4.2 Integration in Two and More Dimensions

4.2 Integration in Two and More Dimensions

  1. Integration in Two Dimensions
  2. Properties of Integration
  3. Integration over Other Sets
  4. Main Conclusion
  5. Integration in Three or More Dimensions
  6. Problems

\(\Leftarrow\)  \(\Uparrow\)  \(\Rightarrow\)

The definitions 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.

Integration in Two Dimensions

A rectangle \(R \subseteq\R^2\) is a set of the form \[ R = [a,b]\times[c,d] = \left\{ (x,y) : a\le x\le b, \ c\le y \le d\right\} \] for some \(a<b\) and \(c<d\).

Such a set is also called the Cartesian product of \([a,b]\) 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 \(\R^3\) 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.

A partition \(P\) of \(R\) is a collection of rectangles obtained by choosing partitions \[ a = x_0 < x_1 <\ldots < x_J = b \text{ of }[a,b] \quad\text{ and } c = y_0 < y_1 <\ldots < y_K=d\text{ of }[c,d]. \] The associated partition \(P\) of \(R\) consists of all the rectangles \[ [x_{j-1}, x_j]\times [y_{k-1}, y_k], \quad\text{ for } j=1,\ldots, J, \ \ k=1,\ldots, K \]

We will suppose that the rectangles in the partition have been labelled \(R_1,\ldots, R_{JK}\), so that “\(\sum_{i=1}^{JK}\cdots \ \)” means “the sum of \(\,\cdots\,\) over all rectangles in the partition.”

If \(P\) and \(P'\) are two partitions, with the rectangles of \(P'\) denoted \(R_{i'}'\) for \(i'=1,\ldots, J'K'\), then we will say that \(P'\) is a refinement of \(P\) if \[ \text{ for every }i\in\left\{ 1,\ldots, JK\right\} \text{ and }i'\in \left\{1,\ldots, J'K'\right\}, \text{ either } R_{i'}' \subseteq R_i\text{ or }(R_{i'}')^{int}\cap R_i^{int} = \emptyset. \]

This is the same as saying that we can obtain \(P'\) from \(P\) by refining the one dimensional 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 \[ m_i = \inf \left\{ f(\mathbf x) : \mathbf x\in R_i\right\}, \qquad M_i = \sup \left\{ f(\mathbf x) : \mathbf x\in R_i\right\}. \] We can also write \(m_i(f)\) or \(M_i(f)\), if we want to indicate explicitly which function we are considering (helpful in settings where we have to consider more than one function simultaneously.)

Given a function \(f\) and partition \(P\), we define \[ L_P f =\sum_{i=1}^{JK} m_i \text{area}(R_i), \qquad U_P f =\sum_{i=1}^{JK} M_i \text{area}(R_i), \] where the area of a rectangle is defined as: \[ \text{area}( [x_{j-1},x_j]\times [y_{k-1}, y_k]) = (x_j-x_{j-1}) \, (y_k-y_{k-1}) \] The sum \(L_Pf\) is called the lower Darboux sum of \(f\) (over the rectangle \(R\), with respect to the partition \(P\)), and \(U_P f\) is called the upper Darboux sum.

Note that if \(m_i>0\), then

Note that Theorem 1 in Section 4.1 applies with no change to \(m_i f\) and \(M_i f\).

The proofs of the next three lemmas are almost exactly the same as in the \(1d\) case.

Given any two partitions \(P\) and \(P'\), there is a partition \(P''\) that is a refinement of both of them, called a common refinement.
If \(P'\) is a refinement of \(P\), then \[ L_{P'}f \ge L_P f\qquad\text{ and }\quad U_{P'}f \le U_P f\ . \]
If \(P\) and \(P'\) are any partitions of \(R\), then \(L_Pf \le U_{P'}f\).

Next, we define \[ \underline {I_R}(f) = \sup_P L_P f,\qquad \overline {I_R}(f) = \inf_P U_P f, %% \] It follows from Lemma 3 that \(\underline {I_R}(f) \le \overline {I_R}(f)\).

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 \[ \iint_R f(x)\,dA = \text{ the integral of }f\text{ over }R = \underline {I_R}(f) = \overline {I_R}(f) \ . \]

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 \(\iint_R f(\mathbf x) d^2\mathbf x\). Also, sometimes people write \(\int\) instead of \(\iint\), etc.

The \(A\) in \(dA\) stands for “area”.

Properties of Integration

As in the single variable case, the following lemma is often useful to prove that a function is integrable.

If \(R\) is a rectangle and \(f\) is a bounded function \(R\to \R\), then \(f\) is integrable if and only if \[\begin{equation}\label{useful} \forall \varepsilon>0, \exists\text{ a partition }P \text{ of } R \text{ such that } U_Pf - L_Pf < \varepsilon. \end{equation}\] In addition, if \(f\) is integrable and \(P\) is a partition such that \(\eqref{useful}\) holds, then \[\begin{equation} 0 \le \iint_R f\,dA - L_Pf < \varepsilon, \qquad 0\le U_P f - \iint_R f\,dA < \varepsilon. \label{useful2}\end{equation}\]

Next we summarize some basic properties of integration.

Suppose that \(f\) and \(g\) are integrable functions on \(R\) and that \(c\in \R\).

  1. \(f+g\) is integrable, and \(\iint_R (f+g)dA = \iint_R f\, dA+ \iint_R g\, dA\).
  2. \(cf\) is integrable, and \(\iint_R c f(x)\,dA = c\iint_R f(x)\,dA\).
  3. If \(f(x)\le g(x)\) for all \(x\), then \(\iint_R f(x)\,dA\le \iint_R g(x)\,dA\).
  4. \(|f|\) is integrable, and \(\left|\iint_R f(x)\,dA\right| \le \iint_R |f(x)|\,dA\).
The proofs of Lemma 4 and Theorem 5 are basically identical to the corresponding results in the previous section.

Integration over Other Sets

In \(\R\), we are happy to only integrate over intervals. In \(\R^n\), only integrating over rectangles would be very restrictive. So we need to generalize to other sets. We start with a
The characteristic function of a set \(S\subseteq\R^n\) is the function \(\chi_S:\R^n\to \R\) \[ \chi_S(\mathbf x) = \begin{cases} 1&\text{ if }\mathbf x\in S \\ 0&\text{ if }\mathbf x\not\in S.\end{cases} \]
Suppose that \(R\) is a rectangle in \(\R^2\) and that \(S\subseteq R\). Then \(f:R\to\R\) is integrable on \(S\) if \(\chi_S f\) is integrable on \(R\).

Note that the definition of \(\chi_S\) implies that \[\chi_S f(\mathbf x) = \begin{cases} f(\mathbf x)&\text{ if }\mathbf x\in S \\ 0&\text{ if }\mathbf x\not\in S.\end{cases} \]

If \(f\) is integrable on \(S\), then we write \[ \iint_S f\, dA = \iint_R \chi_S \,f \, dA \]

Note that for the above definitions, we do not actually require \(f\) to be defined on \(R\setminus S\); the definitions 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 \(S\) does not depend on whether we choose \(R\) or \(R'\).

Even when \(f\) is continuous on \(S\), typically \(\chi_Sf\) is not continuous on \(R\); it is discontinuous at every point \(\mathbf x \in \partial S\) where \(f(\mathbf x)\ne 0\). 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, for single variable functions, the question of integrability of discontinuous functions may seem like a point of 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 single variable case.

A set \(Z\subseteq \R^2\) has zero content if for every \(\varepsilon>0\), there exist finitely many rectangles \(R_1,\ldots, R_{JK}\) such that \[ \begin{cases} Z\subseteq \bigcup\limits_{i=1}^{JK} R_i, \ \ \text{and }& \\ \displaystyle \sum_{i=1}^{JK}\text{area}(R_i)<\varepsilon. & \end{cases} \]
Remark We can check that an equivalent definition has \(Z\subseteq \cup_{i=1}^{JK} R_i^{int}\). That is, \(Z\) is contained in the interiors of the rectangles. This will be useful in the proof of Theorem 2 below.

As in one dimension, zero content informally means small enough to be negligible, for purposes of integration.

We next summarize some properties of sets of zero content. You should try to prove any that seem interesting.

  1. If \(Z'\subseteq Z\) 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 \(\left\{ (x,f(x))\in \R^2 : x\in [a,b]\right\}\) has zero content.
  4. Suppose that \(U\subseteq \R\) is open and that \(\mathbf f:U\to \R^2\) is \(C^1\). Then for any closed interval \([a,b]\subseteq U\), the set \(\left\{\mathbf f(t) : t\in [a,b]\right\}\) has zero content.
A set \(S\) is (Jordan) measurable if it is bounded and \(\partial S\) has zero content.

We will usually say measurable instead of Jordan measurable, but note, this is different from Lebesgue measurable, which is what most mathematicians mean when they say a set is measurable. Every Jordan measurable set is Lebesgue measurable, but the Lebesgue integral can deal with more complicated sets. It is beyond the scope of this class.

The measurable sets are the sets where \(\chi_S\) is integrable on any rectangle containing \(S\). We will expand this idea in the following theorem.

Our main theorem about integrability of discontinuous functions involves zero content sets.

Let \(R\) be a rectangle in \(\R^2\). Then a bounded function \(f:R\to \R\) is integrable if \[\begin{equation}\label{dzc} \left\{ \mathbf x\in R : f\text{ is discontinuous at }\mathbf x\right\} \quad\text{ has zero content}. \end{equation}\]
Optional! Also, may be a little hard to read, due to lack of detail in places.

Fix \(\varepsilon>0\). By Lemma 4, it suffices to show that there exists a partition \(P\) of \(R\) such that \[ U_Pf - L_Pf < \varepsilon. \]

To do this, let \[ Z = \left\{ \mathbf x\in R : f\text{ is discontinuous at }\mathbf x\right\}. \] Also, let $M = _R f $, a finite number, since \(f\) is bounded.

By assumption, \(Z\) has content zero. Thus, in view of the previous Remark, there exist rectangles \(R_1,\ldots, R_{JK}\) such that \[ Z \subseteq \cup_{i=1}^{JK} R_i^{int},\qquad \sum_{i=1}^{JK} \text{area}(R_i) <\frac \varepsilon {4 M}. \] Let \[ T = R\setminus \left( \cup_{i=1}^{JK} R_i^{int}\right) \] 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 \(T\). It follows that \[ \forall \varepsilon_1 >0, \ \exists \delta>0 \text{ such that } \ \ \left. \begin{array}{r} \mathbf x,\mathbf y\in T \text{ and } \\ |\mathbf x-\mathbf y|< \delta \end{array} \right\\right\} \Rightarrow |f(\mathbf x)-f(\mathbf y)|<\varepsilon_1. \] Let \(\delta>0\) be a number such that this holds when \(\varepsilon_1 = \frac \varepsilon{3\,\text{area}(R)}\).

Now fix a partition \(P\) of \(R\) into rectangles \(\widetilde R_k, k=1,\dots, K\) such that \[ \text{ for every }k, \text{ both sides of }\widetilde R_k\text{ have length less than }\delta/2, \text{ and } \] \[ \text{ for every }k,\quad \ \ \text{ EITHER }\widetilde R_k \subseteq T \ \quad\text{ OR } \ \quad \widetilde R_k\subseteq \cup_{i=1}^{JK} R_i. \] 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 \(\widetilde R_k \subseteq T\). 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 \[ \mathbf x, \mathbf y \in \widetilde R_k \quad\Rightarrow |f(\mathbf x) - f(\mathbf y)|< \varepsilon_1. \] It follows that for such a \(k\), \[ \widetilde M_k - \widetilde m_k \le \varepsilon_1 = \frac \varepsilon{3\text{area}(R)} \] Thus \[\begin{align} \nonumber \sum_{ \left\{ k :\widetilde R_k\subseteq T \right\} }(\widetilde M_k - \widetilde m_k)\text{area}(\widetilde R_k) &\le \sum_{ \left\{ k :\widetilde R_k\subseteq T \right\} }\frac \varepsilon{3\text{area}(R)}\text{area}(\widetilde R_k) \nonumber \\ &\le \frac \varepsilon{3\text{area}(R)} \sum_{ \left\{ k :\widetilde R_k\subseteq T \right\} }\text{area}(\widetilde R_k) &\le \frac \varepsilon 3.\label{rc} \end{align}\]

On the other hand, if \(\widetilde R_k\not\subseteq T\), and if \(\mathbf x, \mathbf y\) belong to \(\widetilde R_k\), then \(|f(\mathbf x) - f(\mathbf y)| \le |f(\mathbf x)| + |f(\mathbf y)| \le 2 M\). It follows that \[ \widetilde M_k - \widetilde m_k \le 2M \qquad\text{ if }\widetilde R_k\not\subseteq T. \] In addition, we know that if \(\widetilde R_k\not\subseteq T\) then \(\widetilde R_k\subseteq R_i\) for some \(i\). (This is one of the conditions we placed when choosing the partition. Thus \[ \begin{aligned} \sum_{ \left\{ k :\widetilde R_k\not \subseteq T \right\} }(\widetilde M_k - \widetilde m_k)\text{area}(\widetilde R_k) &\le 2M \sum_{ \left\{ k :\widetilde R_k\not \subseteq T \right\} }\text{area}(\widetilde R_k) \nonumber \\ &\le 2M \sum_{ i =1, \ldots JK }\text{area}(R_i) \nonumber \\ &\le 2M \frac \varepsilon{4M} = \frac \varepsilon 2 \end{aligned} \] using the choice of \(\left\{ R_i\right\}\). Adding this to \(\eqref{rc}\), we finally conclude that \[ \begin{aligned} U_P f - L_Pf & \ = \sum_{k=1}^K (\widetilde M_k - \widetilde m_k)\text{area}(\widetilde R_k) \\ & \ = \ \sum_{ \left\{ k :\widetilde R_k \subseteq T \right\} }(\widetilde M_k - \widetilde m_k)\text{area}(\widetilde R_k) + \sum_{ \left\{ k :\widetilde R_k\not \subseteq T \right\} }(\widetilde M_k - \widetilde m_k)\text{area}(\widetilde R_k) \\ & \ \le \frac 5 6 \varepsilon < \varepsilon. \end{aligned} \]

The Main Conclusion

In Section 4.3, we will start to compute integrals \[\iint_S f\, dA\] where \(S\subseteq \R^2\) is a compact subset defined for example as the set of all points bounded by the curves in \(\mathbb R^2\), and \(f:S\to \R\) is continuous.

The main 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 Theorems 2 and 3.

Integration in dimensions \(n\ge 3\)

The Main Conclusion in \(\R^n\)

The main conclusion of all the theory about integration in \(\R^n\) for \(n\ge 3\) is that if

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.

A Very Brief Summary of the Theory

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 a box using the natural generealization of the \(2\)-d procedure above. Then given a function \(f:R\to \R\) and a partition \(P\), we can define \(m_i, M_i\) as the inf and sup of \(f\) on the \(i\)th box \(R_i\) of the partition, and \[ L_Pf = \sum_{i} m_i \, \text{Vol}^n(R_i), \qquad U_Pf = \sum_{i} M_i \, \text{Vol}^n(R_i), \] where the sum is taken over all boxes, and \(\text{Vol}^n\) denotes the the “\(n\)-dimensional volume” of a box, defined to be the product of the lengths of its sides. When \(n=3\) we write \(\text{Vol}\) instead of \(\text{Vol}^n\).

Finally, 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 \[ \iiint f\, dV \qquad\text{ or sometimes as }\quad \iiint f(\mathbf x) d^3\mathbf x \] and for the general case, we use the notation \[ \idotsint f \ dV^n \qquad\text{ or sometimes as }\quad \idotsint f(\mathbf x) d^n\mathbf x. \]

A set \(S\subseteq \R^n\) has zero content if for any \(\varepsilon>0\), there exist a finite collection of boxes \(R_1,\ldots, R_K\) such that \[ S\subseteq \cup_{k=1}^K R_k, \qquad \sum_{k=1}^K \text{Vol}^n(R_k)<\varepsilon. \]

With these definitions, analogs of Theorem 5, 6 and 7 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 Conclusion.

Problems

Basic

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

Advanced

The questions below are stated about integration in \(\R^2\), but all the conclusions remain valid, with suitable modifications, for integrals in \(\R^n\).

  1. Suppose that \(R\) is a rectangle in \(\R^2\).

  2. Suppose that \(R\) is a rectangle and \(f:R\to \R\) is a function such that \(f(\mathbf x)\ge 0\) for all \(\mathbf x\in R\), and in addition \[ \inf\left\{ \iint_R g \, dA : g(\mathbf x)\ge f(\mathbf x)\text{ for all }\mathbf x\in R\right\} =0 \] then \(f\) is integrable, and \(\iint_R f\, dA=0\).

Hint The previous question may be helpful.
  1. Let \(S\) be a bounded subset of \(\R^2\). Prove that \[ S\text{ has zero content } \quad \iff\qquad \iint_S 1 \, dA = 0 . \] Of course \(\iint_S 1\ dA\) can also be written as \(\iint_R \chi_S \, dA\), where \(R\) is any rectangle such that \(S\subseteq R\).
HintThe previous question may be helpful.
  1. 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.
Another HintAlso, what can we say about the difference between \(f\) and \(\chi_{R_1} f + \chi_{R_2} f\)?
  1. Suppose that \(S, U_1\) and \(U_2\) are compact subsets of \(\R^2\) such that \[\begin{equation} \label{S1S2} S = U_1 \cup U_2, \qquad U_1\cap U_2 \text{ has zero content}. \end{equation}\] Suppose also that \(f\) is a function that is integrable on \(U_1\) and \(U_2\). Prove that \(f\) is integrable on \(S\), and that \[\begin{equation}\label{additive} \iint_S f\, dA = \iint_{U_1}f\, dA + \iint_{U_2} f\, dA. \end{equation}\]
HintThis 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 \(U_1, U_2\) that satisfy \(\eqref{S1S2}\), then computing the integral over each piece and \(\eqref{additive}\). See the next section for an example. The last exercise above shows that this procedure is justified.

\(\Leftarrow\)  \(\Uparrow\)  \(\Rightarrow\)

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