Multiple integral on a bounded closed set

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 $\epsilon >0$, 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 $$\sum _{i=1}^r(b_i-a_i)(d_i-c_i)<\epsilon .$$

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,

$$\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }x,y\in D[\Vert x-y\Vert <\delta ⟹|f(x)-f(y)|<\epsilon ].$$

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

$$\mathrm{\forall }x\in D\mathrm{\forall }\epsilon >0\mathrm{\exists }\delta >0\mathrm{\forall }y\in D[\Vert x-y\Vert <\delta ⟹|f(x)-f(y)|<\epsilon ].$$

Here, the value of $\delta$ may depend on $x$ (i.e., $\delta$ can be a function of $x$: $\delta =\delta (x)$) as $\mathrm{\forall }x\in D$ comes before $\mathrm{\exists }\delta >0$. In other words, how close $x$ and $y$ to each other should be for $|f(x)-f(y)|<\epsilon$ 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 $|f(x)-f(y)|<\epsilon$ wherever the $x$ is in $D$. □

Lemma

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

$$\mathrm{\Gamma }=\{(x,y)\mid y=\phi (x),a\le x\le b\}$$

is a null set.

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

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

$$\begin{array}{rcl}{M}_i& =& max\{\phi (x)\mid x\in [a_i,a_{i+1}]\},\\ m_i& =& min\{\phi (x)\mid x\in [a_i,a_{i+1}]\}.\end{array}$$

Then,

$$M_i-m_i<\frac{\epsilon }{b-a}.$$

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,

$$\mathrm{\Gamma }\subset L_0\cup L_1\cup \cdots \cup L_{r-1}.$$

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

$$\sum _{i=0}^{r-1}(a_{i+1}-a_i)(M_i-m_i)<\sum _{i=0}^{r-1}(a_{i+1}-a_i)\frac{\epsilon }{b-a}=\epsilon .$$

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 $\epsilon$. Thus, $\mathrm{\Gamma }$ 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 $\epsilon >0$, there exists a partition $\mathrm{\Delta }$ of $D$ such that $$S_{\mathrm{\Delta }}-s_{\mathrm{\Delta }}<\epsilon$$ where $S_{\mathrm{\Delta }}$ and $s_{\mathrm{\Delta }}$ are the upper and lower Riemann sums, respectively, of the function $f(x,y)$ with respect to $\mathrm{\Delta }$.

We prove this in three steps.

Step 1. 

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

$$\begin{array}{}\text{(Eq:1M)}& |f(x,y)|<M\end{array}$$

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

$$\begin{array}{}\text{(Eq:muL)}& \sum _{k=1}^r\mu (L_k)<\frac{\epsilon }{4M}\end{array}$$

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

$$D^{\prime }=D\setminus \left(\bigcup _{k=1}^r{L}_k\right).$$

Then, $D^{\prime }$ 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, $D\setminus (\bigcup _k{L}_k)=D\cap (\bigcup _k{L}_k{)}^c$ is closed. □

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

$$\begin{array}{}\text{(Eq:fD)}& |f(x,y)-f(x^{\prime },y^{\prime })|<\frac{\epsilon }{2\mu (D)}\end{array}$$

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

Step 3.

Let $\mathrm{\Delta }=\{D_{ij}\}$ 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 $\mathrm{\Delta }$.

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

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 $\overline{S}$ 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

$$S_{\mathrm{\Delta }}-s_{\mathrm{\Delta }}=\sum _{i=0}^{n-1}\sum _{j=0}^{m-1}(M_{ij}-m_{ij})\mu (D_{ij})$$

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

• $S_1$ 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 $M_{ij}-m_{ij}\le 2M$ (this $M$ is the one in (Eq:1M)). For the terms in $S_1$, we have $\mu (D_{ij})\le \mu (L_k)$ for some $L_k$. Using (Eq:muL),

$$S_1\le 2M\sum _{k=1}^r\mu (L_k)<2M\cdot \frac{\epsilon }{4M}=\frac{\epsilon }{2}.$$

By (Eq:fD), $M_{ij}-m_{ij}<\frac{\epsilon }{2\mu (D)}$ so that

$$S_2<\frac{\epsilon }{2\mu (D)}\cdot \mu (D^{\prime })\le \frac{\epsilon }{2\mu (D)}\cdot \mu (D)=\frac{\epsilon }{2}.$$

Therefore,

$$S_{\mathrm{\Delta }}-s_{\mathrm{\Delta }}=S_1+S_2<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$$

■

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 $\varphi :[0,1]\to {\mathbb{R}}^2$ be a continuous map such that $\varphi (0)=\varphi (1)$ and the restriction to $[0,1)$ is injective. The image of $\varphi$ 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.

• $D=\{(x,y)\mid x^2+y^2\le 1\}$.
• $D=\{(x,y)\mid x^2+y^2\le 1,y\ge 0\}$.

(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:

$$\tilde{f}(x,y)=\{\begin{array}{cc}f(x,y)& ((x,y)\in D),\\ 0& ((x,y)\notin D).\end{array}$$

$\tilde{f}(x,y)$ 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 $D\subset {\mathbb{R}}^3$:

$${\iiint }_{D}f(x,y,z)dxdydz.$$

More generally, we can define the multiple integral on a bounded closed region $D\subset {\mathbb{R}}^n$:

$$\int \cdots {\int }_{D}f(x_1,\cdots ,x_n)d{x}_1\cdots d{x}_n.$$