ge a, \nonumber \] and \(y(a)=y'(a)=\cdots=y^{(n-1)}(a)=0\).

Let \(T_\rho=[0,\rho]\times [0,\rho],\,\rho>0\). By calculating \[ I(a)=\lim_{\rho\to\infty}\int_{T_\rho} e^{-xy}\sin ax\,d(x,y) \nonumber \] in two different ways, show that \[ \int^\infty_0\frac{\sin ax}{ x} dx=\frac{\pi}{2}\mbox{\quad if\quad} a>0. \nonumber \]

\end{exerciselist}

In Section~3.3 we saw that a change of variables may simplify the evaluation of an ordinary integral. We now consider change of variables in multiple integrals.

Prior to formulating the rule for change of variables, we must deal with some rather involved preliminary considerations.

-.4em In Section~ we defined the content of a set \(S\) to be \[\begin{equation}\label{eq:7.3.1} V(S)=\int_S d\mathbf{X} \end{equation} \nonumber \] if the integral exists. If \(R\) is a rectangle containing \(S\), then can be rewritten as \[ V(S)=\int_R\psi_S(\mathbf{X})\,d\mathbf{X}, \nonumber \] where \(\psi_S\) is the characteristic function of \(S\), defined by \[ \psi_S(\mathbf{X})=\left\{\casespace\begin{array}{ll} 1,&\mathbf{X}\in S,\\ 0,&\mathbf{X}\not\in S.\end{array}\right. \nonumber \] From Exercise~, the existence and value of \(V(S)\) do not depend on the particular choice of the enclosing rectangle \(R\). We say that \(S\) is {} if \(V(S)\) exists. Then \(V(S)\) is the {}\(S\).

We leave it to you (Exercise~) to show that \(S\) has zero content according to Definition~ if and only if \(S\) has Jordan content zero.

Let \(R\) be a rectangle containing \(S\). Suppose that \(V(\partial S)=0\). Since \(\psi_{S}\) is bounded on \(R\) and discontinuous only on \(\partial S\) (Exercise~), Theorem~ implies that \(\int_R\psi_S (\mathbf{X})\,d\mathbf{X}\) exists. For the converse, suppose that \(\partial S\) does not have zero content and let \({\bf P}=\{R_1, R_2,\dots, R_k\}\) be a partition of \(R\). For each \(j\) in \(\{1,2,\dots,k\}\) there are three possibilities:

Let \[\begin{equation} \label{eq:7.3.2} {\mathcal U}_1=\set{j}{R_j\subset S} \mbox{\quad and \quad} {\mathcal U}_2=\set{j}{R_j\cap S\ne\emptyset\mbox{ and }R_j\cap S^c\ne\emptyset}. \end{equation} \nonumber \]

Then the upper and lower sums of \(\psi_S\) over \({\bf P}\) are \[\begin{equation}\label{eq:7.3.3} \begin{array}{rcl} S({\bf P})\ar=\dst\sum_{j\in{\mathcal U}_1} V(R_j)+\sum_{j\in{\mathcal U}_2} V(R_j)\\[2\jot] \ar=\mbox{total content of the subrectangles in ${\bf P}$ that intersect $S$} \end{array} \end{equation} \nonumber \] and \[\begin{equation}\label{eq:7.3.4} \begin{array}{rcl} s({\bf P})\ar=\dst\sum_{j\in{\mathcal U}_1} V(R_j) \\ \ar=\mbox{total content of the subrectangles in ${\bf P}$ contained in $S$}. \end{array} \end{equation} \nonumber \] Therefore, \[ S({\bf P})-s({\bf P})=\sum_{j\in {\mathcal U}_2} V(R_j), \nonumber \] which is the total content of the subrectangles in \({\bf P}\) that intersect both \(S\) and \(S^c\). Since these subrectangles contain \(\partial S\), which does not have zero content, there is an \(\epsilon_0>0\) such that \[ S({\bf P})-s({\bf P})\ge\epsilon_0 \nonumber \] for every partition \({\bf P}\) of \(R\). By Theorem~, this implies that \(\psi_S\) is not integrable on \(R\), so \(S\) is not Jordan measurable.

Theorems~ and imply the following corollary.

Since \(V(K)=0\), \[ \int_C\psi_K(\mathbf{X})\,d\mathbf{X}=0 \nonumber \] if \(C\) is any cube containing \(K\). From this and the definition of the integral, there is a \(\delta>0\) such that if \({\bf P}\) is any partition of \(C\) with \(\|{\bf P}\|\le\delta\) and \(\sigma\) is any Riemann sum of \(\psi_K\) over \({\bf P}\), then \[\begin{equation}\label{eq:7.3.6} 0\le\sigma\le\epsilon. \end{equation} \nonumber \]

Now suppose that \({\bf P}=\{C_1,C_2,\dots,C_k\}\) is a partition of \(C\) into cubes with \[\begin{equation}\label{eq:7.3.7} \|{\bf P}\|<\min (\rho,\delta), \end{equation} \nonumber \] and let \(C_1\), \(C_2\), , \(C_k\) be numbered so that \(C_j\cap K\ne \emptyset\) if \(1\le j\le r\) and \(C_j\cap K=\emptyset\) if \(r+1\le j\le k\). Then holds, and a typical Riemann sum of \(\psi_K\) over \({\bf P}\) is of the form \[ \sigma=\sum_{j=1}^r\psi_K(\mathbf{X}_j)V(C_j) \nonumber \] with \(\mathbf{X}_j\in C_j\), \(1\le j\le r\). In particular, we can choose \(\mathbf{X}_j\) from \(K\), so that \(\psi_K(\mathbf{X}_j)=1\), and \[ \sigma=\sum_{j=1}^r V(C_j). \nonumber \] Now and imply that \(C_1\), \(C_2\), , \(C_r\) have the required properties.

To formulate the theorem on change of variables in multiple integrals, we must first consider the question of preservation of Jordan measurability under a regular transformation.

Since \(K\) is a compact subset of the open set \(S\), there is a \(\rho_1>0\) such that the compact set \[ K_{\rho_1}=\set{\mathbf{X}}{\dist(\mathbf{X},K)\le\rho_1} \nonumber \] is contained in \(S\) (Exercise_{5.1.26). From Lemma}, there is a constant \(M\) such that \[\begin{equation}\label{eq:7.3.8} |\mathbf{G}(\mathbf{Y})-\mathbf{G}(\mathbf{X})|\le M|\mathbf{Y}-\mathbf{X}| \mbox{\quad if\quad}\mathbf{X},\mathbf{Y}\in K_{\rho_1}. \end{equation} \nonumber \] Now suppose that \(\epsilon>0\). Since \(V(K)=0\), there are cubes \(C_1\), \(C_2\), , \(C_r\) with edge lengths \(s_1\), \(s_2\), , \(s_r<\rho_1/\sqrt n\) such that \(C_j\cap K\ne\emptyset\), \(1\le j\le r\), \[ K\subset\bigcup_{j=1}^r C_j, \nonumber \] and \[\begin{equation} \label{eq:7.3.9} \sum_{j=1}^r V(C_j)<\epsilon \end{equation} \nonumber \] (Lemma~). For \(1\le j\le r\), let \(\mathbf{X}_j\in C_j\cap K\). If \(\mathbf{X}\in C_j\), then \[ |\mathbf{X}-\mathbf{X}_j|\le s_j\sqrt n<\rho_1, \nonumber \]

so \(\mathbf{X}\in K\) and \(|\mathbf{G}(\mathbf{X})-\mathbf{G}(\mathbf{X}_j)|\le M|\mathbf{X}-\mathbf{X}_j|\le M\sqrt{n}\,s_j\), from . Therefore, \(\mathbf{G}(C_j)\) is contained in a cube \(\widetilde{C}_j\) with edge length \(2M\sqrt{n}\,s_j\), centered at \(\mathbf{G}(\mathbf{X}_j)\). Since \[ V(\widetilde{C}_j)=(2M\sqrt{n})^ns_j^n=(2M\sqrt{n})^nV(C_j), \nonumber \] we now see that \[ \mathbf{G}(K)\subset\bigcup_{j=1}^r\widetilde{C}_j \nonumber \] and \[ \sum_{j=1}^r V(\widetilde{C}_j)\le (2M\sqrt{n})^n\sum_{j=1}^r V(C_j)<(2M\sqrt{n})^n\epsilon, \nonumber \] where the last inequality follows from . Since \((2M\sqrt{n})^n\) does not depend on \(\epsilon\), it follows that \(V(\mathbf{G}(K))=0\).

We leave it to you to prove that \(\mathbf{G}(S)\) is compact (Exercise~6.2.23). Since \(S\) is Jordan measurable, \(V(\partial S)=0\), by Theorem~. Therefore, \(V(\mathbf{G}(\partial S))=0\), by Lemma~. But \(\mathbf{G}(\partial S)= \partial(\mathbf{G}(S))\) (Exercise~), so \(V(\partial(\mathbf{G}(S)))=0\), which implies that \(\mathbf{G}(S)\) is Jordan measurable, again by Theorem~.

-.4em To motivate and prove the rule for change of variables in multiple integrals, we must know how \(V(\mathbf{L}(S))\) is related to \(V(S)\) if \(S\) is a compact Jordan measurable set and \(\mathbf{L}\) is a nonsingular linear transformation. (From Theorem~, \(\mathbf{L}(S)\) is compact and Jordan measurable in this case.) The next lemma from linear algebra will help to establish this relationship. We omit the proof.

Matrices of the kind described in this lemma are called {} matrices. The key to the proof of the lemma is that if \(\mathbf{E}\) is an elementary \(n\times n\) matrix and \(\mathbf{A}\) is any \(n\times n\) matrix, then \(\mathbf{EA}\) is the matrix obtained by applying to \(\mathbf{A}\) the same operation that must be applied to \(\mathbf{I}\) to produce \(\mathbf{E}\) (Exercise~). Also, the inverse of an elementary matrix of type , , or

is an elementary matrix of the same type (Exercise~).

The next example illustrates the procedure for finding the factorization .

Lemma~ and Theorem~

imply that an arbitrary invertible linear transformation \(\mathbf{L}: \R^n\to \R^n\), defined by \[\begin{equation}\label{eq:7.3.12} \mathbf{X}=\mathbf{L}(\mathbf{Y})=\mathbf{AY}, \end{equation} \nonumber \] can be written as a composition \[\begin{equation}\label{eq:7.3.13} \mathbf{L}=\mathbf{L}_k\circ\mathbf{L}_{k-1}\circ\cdots\circ\mathbf{L}_1, \end{equation} \nonumber \] where \[ \mathbf{L}_i(\mathbf{Y})=\mathbf{E}_i\mathbf{Y},\quad 1\le i\le k. \nonumber \]

Theorem~ implies that \(\mathbf{L}(S)\) is Jordan measurable. If \[\begin{equation} \label{eq:7.3.15} V(\mathbf{L}(R))=|\det(\mathbf{A})| V(R) \end{equation} \nonumber \] whenever \(R\) is a rectangle, then holds if \(S\) is any compact Jordan measurable set. To see this, suppose that \(\epsilon>0\), let \(R\) be a rectangle containing \(S\), and let \({\bf P}=\{R_1,R_2,\dots,R_k\}\) be a partition of \(R\) such that the upper and lower sums of \(\psi_S\) over \({\bf P}\) satisfy the inequality \[\begin{equation}\label{eq:7.3.16} S({\bf P})-s({\bf P})<\epsilon. \end{equation} \nonumber \] Let \({\mathcal U}_1\) and \({\mathcal U}_2\) be as in . From and , \[\begin{equation}\label{eq:7.3.17} s({\bf P})=\sum_{j\in{\mathcal U}_1} V(R_j)\le V(S)\le\sum_{j\in{\mathcal U}_1} V(R_j)+\sum_{j\in{\mathcal U}_2} V(R_j)=S({\bf P}). \end{equation} \nonumber \] Theorem~ implies that \(\mathbf{L}(R_1)\), \(\mathbf{L}(R_2)\), , \(\mathbf{L}(R_k)\) and \(\mathbf{L}(S)\) are all Jordan measurable. Since \[ \bigcup_{j\in{\mathcal U}_1}R_j\subset S\subset\bigcup_{j\in{\mathcal S}_1\cup{\mathcal S_2}}R_j, \nonumber \] it follows that \[ L\left(\bigcup_{j\in{\mathcal U}_1}R_j\right)\subset L(S)\subset L\left(\bigcup_{j\in{\mathcal S}_1\cup{\mathcal S_2}}R_j\right). \nonumber \] Since \(L\) is one-to-one on \(\R^n\), this implies that \[\begin{equation} \label{eq:7.3.18} \sum_{j\in{\mathcal U}_1} V(\mathbf{L}(R_j))\le V(\mathbf{L}(S))\le\sum_{j\in{\mathcal U}_1} V(\mathbf{L}(R_j))+\sum_{j\in{\mathcal U}_2} V(\mathbf{L}(R_j)). \end{equation} \nonumber \] If we assume that holds whenever \(R\) is a rectangle, then \[ V(\mathbf{L}(R_j))=|\det(\mathbf{A})|V(R_j),\quad 1\le j\le k, \nonumber \] so implies that \[ s({\bf P})\le \frac{V(\mathbf{L}(S))}{ |\det(\mathbf{A})|}\le S({\bf P}). \nonumber \] This, and imply that \[ \left|V(S)-\frac{V(\mathbf{L}(S))}{ |\det(\mathbf{A})|}\right|<\epsilon; \nonumber \] hence, since \(\epsilon\) can be made arbitrarily small, follows for any Jordan measurable set.

To complete the proof, we must verify for every rectangle \[ R=[a_1,b_1]\times [a_2,b_2]\times\cdots\times [a_n,b_n]=I_1\times I_2\times\cdots\times I_n. \nonumber \] Suppose that \(\mathbf{A}\) in is an elementary matrix; that is, let \[ \mathbf{X}=\mathbf{L}(\mathbf{Y})=\mathbf{EY}. \nonumber \]

{Case 1}. If \(\mathbf{E}\) is obtained by interchanging the \(i\)th and \(j\)th rows of \(\mathbf{I}\), then \[ x_r=\left\{\casespace\begin{array}{ll} y_r&\mbox{if $r\ne i$ and $r\ne j$};\\ y_j&\mbox{if $r=i$};\\ y_i&\mbox{if $r=j$}.\end{array}\right. \nonumber \] Then \(\mathbf{L}(R)\) is the Cartesian product of \(I_1\), \(I_2\), , \(I_n\) with \(I_i\) and \(I_j\) interchanged, so \[ V(\mathbf{L}(R))=V(R)=|\det(\mathbf{E})|V(R) \nonumber \] since \(\det(\mathbf{E})=-1\) in this case (Exercise~

).

{Case 2}. If \(\mathbf{E}\) is obtained by multiplying the \(r\)th row of \(\mathbf{I}\) by \(a\), then \[ x_r=\left\{\casespace\begin{array}{ll} y_r&\mbox{if $r\ne i$},\\ ay_i&\mbox{if $r=i$}.\end{array}\right. \nonumber \] Then \[ \mathbf{L}(R)=I_1\times\cdots\times I_{i-1}\times I'_i\times I_{i+1}\times \cdots\times I_n, \nonumber \] where \(I'_i\) is an interval with length equal to \(|a|\) times the length of \(I_i\), so \[ V(\mathbf{L}(R))=|a|V(R)=|\det(\mathbf{E})|V(R) \nonumber \] since \(\det(\mathbf{E})=a\) in this case (Exercise~

).

{Case 3}. If \(\mathbf{E}\) is obtained by adding \(a\) times the \(j\)th row of \(\mathbf{I}\) to its \(i\)th row (\(j\ne i\)), then \[ x_r=\left\{\casespace\begin{array}{ll} y_r&\mbox{if $r\ne i$};\\ y_i+ay_j&\mbox{if $r=i$}.\end{array}\right. \nonumber \] Then \[ \mathbf{L}(R)=\set{(x_1,x_2,\dots,x_n)}{a_i+ax_j\le x_i\le b_i+ax_j \mbox{ and } a_r\le x_r\le b_r\mbox{if } r\ne i}, \nonumber \] which is a parallelogram if \(n=2\) and a parallelepiped if \(n=3\) (Figure~). Now \[ V(\mathbf{L}(R))=\int_{\mathbf{L}(R)} d\mathbf{X}, \nonumber \] which we can evaluate as an iterated integral in which the first integration is with respect to \(x_i\). For example, if \(i=1\), then \[\begin{equation}\label{eq:7.3.19} V(\mathbf{L}(R))=\int^{b_n}_{a_n} dx_n\int^{b_{n-1}}_{a_{n-1}} dx_{n-1}\cdots\int^{b_2}_{a_2} dx_2\int^{b_1+ax_j}_{a_1+ax_j} dx_1. \end{equation} \nonumber \]

Since \[ \int^{b_1+ax_j}_{a_1+ax_j} dy_1=\int^{b_1}_{a_1} dy_1, \nonumber \] can be rewritten as \[\begin{eqnarray*} V(\mathbf{L}(R))\ar=\int^{b_n}_{a_n} dx_n\int^{b_{n-1}}_{a_{n-1}} dx_{n-1}\cdots\int^{b_2}_{a_2} dx_2\int^{b_1}_{a_1} dx_1\\ \ar=(b_n-a_n)(b_{n-1}-a_{n-1})\cdots (b_1-a_1)=V(R). \end{eqnarray*} \nonumber \] Hence, \(V(\mathbf{L}(R))=|\det(\mathbf{E})|V(R)\), since \(\det(\mathbf{E})=1\) in this case (Exercise~

).

12pt 6pt

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From what we have shown so far, holds if \(\mathbf{A}\) is an elementary matrix and \(S\) is any compact Jordan measurable set. If \(\mathbf{A}\) is an arbitrary nonsingular matrix,

-.0em then we can write \(\mathbf{A}\) as a product of elementary matrices and apply our known result successively to \(\mathbf{L}_1\), \(\mathbf{L}_2\), , \(\mathbf{L}_k\) (see ). This yields \[ V(\mathbf{L}(S))=|\det(\mathbf{E}_k)|\,|\det(\mathbf{E}_{k-1})|\cdots |\det\mathbf{E}_1| V(S)=|\det(\mathbf{A})|V(S), \nonumber \] by Theorem~ and induction.

We now formulate the rule for change of variables in a multiple integral. Since we are for the present interested only in ``discovering’’ the rule, we will make any assumptions that ease this task, deferring questions of rigor until the proof.

Throughout the rest of this section it will be convenient to think of the range and domain of a transformation \(\mathbf{G}:\R^n\to \R^n\) as subsets of distinct copies of \(\R^n\). We will denote the copy containing \(D_{\mathbf{G}}\) as \(\E^n\), and write \(\mathbf{G}: \E^n\to \R^n\) and \(\mathbf{X}=\mathbf{G}(\mathbf{Y})\), reversing the usual roles of \(\mathbf{X}\) and \(\mathbf{Y}\).

If \(\mathbf{G}\) is regular on a subset \(S\) of \(\E^n\), then each \(\mathbf{X}\) in \(\mathbf{G}(S)\) can be identified by specifying the unique point \(\mathbf{Y}\) in \(S\) such that \(\mathbf{X}=\mathbf{G}(\mathbf{Y})\).

Suppose that we wish to evaluate \(\int_T f(\mathbf{X})\,d\mathbf{X}\), where \(T\) is the image of a compact Jordan measurable set \(S\) under the regular transformation \(\mathbf{X}=\mathbf{G}(\mathbf{Y})\). For simplicity, we take \(S\) to be a rectangle and assume that \(f\) is continuous on \(T=\mathbf{G}(S)\).

Now suppose that \({\bf P}=\{R_1,R_2,\dots,R_k\}\) is a partition of \(S\) and \(T_j= \mathbf{G}(R_j)\) (Figure~).

6pt

8pt

Then \[\begin{equation}\label{eq:7.3.20} \int_T f(\mathbf{X})\,d\mathbf{X}=\sum_{j=1}^k\int_{T_j} f(\mathbf{X})\,d\mathbf{X} \end{equation} \nonumber \] (Corollary~ and induction). Since \(f\) is continuous, there is a point \(\mathbf{X}_j\) in \(T_j\) such that \[ \int_{T_j} f(\mathbf{X})\,d\mathbf{X}=f(\mathbf{X}_j)\int_{T_j}\,d\mathbf{X}= f(\mathbf{X}_j) V(T_j) \nonumber \]

(Theorem~), so can be rewritten as \[\begin{equation}\label{eq:7.3.21} \int_T f(\mathbf{X})\,d\mathbf{X}=\sum_{j=1}^k f(\mathbf{X}_j) V(T_j). \end{equation} \nonumber \]

Now we approximate \(V(T_j)\). If \[\begin{equation}\label{eq:7.3.22} \mathbf{X}_j=\mathbf{G}(\mathbf{Y}_j), \end{equation} \nonumber \] then \(\mathbf{Y}_j\in R_j\) and, since \(\mathbf{G}\) is differentiable at \(\mathbf{Y}_j\), \[\begin{equation}\label{eq:7.3.23} \mathbf{G}(\mathbf{Y})\approx\mathbf{G}(\mathbf{Y}_j)+ \mathbf{G}'(\mathbf{Y}_j)(\mathbf{Y}-\mathbf{Y}_j). \end{equation} \nonumber \] Here \(\mathbf{G}\) and \(\mathbf{Y}-\mathbf{Y}_j\) are written as column matrices, \(\mathbf{G}'\) is a differential matrix, and $\approx$'' meansapproximately equal’’ in a sense that we could make precise if we wished (Theorem~).

It is reasonable to expect that the Jordan content of \(\mathbf{G}(R_j)\) is approximately equal to the Jordan content of \(\mathbf{A}(R_j)\), where \(\mathbf{A}\) is the affine transformation \[ \mathbf{A}(\mathbf{Y})=\mathbf{G}(\mathbf{Y}_j)+ \mathbf{G}'(\mathbf{Y}_j)(\mathbf{Y}-\mathbf{Y}_j) \nonumber \] on the right side of ; that is, \[\begin{equation} \label{eq:7.3.24} V(\mathbf{G}(R_j))\approx V(\mathbf{A}(R_j)). \end{equation} \nonumber \] We can think of the affine transformation \(\mathbf{A}\) as a composition \(\mathbf{A}= \mathbf{A}_3\circ\mathbf{A}_2\circ\mathbf{A_1}\), where \[\begin{eqnarray*} \mathbf{A}_1(\mathbf{Y})\ar=\mathbf{Y}-\mathbf{Y}_j,\\ \mathbf{A}_2(\mathbf{Y})\ar=\mathbf{G}'(\mathbf{Y}_j)\mathbf{Y},\\ \arraytext{and}\\ \mathbf{A}_3(\mathbf{Y})\ar=\mathbf{G}(\mathbf{Y}_j)+\mathbf{Y}. \end{eqnarray*} \nonumber \] Let \(R_j'=\mathbf{A}_1(R_j)\). Since \(\mathbf{A}_1\) merely shifts \(R_j\) to a different location, \(R_j'\) is also a rectangle, and \[\begin{equation} \label{eq:7.3.25} V(R_j')=V(R_j). \end{equation} \nonumber \] Now let \(R_j''=\mathbf{A_2}(R_j')\). (In general, \(R_j''\) is not a rectangle.) Since \(\mathbf{A}_2\) is the linear transformation with nonsingular matrix \(\mathbf{G}'(\mathbf{Y}_j)\), Theorem~ implies that \[\begin{equation} \label{eq:7.3.26} V(R_j''))= |\det\mathbf{G}'(\mathbf{Y}_j)|V(R_j')=|J\mathbf{G}(\mathbf{Y}_j)|V(R_j), \end{equation} \nonumber \] where \(J\mathbf{G}\) is the Jacobian of \(\mathbf{G}\). Now let \(R_j'''=\mathbf{A_3}(R_j'')\). Since \(\mathbf{A}_3\) merely shifts all points in the same way, \[\begin{equation} \label{eq:7.3.27} V(R_j''')=V(R_j''). \end{equation} \nonumber \]

Now – suggest that \[ V(T_j)\approx |J\mathbf{G}(\mathbf{Y}_j)|V(R_j). \nonumber \]

(Recall that \(T_j=\mathbf{G}(R_j)\).) Substituting this and into yields \[ \int_T f(\mathbf{X})\,d\mathbf{X}\approx\sum_{j=1}^k f(\mathbf{G}(\mathbf{Y}_j))|J\mathbf{G}(\mathbf{Y}_j)| V(R_j). \nonumber \] But the sum on the right is a Riemann sum for the integral \[ \int_S f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}, \nonumber \] which suggests that \[ \int_T f(\mathbf{X})\,d\mathbf{X}=\int_S f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \nonumber \] We will prove this by an argument that was published in the {} [Vol. 61 (1954), pp. 81-85] by J. Schwartz.

We now prove the following form of the rule for change of variable in a multiple integral.

Since the proof is complicated, we break it down to a series of lemmas. We first observe that both integrals in exist, by Corollary~, since their integrands are continuous. (Note that \(S\) is compact and Jordan measurable by assumption, and \(\mathbf{G}(S)\) is compact and Jordan measurable by Theorem~.) Also, the result is trivial if \(V(S)=0\), since then \(V(\mathbf{G}(S))=0\) by Lemma~, and both integrals in vanish. Hence, we assume that \(V(S)>0\). We need the following definition.

Let \(s\) be the edge length of \(C\). Let \(\mathbf{Y}_0= (c_1,c_2,\dots,c_n)\) be the center of \(C\), and suppose that \(\mathbf{H}=(y_1,y_2,\dots,y_n)\in C\). If \(\mathbf{H}= (h_1,h_2,\dots,h_n)\) is continuously differentiable on \(C\), then applying the mean value theorem (Theorem~) to the components of \(\mathbf{H}\) yields \[ h_i(\mathbf{Y})-h_i(\mathbf{Y}_0)=\sum_{j=1}^n \frac{\partial h_i(\mathbf{Y}_i)}{\partial y_j}(y_j-c_j),\quad 1\le i\le n, \nonumber \] where \(\mathbf{Y}_i\in C\). Hence, recalling that \[ \mathbf{H}'(\mathbf{Y})=\left[\frac{\partial h_i}{\partial y_j}\right]_{i,j=1}^n, \nonumber \] applying Definition~, and noting that \(|y_j-c_j|\le s/2\), \(1\le j\le n\), we infer that \[ |h_i(\mathbf{Y})-h_i(\mathbf{Y}_0)|\le \frac{s}{2} \max\set{\|\mathbf{H}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C},\quad 1\le i\le n. \nonumber \] This means that \(\mathbf{H}(C)\) is contained in a cube with center \(\mathbf{X}_0=\mathbf{H}(\mathbf{Y}_0)\) and edge length \[ s\max\set{\|\mathbf{H}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C}. \nonumber \] Therefore, \[\begin{equation}\label{eq:7.3.30} \begin{array}{rcl} V(\mathbf{H}(C))\ar\le \left[\max\set{\|\mathbf{H}'(\mathbf{Y})\|_\infty\right]^n}{\mathbf{Y}\in C} s^n\\[2\jot] \ar=\left[\max\set{\|\mathbf{H}'(\mathbf{Y})\|_\infty\right]^n}{\mathbf{Y}\in C} V(C). \end{array} \end{equation} \nonumber \] Now let \[ \mathbf{L}(\mathbf{X})=\mathbf{A}^{-1}\mathbf{X} \nonumber \] and set \(\mathbf{H}=\mathbf{L}\circ\mathbf{G}\); then \[ \mathbf{H}(C)=\mathbf{L}(\mathbf{G}(C)) \mbox{\quad and\quad}\mathbf{H}'=\mathbf{A}^{-1}\mathbf{G}', \nonumber \] so implies that \[\begin{equation}\label{eq:7.3.31} V(\mathbf{L}(\mathbf{G}(C)))\le \left[\max\set{\|\mathbf{A}^{-1}\mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C} \right]^nV(C). \end{equation} \nonumber \] Since \(\mathbf{L}\) is linear, Theorem~ with \(\mathbf{A}\) replaced by \(\mathbf{A}^{-1}\) implies that \[ V(\mathbf{L}(\mathbf{G}(C)))=|\det(\mathbf{A})^{-1}|V(\mathbf{G}(C)). \nonumber \] This and imply that \[ |\det(\mathbf{A}^{-1})|V(\mathbf{G}(C)) \le\left[\max\set{\|\mathbf{A}^{-1}\mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C} \right]^nV(C). \nonumber \] Since \(\det(\mathbf{A}^{-1})=1/\det(\mathbf{A})\), this implies .

Let \({\bf P}\) be a partition of \(C\) into subcubes \(C_1\), \(C_2\), , \(C_k\) with centers \(\mathbf{Y}_1\), \(\mathbf{Y}_2\), , \(\mathbf{Y}_k\). Then \[\begin{equation}\label{eq:7.3.33} V(\mathbf{G}(C))=\sum_{j=1}^k V(\mathbf{G}(C_j)). \end{equation} \nonumber \] Applying Lemma~ to \(C_j\) with \(\mathbf{A}=\mathbf{G}'(\mathbf{A}_j)\) yields \[\begin{equation}\label{eq:7.3.34} V(\mathbf{G}(C_j))\le |J\mathbf{G}(\mathbf{Y}_j)| \left[\max\set{\|(\mathbf{G}'(\mathbf{Y}_j))^{-1} \mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C_j} \right]^n V(C_j). \end{equation} \nonumber \] Exercise~ implies that if \(\epsilon>0\), there is a \(\delta>0\) such that \[ \max\set{\|(\mathbf{G}'(\mathbf{Y}_j))^{-1} \mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in C_j} <1+\epsilon,\quad 1\le j\le k,\mbox{\quad if\quad}\|{\bf P}\|<\delta. \nonumber \] Therefore, from , \[ V(\mathbf{G}(C_j))\le (1+\epsilon)^n|J\mathbf{G}(\mathbf{Y}_j)|V(C_j), \nonumber \] so implies that \[ V(\mathbf{G}(C))\le (1+\epsilon)^n\sum_{j=1}^k |J\mathbf{G}(\mathbf{Y}_j)|V(C_j)\mbox{\quad if\quad}\|{\bf P}\|<\delta. \nonumber \] Since the sum on the right is a Riemann sum for \(\int_C |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\) and \(\epsilon\) can be taken arbitrarily small, this implies .

Since \(S\) is Jordan measurable, \[ \int_C\psi_S(\mathbf{X})\,d\mathbf{X}=V(S) \nonumber \] if \(C\) is any cube containing \(S\). From this and the definition of the integral, there is a \(\delta>0\) such that if \({\bf P}\) is any partition of \(C\) with \(\|{\bf P}\|<\delta\) and \(\sigma\) is any Riemann sum of \(\psi_S\) over \({\bf P}\), then \(\sigma>V(S)-\epsilon/2\). Therefore, if \(s(P)\) is the lower sum of \(\psi_S\) over \(\mathbf{P}\), then \[\begin{equation} \label{eq:7.3.36} s(\mathbf{P})>V(S)-\epsilon\mbox{\quad if \quad}\|\mathbf{P}\|<\delta. \end{equation} \nonumber \] Now suppose that \({\bf P}=\{C_1,C_2,\dots,C_k\}\) is a partition of \(C\) into cubes with \(\|{\bf P}\|<\min (\rho,\delta)\), and let \(C_1\), \(C_2\), , \(C_k\) be numbered so that \(C_j\subset S\) if \(1\le j\le r\) and \(C_j\cap S^c\ne\emptyset\) if \(j>r\). From , \(s(\mathbf{P})=\sum_{j=1}^rV(C_k)\). This and imply . Clearly, \(C_i^0\cap C_j^0=\emptyset\) if \(i\ne j\).

From the continuity of \(J\mathbf{G}\) and \(f\) on the compact sets \(S\) and \(\mathbf{G}(S)\), there are constants \(M_1\) and \(M_2\) such that \[\begin{equation}\label{eq:7.3.38} |J\mathbf{G}(\mathbf{Y})|\le M_1\mbox{\quad if\quad}\mathbf{Y}\in S \end{equation} \nonumber \] and \[\begin{equation}\label{eq:7.3.39} |f(\mathbf{X})|\le M_2\mbox{\quad if\quad}\mathbf{X}\in\mathbf{G}(S) \end{equation} \nonumber \] (Theorem~). Now suppose that \(\epsilon>0\). Since \(f\circ\mathbf{G}\) is uniformly continuous on \(S\) (Theorem~), there is a \(\delta>0\) such that \[\begin{equation} \label{eq:7.3.40} |f(\mathbf{G}(\mathbf{Y}))-f(\mathbf{G}(\mathbf{Y}'))|<\epsilon \mbox{\quad if \quad$|\mathbf{Y}-\mathbf{Y}'|<\delta$ and }\mathbf{Y},\mathbf{Y}' \in S. \end{equation} \nonumber \] Now let \(C_1\), \(C_2\), , \(C_r\) be chosen as described in Lemma~, with \(\rho=\delta/\sqrt{n}\). Let \[ S_1=\set{\mathbf{Y}\in S}{\mathbf{Y}\notin\bigcup_{j=1}^r C_j}. \nonumber \] Then \(V(S_1)<\epsilon\) and \[\begin{equation} \label{eq:7.3.41} S=\left(\bigcup_{j=1}^r C_j\right)\cup S_1. \end{equation} \nonumber \]

Suppose that \(\mathbf{Y}_1\), \(\mathbf{Y}_2\), , \(\mathbf{Y}_r\) are points in \(C_1\), \(C_2\), , \(C_r\) and \(\mathbf{X}_j=\mathbf{G}(\mathbf{Y}_j)\), \(1\le j\le r\). From and Theorem~, \[\begin{eqnarray*} Q(S)\ar=\int_{\mathbf{G}(S_1)} f(\mathbf{X})\,d\mathbf{X}-\int_{S_1} f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y} \\ \ar{}+\sum_{j=1}^r\int_{\mathbf{G}(C_j)} f(\mathbf{X})\,d\mathbf{X}- \sum_{j=1}^r\int_{C_j} f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\\ \ar=\int_{\mathbf{G}(S_1)} f(\mathbf{X})\,d\mathbf{X}-\int_{S_1} f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\\ \ar{}+\sum_{j=1}^r\int_{\mathbf{G}(C_j)}(f(\mathbf{X})- f(\mathbf{A}_j))\,d\mathbf{X}\\ \ar{}+\sum_{j=1}^r\int_{C_j}((f(\mathbf{G}(\mathbf{Y}_j))- f(\mathbf{G}(\mathbf{Y})))|J(\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\\ \ar{}+\sum_{j=1}^r f(\mathbf{X}_j)\left(V(\mathbf{G}(C_j))- \int_{C_j} |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\right). \end{eqnarray*} \nonumber \]

Since \(f(\mathbf{X})\ge0\), \[ \int_{S_1}f(\mathbf{G}(\mathbf{Y}))|J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\ge0, \nonumber \] and Lemma~ implies that the last sum is nonpositive. Therefore, \[\begin{equation} \label{eq:7.3.42} Q(S)\le I_1+I_2+I_3, \end{equation} \nonumber \] where \[ I_1=\int_{\mathbf{G}(S_1)} f(\mathbf{X})\,d\mathbf{X},\quad I_2= \sum_{j=1}^r\int_{\mathbf{G}(C_j)}|f(\mathbf{X})-f(\mathbf{X}_j)| \,d\mathbf{X}, \nonumber \] and \[ I_3= \sum_{j=1}^r\int_{C_j}|f(\mathbf{G})(\mathbf{Y}_j))-f(\mathbf{G}(\mathbf{Y}))| |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \nonumber \] We will now estimate these three terms. Suppose that \(\epsilon>0\).

To estimate \(I_1\), we first remind you that since \(\mathbf{G}\) is regular on the compact set \(S\), \(\mathbf{G}\) is also regular on some open set \({\mathcal O}\) containing \(S\) (Definition~). Therefore, since \(S_1\subset S\) and \(V(S_1)<\epsilon\), \(S_1\) can be covered by cubes \(T_1\), \(T_2\), , \(T_m\) such that \[\begin{equation} \label{eq:7.3.43} \sum_{j=1}^r V(T_j)< \epsilon \end{equation} \nonumber \] and \(\mathbf{G}\) is regular on \(\bigcup_{j=1}^m T_j\). Now, \[ \begin{array}{rcll} I_1\ar\le M_2V(\mathbf{G}(S_1))& \mbox{(from \eqref{eq:7.3.39})}\\[2\jot] \ar\le M_2\dst\sum_{j=1}^m V(\mathbf{G}(T_j))&(\mbox{since }S_1\subset\cup_{j=1}^mT_j)\\[2\jot] \ar\le M_2\dst\sum_{j=1}^m\int_{T_j}| J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}& \mbox{(from Lemma~\ref{thmtype:7.3.11})} \\[2\jot] \ar\le M_2M_1\epsilon& \mbox{(from \eqref{eq:7.3.38} and \eqref{eq:7.3.43})}. \end{array} \nonumber \]

To estimate \(I_2\), we note that if \(\mathbf{X}\) and \(\mathbf{X}_j\) are in \(\mathbf{G}(C_j)\) then \(\mathbf{X}=\mathbf{G}(\mathbf{Y})\) and \(\mathbf{X}_j=\mathbf{G}(\mathbf{Y}_j)\) for some \(\mathbf{Y}\) and \(\mathbf{Y}_j\) in \(C_j\). Since the edge length of \(C_j\) is less than \(\delta/\sqrt n\), it follows that \(|\mathbf{Y}-\mathbf{Y}_j|<\delta\), so \(|f(\mathbf{X})-f(\mathbf{X}_j)|<\epsilon\), by . Therefore, \[ \begin{array}{rcll} I_2\ar< \epsilon\dst\sum_{j=1}^r V(\mathbf{G}(C_j))\\[2\jot] \ar\le \epsilon\dst\sum_{j=1}^r\int_{C_j}|J\mathbf{G}(\mathbf{Y})|d\mathbf{Y}& \mbox{(from Lemma~\ref{thmtype:7.3.11})}\\[2\jot] \ar\le \dst\epsilon M_1\sum_{j=1}^r V(C_j)&\mbox{(from \eqref{eq:7.3.38}})\\[2\jot] \ar\le \epsilon M_1 V(S)&(\mbox{since }\dst\cup_{j=1}^rC_j\subset S). \end{array} \nonumber \]

To estimate \(I_3\), we note again from that \(|f(\mathbf{G}(\mathbf{Y}_j))-f(\mathbf{G}(\mathbf{Y}))|< \epsilon\) if \(\mathbf{Y}\) and \(\mathbf{Y}_j\) are in \(C_j\). Hence, \[\begin{eqnarray*} I_3\ar< \epsilon\sum_{j=1}^r \int_{C_j}|J\mathbf{G}(\mathbf{Y})|d\mathbf{Y}\\ \ar\le M_1\epsilon\sum_{j=1}^r V(C_j) \mbox{\quad(from \eqref{eq:7.3.38}}\\ \ar\le M_1 V(S)\epsilon \end{eqnarray*} \nonumber \] because \(\bigcup_{j=1}^r C_j\subset S\) and \(C_i^0\cap C_j^0=\emptyset\) if \(i\ne j\).

From these inequalities on \(I_1\), \(I_2\), and \(I_3\), now implies that \[ Q(S)<M_1(M_2+2V(S))\epsilon. \nonumber \] Since \(\epsilon\) is an arbitrary positive number, it now follows that \(Q(S)\le0\).

Let \[\begin{equation} \label{eq:7.3.44} \mathbf{G} _1=\mathbf{G}^{-1},\quad S_1=\mathbf{G}(S),\quad f_1=(|J\mathbf{G}|)f\circ\mathbf{G}, \end{equation} \nonumber \] and \[\begin{equation} \label{eq:7.3.45} \begin{array} {rcl} Q_1(S_1)\ar=\dst\int_{\mathbf{G}_1(S_1)} f_1(\mathbf{Y})\,d\mathbf{Y}-\int_{S_1} f_1(\mathbf{G}_1(\mathbf{X})) |J\mathbf{G}_1(\mathbf{X})|\,d\mathbf{X}. \end{array} \end{equation} \nonumber \] Since \(\mathbf{G}_1\) is regular on \(S_1\) (Theorem~) and \(f_1\) is continuous and nonnegative on \(\mathbf{G}_1(S_1)=S\), Lemma~ implies that \(Q_1(S_1)\le0\). However, substituting from into and again noting that \(\mathbf{G}_1(S_1)=S\) yields \[\begin{equation} \label{eq:7.3.46} \begin{array} {rcl} Q_1(S_1)\ar=\dst\int_S f(\mathbf{G}(\mathbf{Y}))|J\mathbf{G} (\mathbf{Y})|\,d\mathbf{Y}\\ \ar{}-\dst\int_{\mathbf{G}(S)}f(\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X}))) |J\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X}))| |J\mathbf{G}^{-1}(\mathbf{X})|\,d\mathbf{X}. \end{array} \end{equation} \nonumber \] Since \(\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X}))=\mathbf{X}\), \(f(\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X})))=f(\mathbf{X})\). However, it is important to interpret the symbol \(J\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X}))\) properly. We are not substituting \(\mathbf{G}^{-1}(\mathbf{X})\) into \(\mathbf{G}\) here; rather, we are evaluating the determinant of the differential matrix of \(\mathbf{G}\) at the point \(\mathbf{Y}=\mathbf{G}^{-1}(\mathbf{X})\). From Theorems~ and ~, \[ |J\mathbf{G}(\mathbf{G}^{-1}(\mathbf{X}))||J\mathbf{G}^{-1}(\mathbf{X})|=1, \nonumber \] so can be rewritten as \[ Q_1(S_1)=\int_S f(\mathbf{G}(\mathbf{Y}))|J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y} -\int_{\mathbf{G}(S)}f(\mathbf{X})\,d\mathbf{X} = -Q(S). \nonumber \] Since \(Q_1(S_1)\le0\), it now follows that \(Q(S)\ge0\).

We can now complete the proof of Theorem_{. Lemmas} and imply if \(f\) is nonnegative on \(S\). Now suppose that \[ m=\min\set{f(\mathbf{X})}{\mathbf{X}\in \mathbf{G}(S)}<0. \nonumber \] Then \(f-m\) is nonnegative on \(\mathbf{G}(S)\), so with \(f\) replaced by \(f-m\) implies that \[\begin{equation} \label{eq:7.3.47} \int_{\mathbf{G}(S)} (f(\mathbf{X})-m)\,d\mathbf{X}= \int_S (f(\mathbf{G}(\mathbf{Y})-m) |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}. \end{equation} \nonumber \] However, setting \(f=1\) in yields \[ \int_{\mathbf{G}(S)} \,d\mathbf{X}=\int_S |J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}, \nonumber \] so implies .

The assumptions of Theorem~ are too stringent for many applications. For example, to find the area of the disc \[ \set{(x,y)}{x^2+y^2\le1}, \nonumber \] it is convenient to use polar coordinates and regard the circle as \(\mathbf{G}(S)\), where \[\begin{equation}\label{eq:7.3.48} \mathbf{G}(r,\theta)=\left[\begin{array}{c} r\cos\theta\\ r\sin\theta\end{array} \right] \end{equation} \nonumber \] and \(S\) is the compact set \[\begin{equation}\label{eq:7.3.49} S=\set{(r,\theta)}{0\le r\le1,\ 0\le\theta\le2\pi} \end{equation} \nonumber \] (Figure~).

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Since \[ \mathbf{G}'(r,\theta)= \left[\begin{array}{rr}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta \end{array}\right], \nonumber \] it follows that \(J\mathbf{G}(r,\theta)=r\). Therefore, formally applying Theorem~ with \(f\equiv1\) yields \[ A=\int_{\mathbf{G}(S)}\,d\mathbf{X}=\int_S r\,d(r,\theta)=\int_0^1 r\,dr \int^{2\pi}_0 d\theta=\pi. \nonumber \] Although this is a familiar result, Theorem~ does not really apply here, since \(\mathbf{G}(r,0)=\mathbf{G}(r,2\pi)\), $0r1 $, so \(\mathbf{G}\) is not one-to-one on \(S\), and therefore not regular on \(S\).

The next theorem shows that the assumptions of Theorem~ can be relaxed so as to include this example.

Since \(f\) is continuous on \(\mathbf{G}(S)\) and \((|J\mathbf{G}|) f\circ\mathbf{G}\) is continuous on \(S\), the integrals in both exist, by Corollary~. Now let \[ \rho=\dist\ (\partial S, N^c) \nonumber \] (Exercise~5.1.25), and \[ P=\set{\mathbf{Y}}{\dist(\mathbf{Y}, \partial S)}\le \frac{\rho}{2}. \nonumber \] Then \(P\) is a compact subset of \(N\) (Exercise~5.1.26) and \(\partial S\subset P^0\) (Figure~).

Since \(S\) is Jordan measurable, \(V(\partial S)=0\), by Theorem~. Therefore, if \(\epsilon>0\), we can choose cubes \(C_1\), \(C_2\), , \(C_k\) in \(P^0\) such that \[\begin{equation} \label{eq:7.3.51} \partial S\subset\bigcup_{j=1}^k C_j^0 \end{equation} \nonumber \] and \[\begin{equation}\label{eq:7.3.52} \sum_{j=1}^k V(C_j)<\epsilon \end{equation} \nonumber \]

Now let \(S_1\) be the closure of the set of points in \(S\) that are not in any of the cubes \(C_1\), \(C_2\), , \(C_k\); thus, \[ S_1=\overline{S\cap\left(\cup_{j=1}^k C_j\right)^c}. \nonumber \]

Because of , \(S_1\cap \partial S=\emptyset\), so \(S_1\) is a compact Jordan measurable subset of \(S^0\). Therefore, \(\mathbf{G}\) is regular on \(S_1\), and \(f\) is continuous on \(\mathbf{G}(S_1)\). Consequently, if \(Q\) is as defined in , then \(Q(S_1)=0\) by Theorem~.

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Now \[\begin{equation}\label{eq:7.3.53} Q(S)=Q(S_1)+Q(S\cap S_1^c)=Q(S\cap S_1^c) \end{equation} \nonumber \] (Exercise~) and \[ |Q(S\cap S_1^c)|\le\left|\int_{\mathbf{G}(S\cap S_1^c)} f(\mathbf{X})\,d\mathbf{X}\right|+\left| \int_{S\cap S_1^c} f(\mathbf{G}(\mathbf{Y}))|J\mathbf{G}(\mathbf{Y})|\,d\mathbf{Y}\right|. \nonumber \] But \[\begin{equation} \label{eq:7.3.54} \left|\int_{S\cap S_1^c} f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\, d\mathbf{Y}\right|\le M_1M_2 V(S\cap S_1^c), \end{equation} \nonumber \] where \(M_1\) and \(M_2\) are as defined in and . Since \(S\cap S_1^c\subset \cup_{j=1}^k C_j\), implies that \(V(S\cap S_1^k)<\epsilon\); therefore, \[\begin{equation} \label{eq:7.3.55} \left|\int_{S\cap S_1^c} f(\mathbf{G}(\mathbf{Y})) |J\mathbf{G}(\mathbf{Y})|\, d\mathbf{Y}\right|\le M_1M_2\epsilon, \end{equation} \nonumber \] from . Also \[\begin{equation}\label{eq:7.3.56} \left|\int_{\mathbf{G}(S\cap S_1^c)} f(\mathbf{X})\,d\mathbf{X}\right|\le M_2 V(\mathbf{G}(S\cap S_1^c))\le M_2\sum_{j=1}^k V(\mathbf{G}(C_j)). \end{equation} \nonumber \]

By the argument that led to with \({\bf H}={\bf G}\) and \(C=C_{j}\), \[ V(\mathbf{G}(C_j))\le\left[\max\set{\|\mathbf{G}'(\mathbf{Y})\|_\infty} {\mathbf{Y}\in C_j}\right]^nV(C_j), \nonumber \] so can be rewritten as \[ \left|\int_{\mathbf{G}(S\cap S_1^c)} f(\mathbf{X})\,d\mathbf{X}\right|\le M_2 \left[\max\set{\|\mathbf{G}'(\mathbf{Y})\|_\infty}{\mathbf{Y}\in P} \right]^n\epsilon, \nonumber \] because of . Since \(\epsilon\) can be made arbitrarily small, this and imply that \(Q(S\cap S_1^c)=0\). Now \(Q(S)=0\), from .

The transformation to polar coordinates to compute the area of the disc is now justified, since \(\mathbf{G}\) and \(S\) as defined by and satisfy the assumptions of Theorem~.

If \(\mathbf{G}\) is the transformation from polar to rectangle coordinates \[\begin{equation}\label{eq:7.3.57} \left[\begin{array}{c} x\\ y\end{array}\right]= \mathbf{G}(r,\theta)=\left[\begin{array}{c} r \cos\theta\\ r\sin\theta\end{array}\right], \end{equation} \nonumber \] then \(J\mathbf{G}(r,\theta)=r\) and becomes \[ \int_{\mathbf{G}(S)} f(x,y)\,d(x,y)=\int_S f(r\cos\theta, r\sin\theta)r\, d(r,\theta) \nonumber \] if we assume, as is conventional, that \(S\) is in the closed right half of the \(r\theta\)-plane. This transformation is especially useful when the boundaries of \(S\) can be expressed conveniently in terms of polar coordinates, as in the example preceding Theorem~. Two more examples follow.

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If \(\mathbf{G}\) is the transformation from spherical to rectangular coordinates, \[\begin{equation}\label{eq:7.3.58} \left[\begin{array}{c} x\\ y\\ z\end{array}\right]=\mathbf{G}(r,\theta,\phi)= \left[\begin{array}{c} r\cos\theta\cos\phi\\ r\sin\theta\cos\phi\\ r\sin\phi\end{array}\right], \end{equation} \nonumber \] then \[ \mathbf{G}'(r,\theta,\phi)= \left[\begin{array}{crc} \cos\theta\cos\phi&-r\sin\theta\cos\phi&-r\cos\theta\sin\phi\\ \sin\theta\cos\phi&r\cos\theta\cos\phi&-r\sin\theta\sin\phi\\ \sin\phi&0&r\cos\phi \end{array}\right] \nonumber \] and \(J\mathbf{G}(r,\theta,\phi)=r^2\cos\phi\), so becomes \[\begin{equation}\label{eq:7.3.59} \begin{array}{l} \dst\int_{\mathbf{G}(S)} f(x,y,z)\,d(x,y,z)\\\\ =\dst\int_S f(r\cos\theta\cos\phi, r\sin\theta\cos\phi, r\sin\phi) r^2\cos\phi\,d(r,\theta,\phi) \end{array} \end{equation} \nonumber \] if we make the conventional assumption that \(|\phi|\le\pi/2\) and \(r\ge0\).

We now consider other applications of Theorem~.

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