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The definition and basic properties of integration in $2$ (and higher) dimensions are extremely similar to the familiar $1$-dimensional case, so much so that a lot of this section is copy-pasted from the previous section, with small modifications.
A rectangle $R \subset\R^2$ is a set of the form $$ R = [a,b]\times[c,d] = \{ (x,y) : a\le x\le b, \ c\le y \le d\} $$ for some $a<b$ and $c<d$.
Given a bounded function $f:R\to \R$, we want to define the integral of $f$ over $R$ in such a way that if $f \ge 0$ everywhere in $R$, then its integral should equal the volume in $x$-$y$-$z$ space below the graph of $f$ and above the rectangle $R$, if it is possible to define this volume in an unambiguous way. The basic idea is to approximate the region from the inside and from the outside by unions of $3d$ boxes, and to compute the volume of these approximations.
Definition. A partition $P$ of $R$ is obtained by choosing
partitions
$$
a = x_0 < x_1 <\ldots < x_J = b \mbox{ of }[a,b] \quad\mbox{ and }
c = y_0 < y_1 <\ldots < y_K=d\mbox{ of }[c,d].
$$
The associated partition $P$ of $R$ consists of all the rectangles
$$
[x_{j-1}, x_j]\times [y_{k-1}, y_k], \quad\mbox{ for } j=1,\ldots, J, \ \ k=1,\ldots, K
$$
We will assume that the rectangles in the partition have been labelled
as
$R_1,\ldots, R_L$, where $L=JK$, so that $\sum_{\ell=1}^L\cdots \ $
means the sum of $\,\cdots\,$ over all rectangles in the partition
If $P$ and $P'$ are two partitions, with the rectangles of $P'$ denoted $R_{\ell'}'$ for $\ell'=1,\ldots, L'$, then we will say that $P'$ is a refinement of $P$ if $$ \mbox{ for every }\ell\in\{ 1,\ldots, L\} \mbox{ and }\ell'\in \{1,\ldots, L'\}, \mbox{ either } R_{\ell'}' \subset R_\ell\mbox{ or }(R_{\ell'}')^{int}\cap R_\ell^{int} = \emptyset. $$ This is the same as saying that we can obtain $P'$ from $P$ by refining the (1-d) partitions of $[a,b]$ or $[c,d]$ or both.
Notation. Given a partition $P$ of $R$ and a bounded function $f:R\to \R$, we will write $$ m_\ell := \inf \{ f(\bfx) : \bfx\in R_\ell\}, \qquad M_\ell := \sup \{ f(\bfx) : \bfx\in R_\ell\}. $$ We can also write $m_\ell(f)$ or $M_\ell(f)$, if we want to indicate explicitly which function we are considering (helpful in settings where we have to consider more than one function simultaneously.)
More notation. Given a function $f$ and partition $P$, we define $$ s_P f :=\sum_{\ell=1}^L m_\ell \mbox{area}(R_\ell), \qquad S_P f :=\sum_{\ell=1}^L M_\ell \mbox{area}(R_\ell), $$ where the area of a rectangle is defined in the obvious way: $$ \mbox{area}( [x_{j-1},x_j]\times [y_{k-1}, y_k]) = (x_j-x_{j-1}) \, (y_k-y_{k-1}) $$ Sometimes $s_Pf$ is called the lower Riemann sum of $f$ (over the rectangle $R$, with respect to the partition $P$), and $S_P f$ is called the upper Riemann sum.
Note that if $m_\ell>0$, then
$ m_\ell \mbox{area}(R_\ell)$ and is the volume of the largest $3d$ box whose base is $R_\ell$ and who sits below the graph of $f$.
$ M_\ell \mbox{area}(R_\ell)$ and is the volume of the smallest $3d$ box whose base is $R_\ell$ and whose top sits above the graph of $f$.
Note that Theorem 1 in Section 4.1 applies with absolutely no change to $m_\ell f$ and $M_\ell f$.
The proofs of the next three lemmas are almost exactly the same as in the $1d$ case.
Lemma 1. Given any two partitions $P$ and $P'$, there is a partition $P''$ that is a refinement of both of them, called a common refinement.
Lemma 2. If $P'$ is a refinement of $P$, then $$ s_{P'}f \ge s_P f\qquad\mbox{ and }\quad S_{P'}f \le S_P f\ . $$
Lemma 3. If $P$ and $P'$ are any partitions of $R$, then $s_Pf \le S_{P'}f$.
Next, we define $$ \underline {I_R}(f) := \sup_P s_P f,\qquad \overline {I_R}(f) := \inf_P S_P f, %% $$ It follows from Lemma 3 that $\underline {I_R}(f) \le \overline {I_R}(f)$.
Definition. A function $f:R\to \R$ is integrable if it is bounded and $\underline {I_R}(f) = \overline {I_R}(f)$. When this holds, we define $$ \iint_R f(x)\,dA := \mbox{ the integral of }f\mbox{ over }R := \underline {I_R}(f) = \overline {I_R}(f) \ . $$ We will sometimes write just $f$ instead of $f(x)$ in the integrand, when this does not result in any ambiguity. Other notation is sometimes used, including for example $\iint_R f(\bfx) d^2\bfx$. Also, sometimes people write $\int$ instead of $\iint$, etc.....
The $A$ in $dA$ stands for area
.
As in $1$-d, the following lemma is often (but not always) useful if you need to prove that a function is integrable.
Lemma 4. If $R$ is a rectangle and $f$ is a bounded function $R\to \R$, then $f$ is integrable if and only if \begin{equation}\label{useful} \forall \ep>0, \exists\mbox{ a partition }P \mbox{ of } R \mbox{ such that } S_Pf - s_Pf < \ep. \end{equation} In addition, if $f$ is integrable and $P$ is a partition such that \eqref{useful} holds, then \begin{equation} 0 \le \iint_R f\,dA - s_Pf < \ep, \qquad 0\le S_P f - \iint_R f\,dA < \ep. \label{useful2}\end{equation}
Next we summarize some basic properties of integration.
Theorem 1: Basic Properties of Integration and integrability. Assume that $f$ and $g$ are integrable functions on $R$ and that $c\in \R$.
$f+g$ is integrable, and $\iint_R (f+g)dA = \iint_R f\, dA+ \iint_R g\, dA$.
$cf$ is integrable, and $\iint_R c f(x)\,dA = c\iint_R f(x)\,dA$.
If $f(x)\le g(x)$ for all $x$, then $\iint_R f(x)\,dA\le \iint_R g(x)\,dA$.
$|f|$ is integrable, and $\left|\iint_R f(x)\,dA\right| \le \iint_R |f(x)|\,dA$.
The proofs of Lemma 4 and Theorem 1 are basically identical to the corresponding $1d$ results.
One main difference between integration in $1$ dimensions and in dimensions $n\ge 2$ is that in $1d$ we are normally content to integrate a function whose domain is an interval, whereas in higher dimensions, we very often are interested in integrating a function whose domain is not a rectangle.
First, we define the characteristic function of a set $S\subset\R^n$ to be the function $\chi_S$ such that $$ \chi_S(\bfx) := \begin{cases} 1&\mbox{ if }\bfx\in S \\ 0&\mbox{ if }\bfx\not\in S.\end{cases} $$ Next: Assume that $R$ is a rectangle in $\R^2$, that $S\subset R$ is a bounded set, and that$f:\R\to \R$ is a function.
We define $f$ to be integrable on $S$ if $\chi_S f$ is integrable on $R$. Note, the definition of $\chi_S$ clearly implies that $$\chi_S f(\bfx) := \begin{cases} f(\bfx)&\mbox{ if }\bfx\in S \\ 0&\mbox{ if }\bfx\not\in S.\end{cases} $$
If $f$ is integrable on $S$, then we define $$ \iint_S f\, dA := \iint_R \chi_S \,f \, dA $$
Note that for the above definitions, we do not actually require $f$ to be defined on $R\setminus S$; the definition still make sense as long as the domain of $f$ contains $S$.
(One can also check that if $R$ and $R'$ are two different rectangles that both contain $S$, then the definition of integrable on $S$ and of $\iint_S f\, dA$ does not depend on whether we choose $R$ or $R'$.)
Note that, even if $f$ is continuous everywhere in $S$, typically $\chi_Sf$ is not continuous everywhere in $R$; it is discontinuous at every point $\bfx \in \partial S$ where $f(\bfx)\ne 0$. So to understand when a continuous function $f$ is integrable on a non-rectangular set $S$, we need to understand when a discontinuous function is integrable on a rectangle $R$.
Thus, in $1$-d, the question of integrability of discontinuous functions may seem like a point of possible intellectual interest but no practical importance. In $2$ and more dimensions, however, it is a question we cannot avoid.
With that said, the theory is very similar to the $1$-d case.
Definition. A set $Z\subset \R^2$ has zero content if $$ \forall \ep>0, \mbox{ there exist rectangles } R_1,\ldots, R_L\mbox{ such that } \begin{cases} Z\subset \cup_{\ell=1}^L R_\ell, \ \ \mbox{and }& \\ \sum_{\ell=1}^L\mbox{area}(R_\ell)<\ep. &\ \end{cases} $$
Technical Remark 1.
One can check that if $Z$ has zero content, then in fact $$ \forall \ep>0, \mbox{ there exist rectangles } R_1,\ldots, R_L\mbox{ such that } \begin{cases} Z\subset \cup_{\ell=1}^L R_\ell^{int}, \ \ \mbox{and }& \\ \sum_{\ell=1}^L\mbox{area}(R_\ell)<\ep. &\ \end{cases} $$ That is, $Z$ is contained in the interiors of the rectangles.
This will be useful in the proof of Theorem 2 below. See also Section 4.1, Examples 1 and 2 and Technical Remark 1, for a little more about this.
As in $1$-d, zero content informally means small enough to be negligible, for purposes of integration.
We next summarize some properties/examples of sets of zero content.
Proposition 1.
1. If $Z'\subset Z$ and $Z$ has zero content, then $Z'$ has zero content.
2. If $Z$ is a union of a finite number of sets of zero content, then $Z$ also has zero content.
3. If $f:[a,b]\to \R$ is integrable, then the graph of $f$, that is, the set $\{ (x,f(x))\in \R^2 : x\in [a,b]\}$ has zero content.
4. Assume that $U\subset \R$ is open and that
$\bff:U\to \R^2$ is $C^1$. Then for any closed interval $[a,b]\subset U$,
$$
\{ \bff(t) : t\in [a,b]\}
\quad \mbox{ has zero content.}
$$
Our main theorem about integrability of discontinuous functions involves zero content, as in $1$-d.
Theorem 2. Let $R$ be a rectangle in $\R^2$. Then a bounded function $f:R\to \R$ is integrable if \begin{equation}\label{dzc} \{ \bfx\in R : f\mbox{ is discontinuous at }\bfx\} \quad\mbox{ has zero content}. \end{equation}
The proof (optional!) is described below.
In Section 4.3, we will start to compute integrals $$\iint_S f\, dA$$ where $S\subset \R^2$ is a compact subset defined for example as the set of all points bounded by the curves $\ldots$ and $\ldots$, and $f:S\to \R$ is continuous.
The main practical conclusion of all the above theory is that these integrals make sense, i.e., that $f$ is integrable on $S$ for problems like those described above.
This is a consequence of Theorem 2 and Proposition 1.
Optional! Also, maybe a little hard to read, due to lack of detail in places.
Fix $\ep>0$. By Lemma 4, it suffices to show that there exists a partition $P$ of $R$ such that $$ S_Pf - s_Pf < \ep. $$
To do this, let $$ Z := \{ \bfx\in R : f\mbox{ is discontinuous at }\bfx\}. $$ Also, let $M := \sup_R f $, a finite number, since $f$ is bounded.
By assumption, $Z$ has content zero. Thus, in view of Technical Remark 1 , there exist rectangles $R_1,\ldots, R_L$ such that $$ Z \subset \cup_{\ell=1}^L R_\ell^{int},\qquad \sum_{\ell=1}^L \mbox{area}(R_\ell) <\frac \ep {4 M}. $$ Let $$ T := R\setminus \left( \cup_{\ell=1}^L R_\ell^{int}\right) $$ This is a closed and bounded set, hence compact. Also, $f$ is continuous on $T$, since by definition, $T$ does not contain any of the points of discontinuity of $f$. Thus $f$ is uniformly continuous on $T$. (See Section 1.4 to refresh your memory on uniform continuity.) It follows that $$ \forall \ep_1 >0, \ \exists \delta>0 \mbox{ such that } \ \ \left. \begin{array}{r} \bfx,\bfy\in T \mbox{ and } \\ |\bfx-\bfy|< \delta \end{array} \right\} \Rightarrow |f(\bfx)-f(\bfy)|<\ep_1. $$ Let $\delta>0$ be a number such that this holds when $\ep_1 := \frac \ep{3\,\mbox{area}(R)}$.
Now fix a partition $P$ of $R$ into rectangles $\widetilde R_k, k=1,\dots, K$ such that $$ \mbox{ for every }k, \mbox{ both sides of }\widetilde R_k\mbox{ have length less than }\delta/2, \mbox{ and } $$ $$ \mbox{ for every }k,\quad \ \ \mbox{ EITHER }\widetilde R_k \subset T \ \quad\mbox{ OR } \ \quad \widetilde R_k\subset \cup_{\ell=1}^L R_\ell. $$ Proving that such a partition exists is an exercise. The easiest way to understand why is to draw a picture. We now define $$ \widetilde m_k := \inf_{\widetilde R_k}f, \qquad \widetilde M_k := \sup_{\widetilde R_k}f. $$ Consider a rectangle $\widetilde R_k$ such that $\widetilde R_k \subset T$. The choice of the partition implies that any two points in $\widetilde R_k$ are separated by at most $\frac {\sqrt 2}2\delta < \delta$. Thus the choice of $\delta$ implies that $$ \bfx, \bfy \in \widetilde R_k \quad\Rightarrow |f(\bfx) - f(\bfy)|< \ep_1. $$ It follows that for such a $k$, $$ \widetilde M_k - \widetilde m_k \le \ep_1 = \frac \ep{3\mbox{area}(R)} $$ Thus \begin{align} \nonumber \sum_{ \{ k :\widetilde R_k\subset T \} }(\widetilde M_k - \widetilde m_k)\mbox{area}(\widetilde R_k) &\le \sum_{ \{ k :\widetilde R_k\subset T \} }\frac \ep{3\mbox{area}(R)}\mbox{area}(\widetilde R_k) \\ &\le \frac \ep{3\mbox{area}(R)} \sum_{ \{ k :\widetilde R_k\subset T \} }\mbox{area}(\widetilde R_k) \nonumber \\ &\le \frac \ep 3.\label{rc} \end{align}
On the other hand, if $\widetilde R_k\not\subset T$, and if $\bfx, \bfy$ belong to $\widetilde R_k$, then $|f(\bfx) - f(\bfy)| \le |f(\bfx)| + |f(\bfy)|
\le 2 M$. It follows that
$$
\widetilde M_k - \widetilde m_k \le 2M \qquad\mbox{ if }\widetilde R_k\not\subset T.
$$
In addition, we know that if $\widetilde R_k\not\subset T$
then $\widetilde R_k\subset R_\ell$ for some $\ell$. (This is
one of the conditions we placed when choosing the partition. Thus
$$
\begin{aligned}
\sum_{ \{ k :\widetilde R_k\not \subset T \} }(\widetilde M_k - \widetilde m_k)\mbox{area}(\widetilde R_k)
&\le
2M \sum_{ \{ k :\widetilde R_k\not \subset T \} }\mbox{area}(\widetilde R_k) \nonumber \\
&\le
2M \sum_{ \ell =1, \ldots L }\mbox{area}(R_\ell) \nonumber \\
&\le
2M \frac \ep{4M} = \frac \ep 2
\end{aligned}
$$
using the choice of $\{ R_\ell\}$. Adding this to
\eqref{rc},
we finally conclude that
$$
\begin{aligned}
S_P f - s_Pf
& \ =
\sum_{k=1}^K
(\widetilde M_k - \widetilde m_k)\mbox{area}(\widetilde R_k)
\\
& \ = \
\sum_{ \{ k :\widetilde R_k \subset T \} }(\widetilde M_k - \widetilde m_k)\mbox{area}(\widetilde R_k)
+
\sum_{ \{ k :\widetilde R_k\not \subset T \} }(\widetilde M_k - \widetilde m_k)\mbox{area}(\widetilde R_k)
\\
& \ \le \frac 5 6 \ep < \ep.
\end{aligned}
$$
$\quad \Box$
The main practical conclusion of all the theory about integration in $\R^n$ for $n\ge 3$ is that if
reasonable), and
then $f$ is integrable on $S$, and the technqiues we will develop in Section 4.3 for computing its integral over $S$ are in fact justified.
In short, the theory of integration in sets of dimension $n\ge 3$ is the natural generalization of the $2$-d theory, with the main differences being notational.
We will continue to write $R$ to denote a set of the form
$$
R = [a_1,b_1]\times \cdots \times[a_n,b_n],
$$
which we can call a box, or a hyperbox if you like, when $n\ge 4$.
We can partition such a rectangle via the natural generealization of
the $2$-d procedure use above. Having done this, given a function $f:R\to \R$
and a partition $P$, we can define $m_\ell, M_\ell$ as the inf and sup of $f$
on the $\ell$th box $R_\ell$ of the partition, and
$$
s_Pf := \sum_{\ell=1}^L m_\ell \, \mbox{Vol}^n(R_\ell),
\qquad
S_Pf := \sum_{\ell=1}^L M_\ell \, \mbox{Vol}^n(R_\ell),
$$
where $\mbox{Vol}^n$ denotes the the $n$-dimensional volume
of a box, defined to be the
product of the lengths of its sides. (When $n=3$ we might write simply
$\mbox{Vol}$ instead of $\mbox{Vol}^n$. Although in fact we may not
have many opportunities to write either one.)
Then for $f:R\to \R$ we define integrable exactly as above, and when $f$ is integrable we define its integral exactly as above. For $n=3$ the integral is denoted as $$ \iiint f\, dV \qquad\mbox{ or sometimes as }\quad \iiint f(\bfx) d^3\bfx $$ and for the general case, we use the notation $$ \idotsint f \ dV^n \qquad\mbox{ or sometimes as }\quad \idotsint f(\bfx) d^n\bfx \ . $$ (You may sometimes see other notations....)
A set $S\subset \R^n$ has zero content if for any $\ep>0$, there exist a finite collection of boxes $R_1,\ldots, R_K$ such that $$ S\subset \cup_{k=1}^K R_k, \qquad \sum_{k=1}^K \mbox{Vol}^n(R_k)<\ep. $$
With these definitions, analogs of Theorem 1, Theorem 2 and Proposition 1 are valid for every dimension $n\ge 3$, with more or less the same proofs in every case. These theorems underlie what we summarized vaguely above as the Main Practical Conclusion.
This section is mostly theoretical; the basic skills (which are necessary for almost every problem we will solve for the remainder of the year) appear in Section 4.3
The questions below are stated about integration in $\R^2$, but all the conclusions remain valid, with suitable modifications, for integrals in $\R^n$.
Assume that $R$ is a rectangle in $\R^2$.
Assume that $R$ is a rectangle and $f:R\to \R$ is a function such that
$f(\bfx)\ge 0$ for all $\bfx\in R$, and in addition
$$
\inf\{ \iint_R g \, dA : g(\bfx)\ge f(\bfx)\mbox{ for all }\bfx\in R\}
=0
$$
then $f$ is integrable, and $\iint_R f\, dA=0$.
Hint. The previous question may be helpful.
Let $S$ be a bounded subset of $\R^2$. Prove that
$$
S\mbox{ has zero content } \quad \iff\qquad \iint_S 1 \, dA = 0 .
$$
Of course $\iint_S 1\ dA$ can also be written as $\iint_R \chi_S \, dA$,
where $R$ is any rectangle such that $S\subset R$.
Hint. The previous question may be helpful.
Let $R = [a,b]\times [c,d]$, and for $e\in (a,b)$, define $$ R_1 := [a,e]\times [c,d], \qquad R_2 = [e,b]\times [c,d] $$ Prove that if $f$ is integrable on $R_1$ and $R_2$, then $f$ is integrable on $R$, and $$ \iint_R f\, dA = \iint_{R_1}f\, dA + \iint_{R_2} f\, dA. $$ Hint. The previous question may be helpful. Also, what can we say about the difference between $f$ and $\chi_{R_1} f + \chi_{R_2} f $?
Assume that $S, S_1$ and $S_2$ are compact subsets of $\R^2$ such that
\begin{equation} \label{S1S2}
S = S_1 \cup S_2, \qquad S_1\cap S_2 \mbox{ has zero content}.
\end{equation}
Assume also that $f$ is a function that is integrable on $S_1$ and $S_2$.
Prove that $f$ is integrable on $S$, and that
\begin{equation}\label{additive}
\iint_S f\, dA = \iint_{S_1}f\, dA + \iint_{S_2} f\, dA.
\end{equation}
Hint. This is essentially just an abstract version of the previous question. So the previous hint may be useful.
In practice, it is sometimes easiest to compute an integral $\iint_S f\, dA$ by splitting $S$ up into pieces $S_1, S_2$ that satisfy \eqref{S1S2}, then computing the integral over each piece and \eqref{additive}. See the next section for an example. The last exercise about shows that this procedure is justified.