Multiple integral on a bounded closed set

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In a previous post, we defined the multiple integral on a (closed) rectangular region and saw that continuous functions were integrable on such regions. 

But more functions are integrable. Some functions are not necessarily defined on a rectangular region. Yet other functions may not be continuous everywhere. To describe a wider class of integrable functions, we define the notion of the "null" sets.



Null sets

Definition (Null set)

The subset \(A\) of \(\mathbb{R}^2\) is said to be a null set if the following condition is satisfied:

  • For all \(\varepsilon > 0\), there exist a finite number of open rectangular regions \[L_i = (a_i, b_i)\times (c_i, d_i) ~(a_i < b_i, c_i < d_i, i = 1, 2, \cdots, r)\] such that \[A \subset \bigcup_{i=1}^{r}L_i\] and \[\sum_{i=1}^{r}(b_i - a_i)(d_i - c_i) < \varepsilon.\]

This means that if a set is a null set if it can be "covered" by a finite number of open rectangular regions, and the total area of these rectangular regions is arbitrarily small.

Example. Let \(A = \{(a,b) \}\) where \((a,b) \in \mathbb{R}^2\) is a point. Then \(A\) is a null set. □

In general, if \(A_1, A_2, \cdots, A_N\) are null sets, then their union \(\bigcup_{i=1}^{N}A_i = A_1 \cup A_2 \cup \cdots \cup A_N\) is also a null set. In particular, a set of finitely many points of \(\mathbb{R}^2\) is a null set. Furthermore, the line segment \(PQ\) connecting the two points \(P\) and \(Q\) is a null set. We can also show that the graph of a continuous function on a closed interval is also a null set. But before proving it, we list the following theorem without proof.

Theorem (Th:Unif)

Let \(D\) be a bounded closed set in \(\mathbb{R}^n\). Let \(f(x)\) be a continuous function on \(D\) where \(x = (x_1, x_2, \cdots, x_n)\). Then, \(f(x)\) is uniformly continuous on \(D\). That is,

\[\forall\varepsilon > 0 ~ \exists\delta > 0 ~ \forall x, y \in D ~ [\|x - y\| < \delta\implies |f(x) - f(y)| < \varepsilon].\]

Proof. See the Heine-Cantor theorem (Wikipedia). ■

Remark. This is a natural extension of the univariate version of the theorem. □

Remark. Note the difference between ``continuous'' and ``uniformly continuous.'' The function \(f(x)\) is said to be continuous on \(D\) if

\[\forall x\in D ~ \forall\varepsilon > 0 ~ \exists\delta > 0 ~ \forall y\in D ~ [\|x - y\| < \delta \implies |f(x) - f(y)| < \varepsilon].\]

Here, the value of \(\delta\) may depend on \(x\) (i.e., \(\delta\) can be a function of \(x\): \(\delta = \delta(x)\)) as \(\forall x \in D\) comes before \(\exists \delta > 0\). In other words, how close \(x\) and \(y\) to each other should be for \(|f(x)-f(y)|<\varepsilon\) to be satisfied may depend on where the \(x\) is. That is not the case if the function is uniformly continuous: the same value of \(\delta\) (the closeness between \(x\) and \(y\)) can be used to satisfy \(|f(x) - f(y)|< \varepsilon\) wherever the \(x\) is in \(D\). □

Lemma

Let \(y = \varphi(x)\) be a continuous function on the bounded closed interval \([a,b]\). Then, its graph

\[\Gamma = \{(x,y) \mid y = \varphi(x), a \leq x \leq b\}\]

is a null set.

Proof. Since \(\varphi(x)\) is a continuous function on a bounded closed interval \([a,b]\), it is uniformly continuous (see Theorem (Th:Unif) above). Thus, for all \(\varepsilon > 0\), there exists some \(\delta > 0\) such that, for all \(x, x' \in [a,b]\), \(|x - x'| < \delta\) implies \(|\varphi(x) - \varphi(x')| < \frac{\varepsilon}{b - a}\).

Now, make a partition of \([a,b]\), \(a = a_0 < a_1 < \cdots < a_r = b\), in such a way that \(|a_{i+1} - a_{i}| < \delta\) for all \(i = 0, 1, \cdots, r-1\). Let us define

\[\begin{eqnarray*} M_i &=& \max\{\varphi(x) \mid x \in [a_i, a_{i+1}]\},\\ m_i &=& \min\{\varphi(x) \mid x \in [a_i, a_{i+1}]\}. \end{eqnarray*}\]

Then,

\[M_i - m_i < \frac{\varepsilon}{b - a}.\]

Next, define open multi-intervals \(L_i\) for \(i = 0, 1, \cdots, r-1\) by

\[L_i = (a_i - \rho, a_{i+1}+\rho)\times(m_i - \rho, M_i + \rho)\]

where \(\rho\) is a positive constant. \(L_i\) "covers" the rectangular region \(D_i = [a_i, a_{i+1}]\times [m_i, M_i]\) and its area converges to the area of \(D_i\) as \(\rho \to 0\). Furthermore,

\[\Gamma \subset L_0\cup L_1 \cup \cdots \cup L_{r-1}.\]

(Draw pictures!) The area of \(D_i\) is given by \((a_{i+1}-a_{i})(M_i-m_i)\), and we have

\[\sum_{i=0}^{r-1}(a_{i+1}-a_{i})(M_i-m_i) <\sum_{i=0}^{r-1}(a_{i+1}-a_{i})\frac{\varepsilon}{b-a} = \varepsilon.\]

Thus, if we take a sufficiently small \(\rho\), the sum of the areas of \(L_i ~ (i = 0, 1, \cdots, r-1)\) is less than \(\varepsilon\). Thus, \(\Gamma\) is a null set. ■

We have seen that continuous functions (on a rectangular region) are Riemann-integrable. This condition of continuity can be relaxed as follows.

Theorem (Integrability theorem)

Let \(f(x,y)\) be a bounded function on the rectangular region \(D = [a,b]\times [c,d]\). Let \(A \subset D\) be a null set. If \(f(x,y)\) is continuous at every \((x,y) \in D\setminus A\), then \(f(x,y)\) is Riemann-integrable on \(D\).

Remark. In other words, a function \(f(x,y)\) on \(D\) is Riemann-integrable if it is continuous on \(D\) except for a null set (\(A\)). It does not have to be continuous everywhere on \(D\). □

Proof. It suffices to show the following:

  • For all \(\varepsilon > 0\), there exists a partition \(\Delta\) of \(D\) such that \[S_{\Delta} - s_{\Delta} < \varepsilon\] where \(S_{\Delta}\) and \(s_{\Delta}\) are the upper and lower Riemann sums, respectively, of the function \(f(x,y)\) with respect to \(\Delta\).

We prove this in three steps.

Step 1. 

Since \(f(x,y)\) is bounded on \(D\), there exists a real number \(M\) such that

\[|f(x,y)| < M  \tag{Eq:1M}\]

for all \((x,y) \in D\). Since \(A\) is a null set, there exist open multi-intervals \(L_1, L_2, \cdots, L_r\) such that \(A \subset \bigcup_{k=1}^{r}L_k\) and

\[\sum_{k=1}^{r}\mu(L_k) < \frac{\varepsilon}{4M}\tag{Eq:muL}\]

where \(\mu(L_k) = (b_k - a_k)(d_k - c_k)\) is the area of the open multi-interval \(L_k = (a_k, b_k)\times(c_k, d_k)\). 

Step 2.

Let us define

\[D' = D \setminus \left(\bigcup_{k=1}^{r}L_k\right).\]

Then, \(D'\) is a closed and bounded set.

Remark. Each \(L_k\) is open. The union of open sets is open. The complement of an open set is closed (by definition). Thus, \((\bigcup_{k}L_k)^c\) is closed. \(D\) is closed (by assumption). The intersection of closed sets is closed (why?). Thus, \(D\setminus(\bigcup_kL_k) = D \cap (\bigcup_kL_k)^c\) is closed. □

\(f(x,y)\) is continuous on \(D'\), and hence, it is uniformly continuous on \(D'\) (Theorem (Th:Unif)). Therefore, there exists a \(\delta > 0\) such that, for all \((x, y), (x',y') \in D'\), if \(d((x,y), (x',y')) < \delta\), then

\[|f(x,y) - f(x',y')| < \frac{\varepsilon}{2\mu(D)}\tag{Eq:fD}\]

where \(\mu(D) = (b-a)(d-c)\) is the area of \(D\).

Step 3.

Let \(\Delta = \{D_{ij}\}\) be a partition of \(D\) that is sufficiently fine so that the distance between any two points in any \(D_{ij}\) is less than \(\delta\). Furthermore, we also include \(a_k, b_k\) and \(c_k, d_k\) in the above open multi-intervals \(L_k = (a_k, b_k)\times(c_k, d_k)\) as partitioning points of \(\Delta\).

Then, (the closure of) each \(L_k\) is a finite union of small rectangles in \(\Delta\).

Remark. The "closure" of \(L_k = (a_k,b_k)\times(c_k, d_k)\) is the closed multi-interval \([a_k,b_k]\times[c_k,d_k]\). In general, the closure \(\bar{S}\) of an open set \(S\) is the union of \(S\) and its boundary (which we often denote by \(\partial S\)). □

Using the same notation as when we defined Riemann sums, we have

\[S_{\Delta} - s_{\Delta} = \sum_{i=0}^{n-1}\sum_{j=0}^{m-1}(M_{ij} - m_{ij})\mu(D_{ij})\]

where \(\mu(D_{ij})\) is the area of \(D_{ij}\). We split the sum on the right-hand side into two parts: \(S_{\Delta} - s_{\Delta} = S_1 + S_2\).

  • \(S_1\) is the sum of \((M_{ij} - m_{ij})\mu(D_{ij})\) over all index pairs \((i,j)\) such that \((D_{ij})^{o} \subset L_k\) for some \(L_k\) where \((D_{ij})^o\) is the interior of \(D_{ij}\) (i.e., \(D_{ij}\) except for the boundary).
  • \(S_2\) is all the remaining terms.
Note that \(M_{ij} - m_{ij} \leq 2M\) (this \(M\) is the one in (Eq:1M)). For the terms in \(S_1\), we have \(\mu(D_{ij}) \leq \mu(L_k)\) for some \(L_k\). Using (Eq:muL),
\[S_1 \leq 2M\sum_{k=1}^{r}\mu(L_k) < 2M\cdot \frac{\varepsilon}{4M} = \frac{\varepsilon}{2}.\]
By (Eq:fD), \(M_{ij} - m_{ij} < \frac{\varepsilon}{2\mu(D)}\) so that
\[S_2 < \frac{\varepsilon}{2\mu(D)}\cdot\mu(D')\leq \frac{\varepsilon}{2\mu(D)}\cdot \mu(D) = \frac{\varepsilon}{2}.\]
Therefore,
\[S_{\Delta} - s_{\Delta} = S_1 + S_2 < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\]

■

A closed region bounded by a curve

We now try to define the integral of functions that are not necessarily defined on a rectangular region.

Definition (Jordan curve)

Let \(\phi: [0,1] \to \mathbb{R}^2\) be a continuous map such that \(\phi(0) = \phi(1)\) and the restriction to \([0, 1)\) is injective. The image of \(\phi\) is called a Jordan curve.
Remark. In other words, a Jordan curve is a continuous curve in \(\mathbb{R}^2\) that is non-self-crossing and closed. □

We give the following seemingly trivial theorem without proof (it's not easy to prove).

Theorem (Jordan curve theorem)

A Jordan curve \(C\) divides the plane \(\mathbb{R}^2\) into its interior and exterior.

See also: Jordan curve theorem (Wikipedia).

The union of the interior and the curve \(C\) defines a bounded and closed region. This region is called a closed Jordan region, and the Jordan curve is its boundary. The boundary of a closed Jordan region is a null set.

Example. The following are typical closed Jordan regions.
  • \(D = \{(x,y)\mid x^2 + y^2 \leq 1\}\).
  • \(D = \{(x,y)\mid x^2 + y^2 \leq 1, y \geq 0 \}\).
(Draw pictures!) □

Theorem (A continuous function on a bounded closed region is integrable)

Let \(f(x,y)\) be a continuous function on a bounded closed region \(D\). Then, \(f(x,y)\) is Riemann-integrable.
Proof. Consider a rectangular region \(R = [a,b]\times [c,d]\) such that \(D \subset R\). Let us define a bounded function \(\tilde{f}(x,y)\) on \(R\) as follows:
\[\tilde{f}(x,y) = \left\{ \begin{array}{cc} f(x,y) & ((x,y) \in D),\\ 0 & ((x,y) \not\in D). \end{array} \right.\]
\(\tilde{f}(x,y)\) is continuous on \(R\setminus\partial D\) (where \(\partial D\) is the boundary of \(D\)). Note that \(\partial D\) is a null set. By the Integrability Theorem (see above), \(\tilde{f}(x,y)\) is Riemann-integrable on \(R\), and
\[\iint_{D}f(x,y)dxdy = \iint_R\tilde{f}(x,y)dxdy.\]
■

This theorem can be naturally extended to higher dimensions. We can define the triple integral on a bounded closed region \(D \subset \mathbb{R}^3\):
\[\iiint_Df(x,y,z)dxdydz.\]
More generally, we can define the multiple integral on a bounded closed region \(D \subset \mathbb{R}^n\):
\[\idotsint_Df(x_1, \cdots, x_n)dx_1\cdots dx_n.\]



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