The Riemann Integrals

Emile Munier (1840-1895) was a French painter.

We have sections

• Upper and Lower Riemann Integrals. 
• Existence of Riemann Integrals
• Multiple Riemann Integrals

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Upper and Lower Riemann Integrals

Given a closed interval \([a,b]\), a set of points \({\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}\) satisfying the following inequalities
\[a=x_{0}<x_{1}<x_{2}<\cdots <x_{n-1}<x_{n}=b\]
is called a partition of \([a,b]\). The interval \([x_{k-1},x_{k}]\) is called the \(k\)th subinterval of \([a,b]\). We also write \(\Delta x_{k}=x_{k}-x_{k-1}\). In this case, we have
\[\sum_{k=1}^{n}\Delta x_{k}=b-a.\]
The collection of all possible partitions of \([a,b]\) will be denoted by \(\mathfrak{P}[a,b]\).

Let \({\cal P}\) be a partition of the closed interval \([a,b]\). Another partition \({\cal P}’\) of \([a,b]\) is said to be finer than \({\cal P}\) when \({\cal P}\subseteq {\cal P}’\). The norm of a partition \({\cal P}\) is the length of the largest sub-interval of \([a,b]\), and is denoted by \(\parallel {\cal P}\parallel\). For example, let \({\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}\). Then, we have
\[\parallel{\cal P}\parallel=\max_{k=1,\cdots,n}\left |x_{k}-x_{k-1}\right |.\]
It is clear to see
\[{\cal P}\subseteq {\cal P}’\mbox{ implies }\parallel {\cal P}’\parallel\leq\parallel {\cal P}\parallel;\]
that is, the refinement of a partition decreases its norm.

We need the concepts of infimum and supremum by referring to the page Infimum and Supremum. Let \(f\) be a bounded real-valued function defined on the closed interval \([a,b]\). Give any partition \({\cal P}\in\mathfrak{P}[a,b]\), the upper Riemann sum of \(f\) is defined by
\begin{equation}{\label{ma25}}\tag{1}
U({\cal P},f)=\sum_{i=1}^{n}(x_{i}-x_{i-1})M_{i},\mbox{ where }M_{i}=\sup_{x_{i-1}\leq x\leq x_{i}}f(x),
\end{equation}
and the lower Riemann sum of \(f\) is defined by
\begin{equation}{\label{ma27}}\tag{2}
L({\cal P},f)=\sum_{i=1}^{n}(x_{i}-x_{i-1})m_{i},\mbox{ where }m_{i}=\inf_{x_{i-1}\leq x\leq x_{i}}f(x).
\end{equation}

The upper Riemann integral of \(f\) is defined by
\[\overline{\int}_{a}^{b}f(x)dx=\inf_{{\cal P}\in\mathfrak{P}[a,b]} U({\cal P},f),\]
and the {\bf lower Riemann integral} of \(f\) is defined by
\begin{equation}{\label{ma28}}\tag{3}
\underline{\int}_{a}^{b}f(x)dx=\sup_{{\cal P}\in\mathfrak{P}[a,b]} L({\cal P},f).
\end{equation}
We always have the following inequality
\[\underline{\int}_{a}^{b}f(x)dx\leq\overline{\int}_{a}^{b}f(x)dx.\]

\begin{equation}{\label{ma1}}\tag{4}\mbox{}\end{equation}

Definition \ref{ma1}. Let \(f\) be a real-valued functions defined on the closed interval \([a,b]\). We say that \(f\) is Riemann-integrable on \([a,b]\) when
\[\underline{\int}_{a}^{b}f(x)dx=\overline{\int}_{a}^{b}f(x)dx.\]
In this case, it is denoted by
\[\int_{a}^{b}f(x)dx\]

Another definition of Riemann integral is given below.

\begin{equation}{\label{ma2}}\tag{5}\mbox{}\end{equation}

Definition \ref{ma2}. Let \(f\) be a real-valued functions defined on the closed interval \([a,b]\). Let \({\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}\) be a partition of \([a,b]\), and let \(t_{k}\in [x_{k-1},x_{k}]\). The following sum
\[S({\cal P},f)=\sum_{k=1}^{n}f(t_{k})\Delta x_{k}\]
is called a Riemann sum of \(f\). We say that \(f\) is Riemann-integrable on \([a,b]\) when there exists a real number \(A\) such that, given any \(\epsilon >0\), there exists a partition \({\cal P}(\epsilon )\) of \([a,b]\) satisfying that, for any partition \({\cal P}\) that is finer than \({\cal P}(\epsilon )\) and for every choice \(t_{k}\in [x_{k-1},x_{k}]\), we have
\[\left |S({\cal P},f)-A\right |<\epsilon .\]

Definitions \ref{ma1} and \ref{ma2} are equivalent, which can be realized from Theorem \ref{mat62} that will be given below.

By a step function, we mean a function \(\psi\) which has the form
\[\psi (x)=c_{i}\mbox{ for }x_{i-1}<x<x_{i}\]
for some partition \({\cal P}\in\mathfrak{P}[a,b]\) and some set of constants \(\{c_{1},\cdots ,c_{n}\}\). Suppose that we define
\[\int_{a}^{b}\psi (x)dx=\sum_{i=1}^{n}c_{i}\left (x_{i}-x_{i-1}\right ).\]
Then, we have
\begin{equation}{\label{raeq240}}\tag{6}
\overline{\int}_{a}^{b}f(x)dx=\inf_{\{\psi:\psi\geq f\}}\int_{a}^{b}\psi (x)dx
\end{equation}
and
\begin{equation}{\label{raeq241}}\tag{7}
\underline{\int}_{a}^{b}f(x)dx=\sup_{\{\phi:\phi\leq f\}}\int_{a}^{b}\phi (x)dx.
\end{equation}

Example. We consider the function

\[f(x)=\left\{\begin{array}{ll}
0 & \mbox{if }x\in\mathbb{Q}\\
1 & \mbox{if }x\not\in\mathbb{Q}.
\end{array}\right .\]
Then, we have
\[\overline{\int}_{a}^{b}f(x)dx=b-a\]

and

\[\underline{\int}_{a}^{b}f(x)dx=0.\]
This says that the function \(f\) is not Riemann-integrable. \(\sharp\)

Example. Given \(0<a<b\) and a real-valued function \(f\) defined on \([a,b]\) by

\[f(x)=\left\{\begin{array}{ll}
0 & \mbox{if \(x\in [a,b]\cap\mathbb{Q}\)}\\
x & \mbox{if \(x\in [a,b]\) and \(x\not\in\mathbb{Q}\)},
\end{array}\right .\]
we are going to evaluate the upper and lower Riemann integrals of \(f\) on \([a,b]\). Given a partition \({\cal P}=\{a=x_{0},x_{1},\cdots,x_{n}=b\}\), since \(\mathbb{Q}\) is dense in \(\mathbb{R}\), we have
\[M_{i}=\sup_{x_{i-1}\leq x\leq x_{i}}f(x)=x_{i}\]

and
\[m_{i}=\inf_{x_{i-1}\leq x\leq x_{i}}f(x)=0.\]
Therefore, we have \(L({\cal P},f)=0\) and
\[U({\cal P},f)=\sum_{i=1}^{n}(x_{i}-x_{i-1})x_{i}.\]
We consider function \(h:[a,b]\rightarrow\mathbb{R}\) defined by \(h(x)=x\). Then, we have
\[U({\cal P},f)=U({\cal P},h).\]
Therefore, we obtain
\[\underline{\int}_{a}^{b}f(x)dx=\sup_{{\cal P}\in\mathfrak{P}[a,b]}L({\cal P},f)=0\]
and
\begin{align*}
\overline{\int}_{a}^{b}f(x)dx & =\inf_{{\cal P}\in\mathfrak{P}[a,b]}U({\cal P},f)=\inf_{{\cal P}\in\mathfrak{P}[a,b]} U({\cal P},h)\\
& =\int_{a}^{b}h(x)dx\mbox{ (since \(h\) is Riemann-integrable on \([a,b]\))}\\
& =\int_{a}^{b}xdx=\frac{b^{2}-a^{2}}{2}.
\end{align*}
This also says that \(f\) is not Riemann-integrable on \([a,b]\). \(\sharp\)

Proposition. Let \(f\) be a Riemann-integrable function defined on \([a,b]\), and let \(g\) be a real-valued function defined on \([a,b]\). Suppose that the following set
\[\left\{x\in [a,b]:f(x)\neq g(x)\right\}\]
is finite. Then \(g\) is Riemann-integrable on \([a,b]\) satisfying
\[\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx.\]

Proof. We consider the function \(h=g-f\) on \([a,b]\). Then the following set
\[H=\left\{x\in [a,b]:h(x)\neq 0\right\}\]
is finite. We assume that \(H=\{x_{1},\cdots,x_{n}\) satisfying
\[a\leq x_{1}<x_{2}\cdots<x_{n-1}\leq x_{n}=b.\]
Let
\[M=\max\left\{|h(x_{1})|,\cdots,|h(x_{n})|\right\}.\]
Then \(M>0\). Given any \(\epsilon >0\), we can take a sufficiently small \(\delta>0\) satisfying
\[a<x_{1}-\delta,x_{1}+\delta<x_{2}-\delta,\cdots,x_{i}+\delta<x_{i+1}-\delta,\cdots,x_{n}+\delta<b\]

and

\[\delta<\frac{\epsilon}{4nM}.\]
We consider the partition
\[{\cal P}=\left\{a,x_{1}-\delta,x_{1}+\delta,x_{2}-\delta,x_{2}+\delta,\cdots,x_{i}+\delta,x_{i+1}-\delta,\cdots,x_{n}+\delta,b\right\}.\]
Since \(|h(x)|\leq M\) for all \(x\in [a,b]\), we have
\[\sup_{a\leq x\leq x_{1}-\delta}|f(x)|\leq M\]

and
\[\sup_{x_{n}+\delta\leq x\leq b}|f(x)|\leq M.\]
We also have
\[\sup_{x_{i}-\delta\leq x\leq x_{i}+\delta}|f(x)|\leq M\]

and
\[\sup_{x_{i}+\delta\leq x\leq x_{i+1}-\delta}|f(x)|\leq M\]

for \(i=1,\cdots,n\), which implies
\[\left |U({\cal P},f)\right |\leq 2n\cdot 2\delta\cdot M<\epsilon.\]
Since \(|L({\cal P},f)|\leq |U({\cal P},f)|\), we also have \(|L({\cal P},f)|<\epsilon\). Since \(\epsilon\) can be any positive number, we have
\[\overline{\int}_{a}^{b}h(x)dx=\inf_{{\cal P}\in\mathfrak{P}[a,b]} U({\cal P},h)=0=
\underline{\int}_{a}^{b}h(x)dx=\sup_{{\cal P}\in\mathfrak{P}[a,b]} L({\cal P},h),\]
which also says that \(h\) is Riemann-integrable on \([a,b]\) with \(\int_{a}^{b}h(x)dx=0\). Since \(g=f+h\), it follows
\[\int_{a}^{b}g(x)dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}h(x)dx=\int_{a}^{b}f(x)dx.\]
This completes the proof. \(\blacksquare\)

Theorem.  (Riemann-Lebesgue’s Lemma). Let \(f\) be a real-valued function defined on the compact interval \([a,b]\). Then, we have
\[\lim_{n\rightarrow\infty}\int_{a}^{b}f(x)\sin (nx)dx=0=\lim_{n\rightarrow\infty}\int_{a}^{b}f(x)\cos (nx)dx.\]

Proof. We first have
\[\int_{a}^{b}\sin (nx)dx=\left [-\frac{\cos (nx)}{n}\right ]_{a}^{b}=\frac{\cos (na)}{n}-\frac{\cos (nb)}{n},\]
which implies
\[\lim_{n\rightarrow\infty}\int_{a}^{b}\sin (nx)dx=\lim_{n\rightarrow\infty}\left (
\frac{\cos (na)}{n}-\frac{\cos (nb)}{n}\right )=0.\]
Let \(s\) be a step function defined on \([a,b]\). Then, the linearity of Riemann integral shows
\begin{equation}{\label{ma158}}\tag{8}
\lim_{n\rightarrow\infty}\int_{a}^{b}s(x)\sin (nx)dx=0.
\end{equation}
The continuity of \(f\) on \([a,b]\) also says that \(f\) is Riemann-integrable on \([a.b]\). It means
\[\underline{\int}_{a}^{b}f(x)dx=\int_{a}^{b}f(x)dx.\]
Therefore, given any \(\epsilon >0\), there exists a step function \(s\) satisfying \(s\leq f\) and
\[\int_{a}^{b}\left [f(x)-s(x)\right ]dx<\epsilon.\]
Now, we have
\begin{align*}
& \left |\int_{a}^{b}\left [f(x)\sin (nx)-s(x)\sin (nx)\right ]dx\right |\leq\int_{a}^{b}\left |f(x)\sin (nx)-s(x)\sin (nx)\right |dx\\
& \quad=\int_{a}^{b}\left [f(x)-s(x)\right ]\left |\sin (nx)\right |dx\leq\int_{a}^{b}\left [f(x)-s(x)\right ]<\epsilon,
\end{align*}
which implies
\begin{align*}
\left |\int_{a}^{b}f(x)\sin (nx)dx\right | & \leq\left |\int_{a}^{b}\left [f(x)\sin (nx)-s(x)\sin (nx)\right ]dx\right |
+\left |\int_{a}^{b}s(x)\sin (nx)dx\right |\\
& <\epsilon+\left |\int_{a}^{b}s(x)\sin (nx)dx\right |.
\end{align*}
Using (\ref{ma158}), since \(\epsilon\) can be any positive number, it follows
\[\lim_{n\rightarrow\infty}\int_{a}^{b}f(x)\sin (nx)dx=0.\]
We can similarly obtain
\[\lim_{n\rightarrow\infty}\int_{a}^{b}f(x)\cos (nx)dx=0.\]
This completes the proof. \(\blacksquare\)

Theorem. (Cauchy Criterion). Let the real-valued function \(f:[a,\infty)\rightarrow\mathbb{R}\) be Riemann-integrable on any compact intervals. Then, the improper Riemann integral \(\int_{a}^{\infty}f(x)dx\) is convergent if and only if, given any \(\epsilon>0\), there exists a positive number \(A>a\) such that
\[t_{1},t_{2}>A\mbox{ implies }\left |\int_{t_{1}}^{t_{2}}f(x)dx\right |<\epsilon.\]

Proof. Suppose that the improper Riemann integral is convergent. Then, we have
\[\int_{a}^{\infty}f(x)dx=\lim_{t\rightarrow\infty}\int_{a}^{t}f(x)dx\equiv\lim_{t\rightarrow\infty}F(t),\]
where
\[F(t)=\int_{a}^{t}f(x)dx.\]
The limit says that, given any \(\epsilon>0\), there exists a positive number \(A\) such that
\[t>A\mbox{ implies }\left |F(t)-\int_{a}^{\infty}f(x)dx\right |<\frac{\epsilon}{2}.\]
Therefore, for any \(t_{1},t_{2}>A\), we have
\begin{align*}
\left |\int_{t_{1}}^{t_{2}}f(x)dx\right | & =\left |F(t_{1})-F(t_{2})\right |\\
& \leq\left |F(t_{1})-\int_{a}^{\infty}f(x)dx\right |+\left |\int_{a}^{\infty}f(x)dx-F(t_{2})\right |\\
& <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.
\end{align*}

To prove the converse, we first note that the limit \(\lim_{t\rightarrow\infty}F(t)\) exists when there exists a sequence \(\{t_{n}\}_{n=1}^{\infty}\) such that \(\lim_{t\rightarrow\infty}t_{n}=\infty\) and the sequence \(\{F(t_{n})\}_{n=1}^{\infty}\) is convergent. Using the completeness of \(\mathbb{R}\), we remain to show that \(\{F(t_{n})\}_{n=1}^{\infty}\) is a Cauchy sequence. Given any \(\epsilon>0\), the hypothesis says that there exists a positive number \(A>a\) such that \(\alpha,\beta>A\) implies
\[\left |F(\alpha)-F(\beta)\right |=\left |\int_{\alpha}^{\beta}f(x)dx\right |<\epsilon.\]
Since \(\lim_{t\rightarrow\infty}t_{n}=\infty\), there exists an integer \(N\) such that \(t_{n},t_{m}>A\) for \(n,m\geq N\), which implies
\[\left |F(t_{n})-F(t_{m})\right |=\left |\int_{t_{n}}^{t_{m}}f(x)dx\right |<\epsilon.\]
This shows that \(\{F(t_{n})\}_{n=1}^{\infty}\) is a Cauchy sequence, and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Lebesgue’s Criterion for the Existence of Riemann Integrals.

Since every continuous functions is Riemann-integrable, it is possible to find some Riemann-integrable functions that are not continuous, For example, we can take \(f\) to be monotonic function with a countable set of discontinuities. The following example shows that there are Riemann-integrable functions whose discontinuities are uncountable.

\begin{equation}{\label{maex146}}\tag{9}\mbox{}\end{equation}
Example \ref{maex146}. Let \(I=[0,1]\), and let \(A_{1}=I\setminus (\frac{1}{3},\frac{2}{3})\) be the subset of \(I\) obtained by removing those points which lie in the open middle third sub-interval of \(I\). Let \(A_{2}\) be the subset of \(A_{1}\) obtained by removing the open middle third subintervals of \([0,\frac{1}{3}]\) and \([\frac{2}{3},1]\). We continue this process and define \(A_{3},A_{4},\cdots\). The set \(C=\bigcap_{n=1}^{\infty}A_{n}\) is called the Cantor set. Then, we have the following results.

• \(C\) is a compact set in \(\mathbb{R}\), i.e., a closed and bounded set in \(\mathbb{R}\). It is clear that \(C\) is bounded, since \(C\) is a subset of \([0,1]\). Since each \(A_{n}\) is an union finitely many closed intervals, it means that each \(A_{n}\) is a closed set. Therefore, we conclude that \(C\) is also a closed set, since it is a countable intersection of closed sets.
• The total length of removed intervals is equal to \(1\). To see this, by summing up the length of removed intervals is given by
  \begin{align*} \frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\cdots & =\frac{1}{3}\left [1+\frac{2}{3}+\left (
  \frac{2}{3}\right )^{2}+\cdots\right ]\\ & =\frac{1}{3}\frac{1}{1-\frac{2}{3}}=1.\end{align*}
  This also says that the “total length” of Cantor set \(C\) is zero. In other words, the measure of Cantor set is zero.
• \(x\in C\) if and only if
  \[x=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}},\]
  where each \(a_{n}\) is either \(0\) or \(2\).
• \(C\) is an uncountable set.
• We define the function
  \[f(x)=\left\{\begin{array}{ll}
  1 & \mbox{if \(x\in C\)}\\
  0 & \mbox{if \(x\not\in C\).}
  \end{array}\right .\]
  Since the set of discontinuities of \(f\) is the Cantor set and the Cantor set has measure zero, it follows that \(f\) is Riemann-integrable on \([0,1]\). \(\sharp\)

It is natural to ask “how many” discontinuities a function can be Riemann-integrable. This problem was discovered by Lebesgue using the concept of measure zero. The difference between the upper and lower Riemann sums is given by
\begin{equation}{\label{maeq143}}\tag{10}
\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta x_{k}.
\end{equation}
Therefore, roughly speaking, the function \(f\) will be Riemann-integrable if and only if this sum can be made as small as possible. We split the sum (\ref{maeq143}) into two parts as \(S_{1}+S_{2}\), where \(S_{1}\) comes from sub-intervals containing only points of continuity of \(f\), and \(S_{2}\) contains the remaining terms. In \(S_{1}\), each difference \(M_{k}(f)-m_{k}(f)\) can be as small as possible because of continuity. However, in \(S_{2}\), the differences \(M_{k}(f)-m_{k}(f)\) may not be small. Suppose that \(f\) is bounded by \(M\). Then, we have
\[|S_{2}|\leq M\sum_{k=1}^{n}\Delta x_{k}\]
such that \(S_{2}\) will be small if the sum of lengths of the subintervals corresponding to \(S_{2}\) can be sufficiently small. Therefore, we may expect that the set of discontinuities of an integrable function can be covered by intervals whose total length can be sufficiently small. This is the idea in Lebesgue’s criterion.

\begin{equation}{\label{ma23}}\tag{11}\mbox{}\end{equation}

Definition. \ref{ma23}. A set \(S\) of \(\mathbb{R}\) is said to have measure zero when, for every \(\epsilon >0\), there is a countable covering of \(S\) by open intervals such that the sum of whose lengths is less than \(\epsilon\). More precisely, suppose that the intervals are denoted by \((a_{k},b_{k})\) for \(k=1,2,\cdots\). Then, we have
\[S\subseteq\bigcup_{k=1}^{\infty}(a_{k},b_{k})\mbox{ and }\sum_{k=1}^{\infty}(b_{k}-a_{k})<\epsilon .\]

\begin{equation}{\label{map145}}\tag{12}\mbox{}\end{equation}
Proposition \ref{map145}. Let \(\{F_{1},F_{2},\cdots ,\}\) be a countable collection of subsets of \(\mathbb{R}\) such that each \(F_{k}\) has measure zero for all \(k\). Then, their union
\[S=\bigcup_{k=1}^{\infty}F_{k}\]
has measure zero.

Proof. Given any \(\epsilon >0\), for each \(k\), there is a countable covering of \(F_{k}\) by open intervals such that the sum of whose length is less than \(\epsilon /2^{k}\). Therefore, the union of all these coverings is also a countable covering of \(S\) by open intervals, and the sum of the lengths is less than
\[\sum_{k=1}^{\infty}\frac{1}{2^{k}}\cdot\epsilon =\epsilon .\]
This completes the proof. \(\blacjsquare\)

Example. It is obvious that each set consists of one point in \(\mathbb{R}\) has measure zero. Therefore, Proposition \ref{map145} says that every countable subset of \(\mathbb{R}\) has measure zero. In particular, the set \(\mathbb{Q}\) of rational numbers has measure zero. However, the Cantor set defined in Example \ref{maex146} is an uncountable set having measure zero. \(\sharp\)

Definition. Let the real-valued function \(f\) be defined and bounded on an interval \(I\). Given \(J\subseteq I\), the number
\[\Omega_{f}(J)=\sup_{x,y\in J}\left [f(x)-f(y)\right ]\]
is called the oscillation of \(f\) on \(J\). The oscillation of \(f\) at \(x\) is defined to be the number
\begin{equation}{\label{maeq147}}\tag{13}
\omega_{f}(x)=\lim_{r\rightarrow 0+}\Omega_{f}(B(x;r)\cap I).
\end{equation}

It is clear to see that \(\Omega_{f}(B(x;r)\cap I)\) is a decreasing function of \(r\). Therefore, the limit (\ref{maeq147}) always exists given by
\[\omega_{f}(x)=\lim_{r\rightarrow 0+}\Omega_{f}(B(x;r)\cap I)=\inf_{r\rightarrow 0+}\Omega_{f}(B(x;r)\cap I).\]

\begin{equation}{\label{ma113}}\tag{14}\mbox{}\end{equation}

Remark \ref{ma113}. We have the following observations

• If \(J_{1}\subseteq J_{2}\), then \(\Omega_{f}(J_{1})\leq\Omega_{f}(J_{2})\).
• \(\omega_{f}(x)=0\) if and only if \(f\) is continuous at \(x\). \(\sharp\)

\begin{equation}{\label{map390}}\tag{15}\mbox{}\end{equation}
Proposition \ref{map390}. Let the real-valued function \(f\) be defined and bounded on \([a,b]\). Given any \(\epsilon >0\), we assume that \(\omega_{f}(x)<\epsilon\) for every \(x\in [a,b]\). Then, there exists \(\delta>0\) (which depends only on \(\epsilon\)) such that, for every closed sub-interval \(J\subseteq [a,b]\), we have \(\Omega_{f}(J)<\epsilon\) whenever the length of \(J\) is less than \(\delta\). \(\sharp\)

\begin{equation}{\label{map389}}\tag{16}\mbox{}\end{equation}
Proposition \ref{map389}. Let the real-valued function \(f\) be defined and bounded on \([a,b]\). Given any \(\epsilon >0\), we define the set \(J_{\epsilon}\) as
\[J_{\epsilon}=\left\{x\in [a,b]:\omega_{f}(x)\geq\epsilon\right\}.\]
Then \(J_{\epsilon}\) is a closed subset of \(\mathbb{R}\). \(\sharp\)

\begin{equation}{\label{mat148}}\tag{17}\mbox{}\end{equation}
Theorem \ref{mat148}. (Lebesgue’s Criterion for Riemann Integrability). Let the real-valued function \(f\) be defined and bounded on \([a,b]\), and let \(D\) denote the set of discontinuities of \(f\) on \([a,b]\). Then \(f\) is Riemann-integrable on \([a,b]\) if and only if \(D\) has measure zero.

Proof. To prove the necessity, we assume that \(D\) does not have measure zero and show that \(f\) is not integrable. From Remark \ref{ma113}, we have \(\omega_{f}(x)>0\) for \(x\in D\). Therefore, we can write \(D\) as a countable union of sets given by
\[D=\bigcup_{r=1}^{\infty}D_{r}\mbox{ with }D_{r}=\left\{x:\omega_{f}(x)\geq\frac{1}{r}\right\}.\]
Suppose that \(D\) does not have measure zero. Then, some set \(D_{r}\) does not have measure zero by Proposition \ref{map145}. Therefore, there exists \(\epsilon >0\) such that every countable collection of open intervals covering \(D_{r}\) has a sum of lengths that is larger than or equal to \(\epsilon\). Given any partition \({\cal P}\) of \([a,b]\), we have
\[U({\cal P},f)-L({\cal P},f)=\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta x_{k}=T_{1}+T_{2}\geq T_{1},\]
where \(T_{1}\) contains those terms coming from sub-intervals containing points of \(D\) in their interior, and \(T_{2}\) contains the remaining terms. The open intervals from \(T_{1}\) cover \(D_{r}\), so the sum of their lengths is larger than or equal to \(\epsilon\). Moreover, in these intervals, we have
\[M_{k}(f)-m_{k}(f)\geq\frac{1}{r},\]
which implies \(T_{1}\geq\epsilon /r\). This also means that
\[U({\cal P},f)-L({\cal P},f)\geq\frac{\epsilon}{r}\]
for every partition \({\cal P}\). Therefore, the Riemann’s condition is not satisfied, which says that \(f\) is not Riemann-integrable. In other words, if \(f\) is Riemann-integrable, then \(D\) has measure zero.

To prove the sufficiency, we assume that \(D\) has measure zero and show that the Riemann’s condition is satisfied. Since \(D_{r}\subseteq D\), each \(D_{r}\) has measure zero. Therefore, \(D_{r}\) can be covered by open intervals such that the sum of their lengths is less than \(1/r\). Since \(D_{r}\) is compact (i.e., closed and bounded) by Proposition \ref{map389}, a finite number of these intervals cover \(D_{r}\). The union of these open intervals is an open set which is
denoted by \(A_{r}\). The complement \(B_{r}=[a,b]\setminus A_{r}\) is the union of a finite number of closed intervals of \([a,b]\). Let \(I\) be a sub-interval of \(B_{r}\). For \(x\in I\), we have \(w_{f}(x)<1/r\). Therefore, using Proposition \ref{map390}, there exists \(\delta >0\) (depending only on \(r\)) such that \(I\) can be further subdivided into a finite number of sub-intervals \(T\) of length that is less than \(\delta\) satisfying \(\Omega_{f}(T)<1/r\). The endpoints of all these sub-intervals determine a partition \({\cal P}_{r}\) of \([a,b]\). Given \({\cal P}\) that is finer than \({\cal P}_{r}\), we can write
\[U({\cal P},f)-L({\cal P},f)=\sum_{k=1}^{n}\left [M_{k}(f)-m_{k}(f)\right ]\Delta x_{k}=T_{1}+T_{2},\]
where \(T_{1}\) contains those terms coming from sub-intervals containing points of \(D\), and \(T_{2}\) contains the remaining term. In the \(k\)th term of \(T_{2}\), we have
\[M_{k}(f)-m_{k}(f)<\frac{1}{r},\mbox{ which implie }T_{2}<\frac{b-a}{r}.\]
Since \(A_{r}\) covers all the intervals contributing to \(T_{1}\), we have \(T_{1}<(M-m)/r\), where \(M\) and \(m\) are the supremum and infimum of \(f\) on \([a,b]\). Therefore, we obtain
\[U({\cal P},f)-L({\cal P},f)<\frac{M-m+b-a}{r}.\]
Since this holds true for every \(r\geq 1\), we conclude that the Riemann’s condition is satisfied by taking \(r\rightarrow\infty\). This shows that \(f\) is Riemann-integrable on \([a,b]\), and the proof is complete. \(\blacksquare\)

A property is said to hold true almost everywhere on a subset \(S\) of \(\mathbb{R}\) when it holds true everywhere on \(S\) except for a set of measure zero. Theorem \ref{mat148} says that a bounded real-valued function \(f\) defined on a compact interval \(I\) is Riemann-integrable on \(I\) if and only if \(f\) is continuous almost everywhere on \(I\). We summarize the Riemann integrability as follows.

• Suppose that \(f\) is of bounded variation on \([a,b]\). Then \(f\) is Riemann-integrable on \([a,b]\)
• Suppose that \(f\) is Riemann-integrable on \([a,b]\). Then \(f\) is Riemann-integrable on \([c,d]\) for every subinterval \([c,d]\subseteq [a,b]\), \(|f|\) and \(f^{2}\) are Riemann-integrable on \([a,b]\).
• Suppose that \(f\) and \(g\) are Riemann-integrable functions defined on \([a,b]\). Then, the product \(f\cdot g\) is Riemann-integrable on \([a,b]\).
• Suppose that \(f\) and \(g\) are Riemann-integrable on \([a,b]\). Then \(f/g\) is Riemann-integrable on \([a,b]\) whenever \(g\) is bounded away from zero.
• Suppose that \(f\) and \(g\) are bounded real-valued functions having the same discontinuities on \([a,b]\). Then \(f\) is Riemann-integrable on \([a,b]\) if and only if \(g\) is Riemann-integrable on \([a,b]\).
• Suppose that the bounded real-valued function \(g\) is Reimann-integrable on \([a,b]\) with \(m\leq g(x)\leq M\) for all \(x\in [a,b]\), and that \(f\) is continuous on \([m,M]\). Then, the composite function \(f(g(x))\) is Riemann-integrable on \([a,b]\).

Example. Consider the function

\[f(x)=\left\{\begin{array}{ll}
0 & \mbox{if \(x=\frac{1}{n}\) for \(n=2,3,4,5\cdots\)}\\
1 & \mbox{if \(x\in [0,1]\) and \(x\neq\frac{1}{n}\) for \(n=2,3,4,5\cdots\)}.
\end{array}\right .\]
We are going to claim that \(f\) is continuous almost everywhere on \([0,1]\). Let
\[D=\left\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\cdots\right\}.\]
Then \(D\) is a countable set and has measure zero. We are going to show that \(f\) is continuous at \(x_{0}\not\in D\), i.e., \(x_{0}\neq\frac{1}{n}\) for \(n=2,3,4,5\cdots\). It is clear to see that there exists \(n_{0}\in\mathbb{N}\) satisfying
\[\frac{1}{n_{0}+1}<x_{0}<\frac{1}{n_{0}}.\]
Let us take
\[\delta<\frac{1}{2}\cdot\left (\frac{1}{n_{0}}-\frac{1}{n_{0}+1}\right ).\]
We see that if \(|x-x_{0}|<\delta\) then \(x\not\in D\). Therefore, given any \(\epsilon >0\), there exists \(\delta >0\) as given above such that
\[|x-x_{0}|<\delta\mbox{ implies }|f(x)-f(x_{0})|=|1-1|<\epsilon,\]

which says that \(f\) is continuous at \(x_{0}\). From Theorem \ref{mat148}, we see that \(f\) is also Riemann-integrable on \([0,1]\). Now, we want to find its Riemann integral by evaluating its upper Riemann integral. Given any partition \({\cal P}\), we have
\[M_{k}=\sup_{x\in [x_{k-1},x_{k}]}f(x)=1,\]
which says
\[U({\cal P},f)=\sum_{k=1}^{n}M_{k}\Delta x_{k}=\sum_{k=1}^{n}\Delta x_{k}=1.\]
Therefore, we obtain the upper Riemann integral
\[\overline{\int}_{0}^{1}f(x)dx=\inf_{{\cal P}\in\mathfrak{P}[0,1]}U({\cal P},f)=1.\]
Since \(f\) is Riemann-integrable on \([0,1]\), it says that its Riemann integral is \(1\).

Alternatively, we can evaluate the lower Riemann integral as follows. We first note that \(L({\cal P},f)\leq 1\) for any partition \({\cal P}\) of \([0,1]\). Since the set \(D\) has measure zero, given any \(\epsilon >0\), there exists a countable collection of open intervals \([a_{k},b_{k}]\) satisfying \(1/k\in[a_{k},b_{k}]\) and \(\sum_{k=1}^{\infty}(b_{k}-a_{k})<\epsilon/2\). We also see that there exists \(k_{0}\in\mathbb{N}\) satisfying \(1/k_{0}<\epsilon/2\), where
\[\frac{1}{k_{0}}\in\left (a_{k_{0}},b_{k_{0}}\right ).\]
Now, we define a partition \({\cal P}_{\epsilon}\) as follows. Let \(x_{0}=0\), \(x_{1}=b_{k_{0}+1}<\epsilon/2\), \(x_{2}=a_{k_{0}}\), \(x_{3}=b_{k_{0}}\), \(x_{4}=a_{k_{0}-1}\), \(x_{5}=b_{k_{0}-1}\), \(x_{6}=a_{k_{0}-2}\), \(x_{7}=b_{k_{0}-2}\). In general, we define \(x_{2k}=a_{k_{0}-k+1}\) and \(x_{2k+1}=b_{k_{0}-k+1}\) for \(k=1,2,\cdots\), \(x_{n-2}=a_{2}\), \(x_{n-1}=b_{2}\) and \(x_{n}=1\). Without the loss of generality, we may assume that \(n\) is an even integer. It is clear to see that \([x_{k},x_{k+1}]\) contains an element of \(D\) for \(k=2,4,6,\cdots,n\) and that \([x_{k},x_{k+1}]\) does not contain any elements of \(D\) for \(k=3,5,7,\cdots,n\). Then, we have
\[m_{1}=\inf_{x\in [x_{0},x_{1}]}f(x)=0\]
and
\[m_{k}=\left\{\begin{array}{ll}
{\displaystyle \inf_{x\in [x_{k},x_{k+1}]}f(x)=0} & \mbox{if \(k=2,4,6,\cdots,n\)}\\
{\displaystyle \inf_{x\in [x_{k},x_{k+1}]}f(x)=1} & \mbox{if \(k=3,5,7,\cdots,n-1\)}.
\end{array}\right .\]
We also see that \(\Delta x_{1}=x_{1}-x_{0}<\epsilon /2\) and
\begin{align*} \sum_{k=2,4,6,\cdots,n}\Delta x_{k} & =\sum_{k=2,4,6,\cdots,n}\left (x_{k}-x_{k-1}\right )
\\ & <\sum_{k=1}^{\infty}(b_{k}-a_{k})<\frac{\epsilon}{2},\end{align*}
which says
\[\sum_{k=3,5,7,\cdots,n-1}\Delta x_{k}=\sum_{k=3,5,7,\cdots,n-1}\left (x_{k}-x_{k-1}\right )\geq 1-\epsilon.\]
Therefore, we obtain the lower Riemann sum
\begin{align*}
1\geq L({\cal P}_{\epsilon},f) & =\sum_{k=1}^{n}m_{k}\Delta_{k}=m_{1}\Delta_{1}+
\sum_{k=2,4,6,\cdots,n}m_{k}\Delta x_{k}+\sum_{k=3,5,7,\cdots,n-1}m_{k}\Delta x_{k}\\
& =\sum_{k=3,5,7,\cdots,n-1}\Delta x_{k}\geq 1-\epsilon.
\end{align*}
Since \(\epsilon\) can be any positive number and
\[\underline{\int}_{0}^{1}f(x)dx=\sup_{{\cal P}\in\mathfrak{P}[0,1]}L({\cal P},f),\]
it follows
\[\underline{\int}_{0}^{1}f(x)dx=1.\]
This shows that its Riemann integral is \(1\). \(\sharp\)

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Multiple Riemann Integrals.

We are going to study the multiple Riemann integrals. In this case, we need to consider the \(n\)-dimensional intervals. For example, in \(\mathbb{R}^{2}\), the \(2\)-dimensional intervals are the rectangles, and in \(\mathbb{R}^{3}\), the \(3\)-dimensional intervals are the parallelepipeds. Let \(I_{1},\cdots ,I_{n}\) be \(n\) intervals in \(\mathbb{R}\); that is, each \(I_{k}\) may be bounded, unbounded, open, closed or half-open interval in \(\mathbb{R}\). The general \(n\)-dimensional interval \(A\) in \(\mathbb{R}^{n}\) has the form
\begin{equation}{\label{maeq257}}\tag{18}
{\bf I}=I_{1}\times\cdots\times I_{n}=\left\{(x_{1},\cdots ,x_{n}):x_{k}\in I_{k}\mbox{ for }k=1,\cdots ,n\right\}.
\end{equation}
We also allow the degenerate case in which one or more of the intervals \(I_{k}\) consists of single point.

We denote by \(\mu (I_{k})\) the length of \(I_{k}\). If each \(I_{k}\) is bounded, then the \(n\)-dimensional measure of \({\bf I}\) given in (\ref{maeq257}) is denoted and defined by
\[\mu ({\bf I})=\prod_{k=1}^{n}\mu (I_{k}).\]
When \(n=2\), we see that \(\mu ({\bf I})\) is the area of \({\bf I}\), and when \(n=3\), we also see that \(\mu ({\bf I})\) is the volume of \({\bf I}\). We also note that \(\mu ({\bf I})=0\) when \(\mu (I_{k})=0\) for some \(k=1,\cdots ,n\).

The only essential difference of Riemann integral between the case \(\mathbb{R}\) and the case \(\mathbb{R}^{n}\) with \(n>1\) is that the quantity \(\Delta x_{k}=x_{k}-x_{k-1}\) is replaced by the measure \(\mu ({\bf I}_{k})\) of an \(n\)-dimensional interval.

Definition. Let \({\bf I}=I_{1}\times\cdots\times I_{n}\) be a compact interval in \(\mathbb{R}^{n}\). If \({\cal P}_{k}\) is a partition of the one-dimensional interval \(I_{k}\), the Cartesian product \({\cal P}={\cal P}_{1}\times\cdots\times {\cal P}_{n}\) is said to be a partition of \({\bf I}\). A partition \({\cal P}’\) is said to be finer than \({\cal P}\) when \({\cal P}\subseteq {\cal P}’\). The set of all partitions of \({\bf I}\) is denoted by \(\mathfrak{P}({\bf I})\).

Definition. Let \(f\) be a bounded real-valued function defined on a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). If \({\cal P}\) is a partition of \({\bf I}\) into \(m\) sub-intervals \({\bf I}_{1},\cdots ,{\bf I}_{m}\) satisfying \({\bf t}_{k}\in {\bf I}_{k}\) for \(k=1,\cdots,m\), the following sum
\[S({\cal P},f)=\sum_{k=1}^{m}f({\bf t}_{k})\mu ({\bf I}_{k})\]
is called a Riemann sum. We say that \(f\) is Riemann-integrable on \({\bf I}\) with integral \(A\) when, given any \(\epsilon >0\), there exists a partition \({\cal P}(\epsilon )\) of \({\bf I}\) such that
\[{\cal P}\supset {\cal P}(\epsilon)\mbox{ implies }\left |S({\cal P},f)-A\right |<\epsilon\]
for all Riemann sums \(S({\cal P},f)\). When such a number \(A\) exists, it is uniquely determined and is denoted by
\begin{equation}{\label{maeq258}}\tag{19}
A=\int_{\bf I}f({\bf x})d{\bf x}\mbox{ or }\int\!\!\!\!\!\int\cdots\int_{\bf I}f(x_{1},\cdots ,x_{n})d(x_{1}\cdots x_{n}).
\end{equation}

For \(n=2\), the multiple integrals in (\ref{maeq258}) is also called the double integral and is denoted by
\[\int\!\!\!\!\!\int f(x,y)d(x,y).\]
For \(n=3\), the multiple integrals in (\ref{maeq258}) is also called the triple integral and is denoted by
\[\int\!\!\!\!\!\int\!\!\!\!\!\int f(x,y,z)d(x,y,z).\]
Let \(E\) be a subset of \(\mathbb{R}^{n}\). The diameter of \(E\) is denoted and defined by
\[\mbox{dia}(E)=\sup\left\{\parallel {\bf x}-{\bf y}\parallel :{\bf x},{\bf y}\in E\right\}.\]
Let \({\cal P}\) be a partition of \({\bf I}\) into \(m\) sub-intervals \({\bf I}_{1},\cdots ,{\bf I}_{m}\). We also define
\[\parallel{\cal P}\parallel=\max_{k=1,\cdots ,m}\mbox{dia}({\bf I}_{k}).\]
Then, we have
\[A=\int_{\bf I}f({\bf x})d{\bf x}\]
when
\[\lim_{\parallel{\cal P}\parallel\rightarrow 0}S({\cal P},f)=A.\]
In other words, given \(\epsilon >0\), there exists \(\delta >0\) such that \(\parallel{\cal P}\parallel <\delta\) implies \(|S({\cal P},f)-A|<\epsilon\) for any \({\cal P}\) and any chosen points \(\{{\bf t}_{k}\}_{k=1}^{n}\).

Definition. Let \(f\) be a bounded real-valued function defined on a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). If \({\cal P}\) is a partition of \({\bf I}\) into \(m\) sub-intervals \({\bf I}_{1},\cdots ,{\bf I}_{m}\), we define
\[m_{k}(f)=\inf_{{\bf x}\in {\bf I}_{k}}f({\bf x})\mbox{ and }M_{k}(f)=\sup_{{\bf x}\in {\bf I}_{k}}f({\bf x}).\]
The numbers
\[U({\cal P},f)=\sum_{k=1}^{m}M_{k}(f)\mu ({\bf I}_{k})\mbox{ and }L({\cal P},f)=\sum_{k=1}^{m}m_{k}(f)\mu ({\bf I}_{k})\]
are called upper Riemann sum and lower Riemann sum, respectively. The upper Riemann integral and lower Riemann integral of \(f\) over \({\bf I}\) are defined as
\[\overline{\int}_{\bf I}f({\bf x})d{\bf x}=\inf_{{\cal P}\in\mathfrak{P}({\bf I})}U({\cal P},f)\mbox{ and }\underline{\int}_{\bf I}f({\bf x})d{\bf x}
=\sup_{{\cal P}\in\mathfrak{P}({\bf I})}L({\cal P},f).\]

The function \(f\) is said to satisfy Riemann’s condition on \({\bf I}\) when, for every \(\epsilon >0\), there exists a partition \({\cal P}(\epsilon )\) of \({\bf I}\) such that \({\cal P}\supset {\cal P}(\epsilon )\) implies \(U({\cal P},f)-L({\cal P},f)<\epsilon\). \(\sharp\)

Proposition. Let \(f\) and \(g\) be a bounded real-valued function defined on a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). Then, we have the following properties.

(i) We have
\[\overline{\int}_{\bf I}(f({\bf x})+g({\bf x}))d{\bf x}\leq\overline{\int}_{\bf I}f({\bf x})d{\bf x}+\overline{\int}_{\bf I}g({\bf x})d{\bf x}.\]

(ii) We have
\[\underline{\int}_{\bf I}(f({\bf x})+g({\bf x}))d{\bf x}\geq\underline{\int}_{\bf I}f({\bf x})d{\bf x}+\underline{\int}_{\bf I}g({\bf x})d{\bf x}.\]

(iii) If the interval \({\bf I}\) is decomposed into a union of nonoverlapping intervals \({\bf I}_{1}\) and \({\bf I}_{2}\), then we have
\[\overline{\int}_{\bf I}f({\bf x})d{\bf x}=\overline{\int}_{{\bf I}_{1}}f({\bf x})d{\bf x}+\overline{\int}_{{\bf I}_{2}}f({\bf x})d{\bf x}\] and
\[\underline{\int}_{\bf I}f({\bf x})d{\bf x}=\underline{\int}_{{\bf I}_{1}}f({\bf x})d{\bf x}+\underline{\int}_{{\bf I}_{2}}f({\bf x})d{\bf x}.\]

Proof. The arguments as in the one-dimensional case are still applicable and they are left as exercises. \(\blacksquare\)

Theorem. Let \(f\) be a bounded real-valued function defined on a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). Then, the following statements are equivalent.

(a) \(f\) is Riemann-integrable on \({\bf I}\).

(b) \(f\) satisfies the Riemann’s condition on \({\bf I}\).

(c) We have
\[\overline{\int}_{\bf I}f({\bf x})d{\bf x}=\underline{\int}_{\bf I}f({\bf x})d{\bf x}.\]

Definition. A subset \(T\) of \(\mathbb{R}^{n}\) is said to be measure zero when, for every \(\epsilon >0\), the set \(T\) can be covered by a countable collection of \(n\)-dimensional intervals such that their sum of measures is less than \(\epsilon\). \(\sharp\)

We have the following observations.

• The union of a countable collection of sets of measure zero has also measure zero.
• If \(m<n\), then every subset of \(\mathbb{R}^{m}\), when it is considered as a subset of \(\mathbb{R}^{n}\), has measure zero. \(\sharp\)

A property is said to be hold almost everywhere on a set \(T\) in \(\mathbb{R}^{n}\) when it holds everywhere on \(T\) except for a subset of measure zero.

Theorem. (Lebesgue’s Criterion). Let \(f\) be a bounded real-valued function defined on a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). Then \(f\) is Riemann-integrable on \({\bf I}\) if and only if the set of discontinuity of \(f\) in \({\bf I}\) has measure zero.

Proof. The proof is essentially the same as that of Theorem \ref{mat148}. \(\blacksquare\)

Let \(Q=\{(x,y):a\leq x\leq b\mbox{ and }c\leq y\leq d\}\) be a rectangle in \(\mathbb{R}^{2}\). The following iterated integral
\[\int_{c}^{d}\int_{a}^{b}f(x,y)dxdy=\int_{c}^{d}\left [\int_{a}^{b}f(x,y)dx\right ]dy\]
is not necessary equal to the double integral
\[\int\!\!\!\!\!\int_{Q}f(x,y)d(x,y).\]
For example, if \(f\) is discontinuous at every point of the line segment \(y=y_{0}\) for \(a\leq x\leq b\), then the integral \(\int_{a}^{b}f(x,y_{0})dx\) will fail to exist. However, this line segment has measure zero when it is considered as a subset of \(\mathbb{R}^{2}\). This says that it does not affect the integrability of \(f\) on \(Q\). We are going to present the sufficient conditions to guarantee the equality of double integrals and iterated integrals.

\begin{equation}{\label{mat260}}\tag{20}\mbox{}\end{equation}
Theorem \ref{mat260}. Let \(f\) be a bounded real-valued function defined on a compact rectangle \(Q=[a,b]\times [c,d]\subseteq\mathbb{R}^{2}\). Then, we have the following properties.

(i) We have the following inequalities
\begin{align*} \underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y) & \leq\underline{\int}_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx\\ & \leq
\overline{\int}_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx\\ & \leq\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\end{align*}

(ii) We have the following inequalities
\begin{align*} \underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y) & \leq\underline{\int}_{a}^{b}\left [\underline{\int}_{c}^{d}f(x,y)dy\right ]dx\\ & \leq
\overline{\int}_{a}^{b}\left [\underline{\int}_{c}^{d}f(x,y)dy\right ]dx\\ & \leq\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\end{align*}

(iii) We have the following inequalities
\begin{align*} \underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y) & \leq\underline{\int}_{c}^{d}\left [\overline{\int}_{a}^{b}f(x,y)dx\right ]dy\\ & \leq
\overline{\int}_{c}^{d}\left [\overline{\int}_{a}^{b}f(x,y)dx\right ]dy\\ & \leq\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\end{align*}

(iv) We have the following inequalities
\begin{align*} \underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y) & \leq\underline{\int}_{c}^{d}\left [\underline{\int}_{a}^{b}f(x,y)dx\right ]dy\\ & \leq
\overline{\int}_{c}^{d}\left [\underline{\int}_{a}^{b}f(x,y)dx\right ]dy\\ & \leq\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\end{align*}

(v) Suppose that the double integral
\[\int\!\!\!\!\!\int_{Q}f(x,y)d(x,y)\]
exists. Then, we have the following equalities
\begin{align*}
\int\!\!\!\!\!\int_{Q}f(x,y)d(x,y) & =\int_{a}^{b}\left [\underline{\int}_{c}^{d}f(x,y)dy\right ]dx=
\int_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx\\
& =\int_{c}^{d}\left [\underline{\int}_{a}^{b}f(x,y)dx\right ]dy=\int_{c}^{d}\left [\overline{\int}_{a}^{b}f(x,y)dx\right ]dy.
\end{align*}

Proof. To prove part (i), we define a new function \(F\) by
\[F(x)=\overline{\int}_{c}^{d}f(x,y)dy\mbox{ for }x\in [a,b].\]
Then, we have
\[|F(x)|\leq M(d-c)\mbox{ for }M=\sup_{(x,y)\in Q}|f(x,y)|.\]
Now, we can consider
\[\overline{I}=\overline{\int}_{a}^{b}F(x)=\overline{\int}_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx\] and
\[\underline{I}=\underline{\int}_{a}^{b}F(x)=\underline{\int}_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx.\]
Let \({\cal P}_{1}=\{x_{0},x_{1},\cdots ,x_{n}\}\) be a partition of \([a,b]\), and let \({\cal P}_{2}=\{y_{0},y_{1},\cdots ,y_{m}\}\) be a partition of \([c,d]\). Then, we see that \({\cal P}={\cal P}_{1}\times {\cal P}_{2}\) is a partition of \(Q\) into \(mn\) rectangles \(Q_{ij}\). Therefore, we can consider
\[\overline{I}_{ij}=\overline{\int}_{x_{i-1}}^{x_{i}}\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx\] and
\[\underline{I}_{ij}=\underline{\int}_{x_{i-1}}^{x_{i}}\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx.\]
Sine we have
\[\overline{\int}_{c}^{d}f(x,y)dy=\sum_{j=1}^{m}\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy,\]
we can write
\begin{align*}
\overline{I} & =\overline{\int}_{a}^{b}\left [\overline{\int}_{c}^{d}f(x,y)dy\right ]dx
=\sum_{j=1}^{m}\overline{\int}_{a}^{b}\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx\\
& =\sum_{i=1}^{n}\sum_{j=1}^{m}\overline{\int}_{x_{i-1}}^{x_{i}}
\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx=\sum_{i=1}^{n}\sum_{j=1}^{m}\overline{I}_{ij}.
\end{align*}
We can similarly obtain
\[\underline{I}=\sum_{j=1}^{m}\sum_{i=1}^{n}\underline{I}_{ij}.\]
We write
\[m_{ij}=\inf_{(x,y)\in Q_{ij}}f(x,y)\mbox{ and }M_{ij}=\sup_{(x,y)\in Q_{ij}}f(x,y),\]
Then, from the inequality \(m_{ij}\leq f(x,y)\leq M_{ij}\) for \((x,y)\in Q_{ij}\), we have
\[m_{ij}(y_{j}-y_{j-1})\leq\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\leq M_{ij}(y_{j}-y_{j-1}),\]
which implies
\begin{align*} m_{ij}\mu (Q_{ij}) & \leq\underline{\int}_{x_{i-1}}^{x_{i}}\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx\\ & \leq
\overline{\int}_{x_{i-1}}^{x_{i}}\left [\overline{\int}_{y_{j-1}}^{y_{j}}f(x,y)dy\right ]dx\\ & \leq M_{ij}\mu (Q_{ij}),\end{align*}
i.e.,
\[m_{ij}\mu (Q_{ij})\leq\underline{I}_{ij}\leq\overline{I}_{ij}\leq M_{ij}\mu (Q_{ij}).\]
Therefore, we obtain
\begin{align*} L({\cal P},f) & =\sum_{i=1}^{n}\sum_{j=1}^{m}m_{ij}\mu (Q_{ij})\\\ & leq\sum_{i=1}^{n}\sum_{j=1}^{m}\underline{I}_{ij}
\leq\sum_{i=1}^{n}\sum_{j=1}^{m}\overline{I}_{ij}\\ & \leq\sum_{i=1}^{n}\sum_{j=1}^{m}M_{ij}\mu (Q_{ij})\\ & =U({\cal P},f),\end{align*}
i.e.,
\[L({\cal P},f)\leq\underline{I}\leq\overline{I}\leq U({\cal P},f).\]
Since this holds true for all partition \({\cal P}\) of \(Q\), we obtain
\[\underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y)\leq\underline{I}\leq\overline{I}\leq\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\]
This proves part (i).

It is clear that the preceding proof is also valid when the function \(F\) is taken to be
\[F(x)=\underline{\int}_{c}^{d}f(x,y)dy,\]
which proves part (ii). Parts (iii) and (iv) can be similarly obtained by interchanging the roles of \(x\) and \(y\). To prove part (v), since the double integral
\[\int\!\!\!\!\!\int_{Q}f(x,y)d(x,y)\]
exists if and only if
\[\underline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y)=\overline{\int\!\!\!\!\!\int}_{Q}f(x,y)d(x,y).\]
The desired results can be obtained from parts (i) through (iv). This completes the proof. \(\blacksquare\)

Corollary. (Fubini’s Theorem). Let \(f\) be a bounded real-valued function defined on a compact rectangle \(Q=[a,b]\times [c,d] \subseteq\mathbb{R}^{2}\). Suppose that \(f\) is continuous on \(Q\). Then, we have
\begin{align*} \int\!\!\!\!\!\int_{Q}f(x,y)d(x,y) & =\int_{a}^{b}\left [\int_{c}^{d}f(x,y)dy\right ]dx\\ & =\int_{c}^{d}\left [\int_{a}^{b}f(x,y)dx\right ]dy.\end{align*}

Proof. The result follows immediately from Theorem \ref{mat260}. \(\blacksquare\)

We also remark that the existence of the iterated integrals
\[\int_{a}^{b}\left [\int_{c}^{d}f(x,y)dy\right ]dx\mbox{ and }\int_{c}^{d}\left [\int_{a}^{b}f(x,y)dx\right ]dy\]
does not necessarily imply the existence of the double integral
\[\int\!\!\!\!\!\int_{Q}f(x,y)d(x,y).\]

We are going to obtain the analog of Theorem \ref{mat260} in \(\mathbb{R}^{n}\). We first introduce some notation and terminology. Given any \(k\in\{1,\cdots ,n\}\), the set of \({\bf x}\in\mathbb{R}^{n}\) for which \(x_{k}=0\) is called the coordinate hyperplane and is denoted by \(H_{k}\). Given a subset \(Q\) of \(\mathbb{R}^{n}\), the projection \(Q_{k}\) of \(Q\) on \(H_{k}\) is defined to be the image of \(Q\) under the mapping whose value at each point \((x_{1},\cdots ,x_{n})\) in \(Q\) is \((x_{1},\cdots ,x_{k-1},0,x_{k+1},\cdots ,x_{n})\). We consider the analog of Theorem \ref{mat260} when \(n=3\). Let \(f\) be a bounded real-valued function defined on a compact interval
\[Q=[a_{1},b_{1}]\times [a_{2},b_{2}]\times [a_{3},b_{3}]\subseteq\mathbb{R}^{3}.\]
Let \(Q_{1}\) be the projection of \(Q\) on the coordinate hyperplane \(H_{1}\). Part (i) of Theorem \ref{mat260} can be replaced by
\begin{align*}
\underline{\int\!\!\!\!\!\int\!\!\!\!\!\int}_{Q}f(x_{1},x_{2},x_{3})d(x_{1},x_{2},x_{3})
& \leq\underline{\int}_{a_{1}}^{b_{1}}\left [\overline{\int\!\!\!\!\!\int}_{Q_{1}}f(x_{1},x_{2},x_{3})d(x_{2},x_{3})\right ]dx_{1}\\
& \leq\underline{\int}_{a_{1}}^{b_{1}}\left [\overline{\int\!\!\!\!\!\int}_{Q_{1}}f(x_{1},x_{2},x_{3})d(x_{2},x_{3})\right ]dx_{1}\\
& \leq\overline{\int\!\!\!\!\!\int\!\!\!\!\!\int}_{Q}f(x_{1},x_{2},x_{3})d(x_{1},x_{2},x_{3}).
\end{align*}
When the triple integral
\[\int\!\!\!\!\!\int\!\!\!\!\!\int_{Q}f(x_{1},x_{2},x_{3})d(x_{1},x_{2},x_{3})\]
exists, the analog of part (v) of Theorem \ref{mat260} is given by
\begin{align*}
\int\!\!\!\!\!\int\!\!\!\!\!\int_{Q}f(x_{1},x_{2},x_{3})d(x_{1},x_{2},x_{3})
& =\int_{a_{1}}^{b_{1}}\left [\overline{\int\!\!\!\!\!\int}_{Q_{1}}f(x_{1},x_{2},x_{3})d(x_{2},x_{3})\right ]dx_{1}\\
& =\int\!\!\!\!\!\int_{Q_{1}}\left [\overline{\int}_{a_{1}}^{b_{1}}f(x_{1},x_{2},x_{3})dx_{1}\right ]d(x_{2},x_{3}).
\end{align*}
There are also analogous formulas for \(Q_{2}\) and \(Q_{3}\). The general case is given below.

Theorem. Let \(f\) be a bounded real-valued function defined on a compact interval
\[Q=[a_{1},b_{1}]\times\cdots\times [a_{n},b_{n}]\subseteq\mathbb{R}^{n}.\]
Suppose that the following multiple integral
\[\int\!\!\cdots\!\!\int\!\!\!\!\!\int_{Q}f(x_{1},\cdots ,x_{n})d(x_{1},\cdots ,x_{n})\]
exists. Then, we have
\begin{align*}
\int\!\!\cdots\!\!\int\!\!\!\!\!\int_{Q}f(x_{1},\cdots ,x_{n})d(x_{1},\cdots ,x_{n})
& =\int_{a_{1}}^{b_{1}}\left [\overline{\int\!\!\cdots\!\!\int\!\!\!\!\!\int}_{Q_{1}}f(x_{1},\cdots,x_{n})d(x_{2},x_{3},\cdots ,x_{n})\right ]dx_{1}\\
& =\int\!\!\cdots\!\!\int\!\!\!\!\!\int_{Q_{1}}\left [\overline{\int}_{a_{1}}^{b_{1}}f(x_{1},\cdots ,x_{n})dx_{1}\right ]d(x_{2},x_{3},\cdots ,x_{n}).
\end{align*}
Similar formulas can be obtained when upper integrals are replaced by lower integrals and \(Q_{1}\) is replaced by \(Q_{k}\). \(\sharp\)

The multiple integrals have been defined only for intervals \({\bf I}\). However, this is too restrictive for the applications of integration. We are going to extend the definition to cover more general sets called Jordan-measurable sets. Recall that a point \({\bf x}\in\mathbb{R}^{n}\) is called a boundary point of \(S\) when every \(n\)-dimensional open ball \(B({\bf x})\) contains a point in \(S\) and also a point not in \(S\). The set of all boundary points of \(S\) is called the boundary of \(S\) and is denoted by \(\partial S\).

Definition. Let \(S\) be a subset of a compact interval \({\bf I}\) in \(\mathbb{R}^{n}\). For every partition \({\cal P}\) of \({\bf I}\), we define \(\underline{J}({\cal P},S)\) to be the sum of measures of those sub-intervals determined by \({\cal P}\) which contain only interior points of \(S\). We also define \(\overline{J}({\cal P},S)\) to be the sum of measures of those sub-intervals determined by \({\cal P}\) which contains points of \(S\cup\partial S\). The following numbers
\[\underline{c}(S)=\sup_{{\cal P}\in\mathfrak{P}({\bf I})}\underline{J}({\cal P},S)\mbox{ and }
\overline{c}(S)=\inf_{{\cal P}\in\mathfrak{P}({\bf I})}\overline{J}({\cal P},S)\]
are called the inner and outer Jordan content of \(S\), respectively. The set \(S\) is said to be Jordan-measurable when \(\underline{c}(S)=\overline{c}(S)\). In this case, the common value is called the Jordan content of \(S\) and is denoted by \(c(S)\). \(\sharp\)

We have the following observations.

• The values \(\underline{c}(S)\) and \(\overline{c}(S)\) depend only on \(S\) and not on the interval \({\bf I}\) which contains \(S\).
• We have \(0\leq\underline{c}(S)\leq\overline{c}(S)\).
• Suppose that = \(S\) has content zero. Then, we have \(0=\underline{c}(S)=\overline{c}(S)\). This says that, given any \(\epsilon >0\), the set \(S\) can be covered by a finite collection of intervals such that the sum of their measures is less than \(\epsilon\). We remark that content zero is described in terms of finite coverings. However, measure is described in terms of countable coverings. Therefore, any set with content zero has also measure zero. However, the converse is not necessarily true.
• Every compact interval \(Q\) is Jordan-measurable and its content \(c(Q)\) is equal to its measure \(\mu (Q)\).
• For \(k<n\), the \(n\)-dimensional content of every bounded subset of \(\mathbb{R}^{k}\) is zero.

Suppose that \(S\) is a subset of \(\mathbb{R}^{2}\). The numbers \(\underline{J}({\cal P},S)\) and \(\overline{J}({\cal P},S)\) represent the approximations regarding the area of \(S\) from the “inside” and the “outside” of \(S\), respectively. In this case, the content \(c(S)\) is the area of \(S\). For the subset \(S\) of \(\mathbb{R}^{3}\), the content \(c(S)\) of \(S\) will be the volume of \(S\).

\begin{equation}{\label{map391}}\tag{21}\mbox{}\end{equation}
Proposition \ref{map391}. Let \(S\) be a bounded subset of \(\mathbb{R}^{n}\), and let \(\partial S\) be its boundary. Then, we have
\[\overline{c}(\partial S)=\overline{c}(S)-\underline{c}(S).\]
This says that \(S\) is Jordan-measurable if and only if \(\partial S\) has content zero.

Proof. Let \(I\) be a compact interval containing \(S\) and \(\partial S\). Then, for each partition \({\cal P}\) of \(I\), we have
\[\overline{J}({\cal P},\partial S)=\overline{J}({\cal P},S)-\underline{J}({\cal P},S).\]
Therefore, we obtain
\begin{align*}
\overline{J}({\cal P},\partial S) & \geq\overline{J}({\cal P},S)-\sup_{{\cal P}\in\mathfrak{P}({\bf I})}\underline{J}({\cal P},S)\\
& \geq\inf_{{\cal P}\in\mathfrak{P}({\bf I})}\overline{J}({\cal P},S)-\sup_{{\cal P}\in\mathfrak{P}({\bf I})}\underline{J}({\cal P},S)
\\ & =\overline{c}(S)-\underline{c}(S),
\end{align*}
which implies
\[\overline{c}(\partial S)=\inf_{{\cal P}\in\mathfrak{P}({\bf I})}
\overline{J}({\cal P},\partial S)\geq\overline{c}(S)-\underline{c}(S).\]

To obtain the reverse inequality, given any \(\epsilon >0\), according to the concepts of supremum and infimum, we can take a partition \({\cal P}_{1}\) satisfying
\[\overline{J}({\cal P}_{1},S)<\overline{c}(S)+\frac{\epsilon}{2}\]
and a partition \({\cal P}_{2}\) satisfying
\[\underline{J}({\cal P}_{2},S)>\underline{c}(S)-\frac{\epsilon}{2}.\]
Let \({\cal P}={\cal P}_{1}\cup {\cal P}_{2}\). Since refinement increases the inner sum \(\underline{J}\) and decreases the outer sum \(\overline{J}\), i.e.,
\[\overline{J}({\cal P},S)\leq\overline{J}({\cal P}_{1},S)\mbox{ and }\underline{J}({\cal P},S)\geq\underline{J}({\cal P}_{2},S),\]
we obtain
\begin{align*}
\overline{c}(\partial S) & \leq\overline{J}({\cal P},\partial S)=\overline{J}({\cal P},S)-\underline{J}({\cal P},S)\\
& \leq\overline{J}({\cal P}_{1},S)-\underline{J}({\cal P}_{2},S)<\overline{c}(S)-\underline{c}(S)+\epsilon .
\end{align*}
Since \(\epsilon\) is an arbitrary positive number, it follows
\[\overline{c}(\partial S)\leq\overline{c}(S)-\underline{c}(S).\]
This completes the proof. \(\blacksquare\)

The above proposition says that a bounded set has Jordan content if and only if its boundary is not too “thick”. Now, we are going to define the multiple integrals over the Jordan-measurable sets.

Definition. Let \(S\) be a bounded Jordan-measurable set in \(\mathbb{R}^{n}\), and let \(f\) be a real-valued function defined and bounded on \(S\).
Let \({\bf I}\) be a compact interval containing \(S\). We define a real-valued function \(g\) on \({\bf I}\) as
\[g({\bf x})=\left\{\begin{array}{ll}
f({\bf x}) & \mbox{if \({\bf x}\in S\).}\\
0 & \mbox{if \({\bf x}\in {\bf I}\setminus S\).}
\end{array}\right .\]
Then \(f\) is said to be Riemann-integrable on \(S\) when the integral \(\int_{\bf I}g({\bf x})d{\bf x}\) exists. In this case, we also write
\[\int_{S}f({\bf x})d{\bf x}=\int_{\bf I}g({\bf x})d{\bf x}.\]
The upper and lower integrals
\[\overline{\int}_{S}f({\bf x})d{\bf x}\mbox{ and }\underline{\int}_{S}f({\bf x})d{\bf x}\]
can be similarly defined. \(\sharp\)

\begin{equation}{\label{mat392}}\tag{22}\mbox{}\end{equation}
Theorem \ref{mat392}. Let \(S\) be a bounded Jordan-measurable set in \(\mathbb{R}^{n}\), and let \(f\) be a real-valued function defined and bounded on \(S\). The function \(f\) is Riemann-integrable on \(S\) if and only if the set of discontinuities of \(f\) in \(S\) has measure zero.

Proof. Let \(I\) be a compact interval containing \(S\) and let
\[g({\bf x})=\left\{\begin{array}{ll}
f({\bf x}) & \mbox{if \({\bf x}\in S\)}\\
0 & \mbox{if \({\bf x}\in I\setminus S\)}.
\end{array}\right .\]
The discontinuities of \(f\) will be the discontinuities of \(g\). However, the function \(g\) may also have discontinuities at some or all of the boundary points of \(S\). Since \(S\) is Jordan measurable, Proposition \ref{map391} says that \(c(\partial S)=0\). Therefore, the function \(g\) is Riemann-integrable on \(I\) if and only if the discontinuities of \(f\) form a set of measure zero. This completes the proof. \(\blacksquare\)

Proposition. Let \(S\) be a compact Jordan-measurable set in \(\mathbb{R}^{n}\). Then, the integral \(\int_{S}1({\bf x})dx\) exists and we have
\[c(S)=\int_{S}1({\bf x})d{\bf x}.\]

Proof. Let \(I\) be a compact interval containing \(S\), and let \(\chi_{S}\) denote the characteristic function of \(S\), i.e.,
\[\chi_{S}({\bf x})=\left\{\begin{array}{ll}
1 & \mbox{if \({\bf x}\in S\)}\\
0 & \mbox{if \({\bf x}\in I\setminus S\)}.
\end{array}\right .\]
The discontinuities of \(\chi_{S}\) in \(I\) are the boundary points of \(S\) and form a set of content zero, which says that the integral \(\int_{I}\chi_{S}({\bf x})d{\bf x}\) exists, i.e., the integral \(\int_{S}1({\bf x})dx\) exists. Let \({\cal P}\) be a partition of \(I\) into sub-intervals \(I_{1},\cdots ,I_{m}\), and let
\[\Gamma =\left\{k:I_{k}\cap S\neq\emptyset\right\}.\]
For \(k\in\Gamma\), we have
\[M_{k}(\chi_{S})=\sup_{{\bf x}\in I_{k}}\chi_{S}({\bf x})=1.\]
We also have \(M_{k}(\chi_{S})=0\) for \(k\not\in\Gamma\). Therefore, we obtain
\[U({\cal P},\chi_{S})=\sum_{k=1}^{m}M_{k}(\chi_{S})\mu (I_{k})=\sum_{k\in\Gamma}\mu (I_{k})=\overline{J}({\cal P},\chi_{S}).\]
Since this holds true for all partitions, we have
\[\overline{\int}_{I}\chi_{S}({\bf x})d{\bf x}=\overline{c}(S)=c(S).\]
Since \(\overline{\int}_{I}\chi_{S}({\bf x})d{\bf x}=\int_{I}\chi_{S}({\bf x})d{\bf x}\), we obtain
\[c(S)=\int_{I}\chi_{S}({\bf x})d{\bf x}=\int_{S}1({\bf x})d{\bf x},\]
and the proof is complete. \(\blacksquare\)

Theorem. Suppose that the real-valued function \(f\) is Riemann-integrable on a Jordan-measurable set \(S\) in \(\mathbb{R}^{n}\). If \(S=A\cup B\), where \(A\) and \(B\) are Jordan-measurable satisfying \(A\cap B=\emptyset\), then \(f\) is Riemann-integrable on \(A\) and \(B\), and we also have
\begin{equation}{\label{maeq393}}\tag{23}
\int_{S}f({\bf x})d{\bf x}=\int_{A}f({\bf x})d{\bf x}+\int_{B}f({\bf x})d{\bf x}.
\end{equation}
The above formula also holds true for upper and lower integrals.

Proof. Let \(I\) be a compact interval containing \(S\). We define a function \(g\) as follows
\[g({\bf x})=\left\{\begin{array}{ll}
f({\bf x}) & \mbox{if \({\bf x}\in S\)}\\
0 & \mbox{if \({\bf x}\in I\setminus S\)}.
\end{array}\right .\]
The existence of \(\int_{A}f({\bf x})d{\bf x}\) and \(\int_{B}f({\bf x})d{\bf x}\) is an easy consequence of Theorem \ref{mat392}. Let \({\cal P}\) be a partition of \(I\) into \(m\) subintervals \(I_{1},\cdots ,I_{m}\). Then, we have the Riemann sum
\[S({\cal P},g)=\sum_{k=1}^{m}g({\bf t}_{k})\mu (I_{k}).\]
Let \(S_{A}\) denote the part of the sum arising from those subintervals containing points of \(A\), let \(S_{B}\) denote the part of the sum arising from those sub-intervals containing points of \(A\), and let \(S_{C}\) denote the part of the sum coming from sub-intervals which contain both points of \(A\) and points of \(B\). Then, we can write
\[S({\cal P},g)=S_{A}+S_{B}-S_{C}.\]
In particular, all points common to the two boundaries \(\partial A\) and \(\partial B\) will fall in the case of \(S_{C}\). We see that \(S_{A}\) is a Riemann sum that approximates the integral \(\int_{A}f({\bf x})d{\bf x}\), and that \(S_{B}\) is a Riemann sum that approximates the integral \(\int_{B}f({\bf x})d{\bf x}\). Since \(c(\partial A\cap\partial B)=0\), it follows that \(|S_{C}|\) can be made arbitrarily small when \({\cal P}\) is sufficiently fine. The equation (\ref{maeq393}) is an easy consequence of these arguments. This completes the proof. \(\blacksquare\)

Proposition. Let \(\phi_{1}\) and \(\phi_{2}\) be two continuous functions defined on \([a,b]\) satisfying \(\phi_{1}(x)\leq\phi_{2}(x)\) for each \(x\in [a,b]\). Let \(S\) be a compact set in \(\mathbb{R}^{2}\) given by
\[S=\left\{(x,y):a\leq x\leq b\mbox{ and }\phi_{1}(x)\leq y\leq\phi_{2}(x)\right\}.\]
Then \(S\) is Jordan-measurable. Suppose that \(f\) is Riemann-integrable on \(S\). Then, we have
\[\int_{S}f(x,y)d(x,y)=\int_{a}^{b}\left [\overline{\int}_{\phi_{1}(x)}^{\phi_{2}(x)}f(x,y)dy\right ]dx.\]

Proof. Since the boundary of \(S\) has content zero, it means that \(S\) is Jordan-measurable. Therefore, the result follows immediately from Theorem \ref{mat260}. \(\blacksquare\)

\begin{equation}{\label{map394}}\tag{24}\mbox{}\end{equation}
Proposition \ref{map394}. Suppose that the real-valued functions \(f\) and \(g\) are Riemann-integrable on a Jordan-measurable set \(S\) in \(\mathbb{R}^{n}\). If \(f({\bf x})\leq g({\bf x})\) for each \({\bf x}\in S\), then we have
\[\int_{S}f({\bf x})d{\bf x}\leq\int_{S}g({\bf x})d{\bf x}.\]

Theorem. (Mean-Value Theorem for Differentiation). Suppose that the real-valued functions \(f\) and \(g\) are Riemann-integrable on a Jordan-measurable set \(S\) in \(\mathbb{R}^{n}\). We also assume \({\bf g}({\bf x})\geq 0\) for each \({\bf x}\in S\). Let
\[m=\inf_{{\bf x}\in S}f({\bf x})\mbox{ and }M=\sup_{{\bf x}\in S}f({\bf x}).\]
Then, there exists a real number \(\lambda\in [m,M]\) satisfying
\begin{equation}{\label{maeq395}}\tag{25}
\int_{S}f({\bf x})g({\bf x})d{\bf x}=\lambda\int_{S}g({\bf x})d{\bf x}.
\end{equation}
In particular, we have
\begin{equation}{\label{maeq396}}\tag{26}
m\cdot c(S)\leq\int_{S}f({\bf x})d{\bf x}\leq M\cdot c(S).
\end{equation}
We further assume that \(S\) is connected and \(f\) is continuous on \(S\). Then, we have \(\lambda =f({\bf x}_{0})\) for some \({\bf x}_{0}\in S\) and
\[\int_{S}f({\bf x})g({\bf x})d{\bf x}=f({\bf x}_{0})\int_{S}g({\bf x})d{\bf x}.\]
In particular, we obtain
\[\int_{S}f({\bf x})d{\bf x}=f({\bf x}_{0})\cdot c(S).\]

Proof. Since \(g({\bf x})\geq 0\), we have
\[mg({\bf x})\leq f({\bf x})g({\bf x})\leq Mg({\bf x})\]
for each \({\bf x}\in S\). By Proposition \ref{map394}, we can write
\[m\int_{S}g({\bf x})d{\bf x}\leq\int_{S}f({\bf x})g({\bf x})d{\bf x}\leq M\int_{S}g({\bf x})d{\bf x}.\]
If \(\int_{S}g({\bf x})d{\bf x}=0\), then (\ref{maeq395}) holds true. If \(\int_{S}g({\bf x})d{\bf x}>0\), then (\ref{maeq395}) holds true with
\[\lambda =\frac{\int_{S}f({\bf x})g({\bf x})d{\bf x}}{\int_{S}g({\bf x})d{\bf x}}.\]
By taking \(g({\bf x})=1\), we obtain (\ref{maeq396}) and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{map261}}\tag{27}\mbox{}\end{equation}
Proposition \ref{map261}. Suppose that the real-valued function \(f\) is Riemann-integrable on a Jordan-measurable set \(S\) in \(\mathbb{R}^{n}\). Let \(T\) be a subset of \(S\) having Jordan content zero. Let \(g\) be a real-valued function defined and bounded on \(S\) satisfying \(g({\bf x})=f({\bf x})\) for \({\bf x}\in S\setminus T\). Then \(g\) is Riemann-integrable on \(S\) and we have
\[\int_{S}f({\bf x})d{\bf x}=\int_{S}g({\bf x})d{\bf x}.\]

Proof. Let \(h=f-g\). Then, we have
\[\int_{S}h({\bf x})d{\bf x}=\int_{T}h({\bf x})d{\bf x}+\int_{S\setminus T}h({\bf x})d{\bf x}.\]
Using (\ref{maeq396}), we have \(\int_{T}h({\bf x})d{\bf x}=0\). Since \(h({\bf x})=0\) for each \({\bf x}\in S\setminus T\), we also have \(\int_{S\setminus T}h({\bf x})d{\bf x}=0\). This completes the proof. \(\blacksquare\)

Proposition\ref{map261} suggests to extend the definition of the Riemann integral for functions which may not be defined and bounded on the whole domain \(S\). In fact, let \(S\) be a bounded set in \(\mathbb{R}^{n}\) having Jordan content \(c(S)\) and let \(T\) be a subset of \(S\) having Jordan content zero. If \(f\) is defined on bounded on \(S\setminus T\) and the integral \(\int_{S\setminus T}f({\bf x})d{\bf x})\) exists, we can write
\[\int_{S}f({\bf x})d{\bf x}=\int_{S\setminus T}f({\bf x})d{\bf x}\]
and say that \(f\) is Riemann-integrable on \(S\).