The Lebesgue Integrals without Using Measure

Jules Girardet (1856-1938) was a French painter.

We have sections

• The Integrals of Step Functions 
• Lebesgue Functions and Integrals
• The Lebesgue-Integrable Functions
• The Improper Riemann Integrals
• Measurable Functions and Measurable Sets
• Properties of Lebesgue Integrals
• Multiple Lebesgue Integrals

The Lebesgue integral studied here will extend the Riemann integral in the sense that some of functions that are not Riemann-integrable can be Lebesgue-integrable. The early work on the Lebesgue integral is based on the measure theory. Here, the Lebesgue integral  does not depend on measure.

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

The Integrals of Step Functions.

We recall that a function \(s\) defined on a closed interval \([a,b]\) is called a step function when there is a partition \({\cal P}=\{x_{0},x_{1},\cdots ,x_{n}\}\) of \([a,b]\) such that \(s\) is constant on every open sub-interval, i.e.,

\[s(x)=c_{k}\mbox{ for }x\in (x_{k-1},x_{k}).\]

A step function \(s\) is Riemann-integrable on each sub-interval \([x_{k-1},x_{k}]\) and is given by

\[\int_{x_{k-1}}^{x_{k}}s(x)dx=c_{k}\cdot (x_{k}-x_{k-1})\]

regardless of the values of \(s\) at the endpoints \(x_{k}\) and \(x_{k-1}\) of sub-interval \([x_{k-1},x_{k}]\). Therefore, the Riemann integral of \(s\) over \([a,b]\) is given by

\[\int_{a}^{b}s(x)dx=\sum_{k=1}^{n}c_{k}\cdot (x_{k}-x_{k-1}).\]

We shall extend the concept of step function without considering closed intervals. In what follows, when we say that \(I\) is an interval, it means that \(I\) can be unbounded with half-open or half-closed, unless we emphasize that \(I\) is a bounded interval.

Definition. Let \(I\) be an interval. A function \(s\) is called a generalized step function of \(I\) when there is a closed sub-interval \([a,b]\) of \(I\) such that \(s\) is a step function on \([a,b]\) and \(s(x)=0\) for \(x\in I\setminus [a,b]\). In this case, the integral is defined by

\[\int_{I}s(x)dx=\int_{a}^{b}s(x)dx=\sum_{k=1}^{n}c_{k}\cdot (x_{k}-x_{k-1}).\]

The sum of two generalized step functions is also a generalized step function. The following properties of the integrals for generalized step functions can be
easily derived by definition:

\begin{align*}
& \int_{I}(s+t)(x)dx=\int_{I}s(x)dx+\int_{I}t(x)dx\\
& \int_{I}cs(x)dx=c\int_{I}s(x)dx\mbox{ for any constant \(c\)}\\
& \mbox{if \(s(x)\leq t(x)\) for all \(x\in I\), then }\int_{I}s(x)dx\leq\int_{I}t(x)dx.
\end{align*}

Suppose that \(I\) is the disjoint union of a finite set of subintervals given by \(I=\bigcup_{k=1}^{n}[a_{k},b_{k}]\). Then, we have

\begin{equation}{\label{ma26}}\tag{1}
\int_{I}s(x)dx=\sum_{k=1}^{n}\int_{a_{k}}^{b_{k}}s(x)dx.
\end{equation}

Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of real-valued functions defined on a set \(S\).

• We say that the sequence \(\{f_{n}\}_{n=1}^{\infty}\) is {\bf increasing} on \(S\) when \(f_{n}(x)\leq f_{n+1}(x)\) for all \(x\in S\) and all \(n=1,2,\cdots\).
• We say that the sequence \(\{f_{n}\}_{n=1}^{\infty}\) is {\bf decreasing} on \(S\) when \(f_{n}(x)\geq f_{n+1}(x)\) for all \(x\in S\) and all \(n=1,2,\cdots\).

Let us recall the concept of measure zero in Definition \ref{ma23}. A property is said to be almost everywhere on a set \(S\) and is written as a.e. on \(S\) when it holds true everywhere on \(S\) except for a set of measure zero. For example, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) converges to \(f\) almost everywhere on \(S\) when the set \(N\subseteq S\) in which \(f_{n}(x)\) does not converge to \(f(x)\) for \(x\in N\) has measure zero.

\begin{equation}{\label{map305}}\tag{2}\mbox{}\end{equation}

Proposition \ref{map305}. Let \(\{s_{n}\}_{n=1}^{\infty}\) be a decreasing sequence of nonnegative step functions such that \(\{s_{n}\}_{n=1}^{\infty}\) decreases to zero a.e. on an interval \(I\). Then, we have

\[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx=0.\]

\begin{equation}{\label{map306}}\tag{3}\mbox{}\end{equation}

Proposition \ref{map306}. Let \(\{s_{n}\}_{n=1}^{\infty}\) be an increasing sequence of step functions on an interval \(I\) such that the following conditions are satisfied.

• There is a function \(f\) such that \(\{s_{n}\}_{n=1}^{\infty}\) converges to \(f\) a.e. on \(I\);
• The sequence \(\{\int_{I}s_{n}(x)dx\}_{n=1}^{\infty}\) is convergent.

Then, for any step function \(s\) satisfying \(s(x)\leq f(x)\) a.e. on \(I\), we have

\begin{equation}{\label{maeq307}}\tag{4}
\int_{I}s(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.
\end{equation}

Proof. We define a new sequence of nonnegative step function \(\{t_{n}\}_{n=1}^{\infty}\) on \(I\) as follows

\[t_{n}(x)=\left\{\begin{array}{ll}
s(x)-s_{n}(x) & \mbox{if \(s(x)\geq s_{n}(x)\)}\\
0 & \mbox{if \(s(x)<s_{n}(x)\)}.
\end{array}\right .\]

Then, we have \(t_{n}(s)=\max\{s(x)-s_{n}(x),0\}\). Since \(\{s_{n}\}_{n=1}^{\infty}\) is increasing on \(I\), it follows that \(\{t_{n}\}_{n=1}^{\infty}\) is decreasing on \(I\) and

\begin{align*}
\lim_{n\rightarrow\infty}t_{n}(x) & =\lim_{n\rightarrow\infty}\max\{s(x)-s_{n}(x),0\}\\ & =\max\left\{s(x)-\lim_{n\rightarrow\infty}s_{n}(x),0\right\}\\
& =\max\left\{s(x)-f(x),0\right\}\mbox{ a.e. on \(I\)}\\
& =0\mbox{ a.e. on \(I\)}\mbox{ (since \(s(x)\leq f(x)\) a.e. on \(I\))}.
\end{align*}

This shows that \(\{t_{n}\}_{n=1}^{\infty}\) is decreasing to \(0\) a.e. on \(I\). Therefore, using Proposition \ref{map305}, we obtain

\begin{equation}{\label{maeq306}}\tag{5}
\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx=0.
\end{equation}

Since

\begin{align*} t_{n}(x) & =\max\{s(x)-s_{n}(x),0\}\\ & \geq s(x)-s_{n}(x)\mbox{ for all }x\in I,\end{align*}

we also have

\begin{align}
\int_{I}t_{n}(x)dx & \geq\int\left [s(x)-s_{n}(x)\right ]dx\nonumber\\ =\int_{I}s(x)dx-\int_{I}s_{n}(x)dx\label{ma24}\tag{6}
\end{align}

Therefore, using (\ref{ma24}) and (\ref{maeq306}), we obtain

\begin{align*}
0 & =\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\\
& \geq\lim_{n\rightarrow\infty}\left (\int_{I}s(x)dx-\int_{I}s_{n}(x)dx\right )\\
& =\int_{I}s(x)dx-\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.
\end{align*}

which implies (\ref{maeq307}), and the proof is complete. \(\blacksquare\)

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

Lebesgue Functions and Integrals.

In order to mimic the lower Riemann integral, we are going to introduce the concept of Lebesgue functions. Recall that the lower Riemann sum is given by

\[L({\cal P},f)=\sum_{k=1}^{n}(x_{k}-x_{k-1})m_{k}.\]

By referring to (\ref{ma26}), the lower Riemann sum can be treated as a Riemann integral of step function \(s_{\cal P}\) given by

\[U({\cal P},f)=\int_{I}s_{\cal P}(x)dx=\sum_{k=1}^{n}\int_{a_{k}}^{b_{k}}s_{\cal P}(x)dx,\]

where the step function \(c_{\cal P}\) that depends on the choice of partition \({\cal P}\) takes values \(m_{k}\) for \(k=1,\cdots,n\) satisfying \(s_{\cal P}(x)\leq f(x)\) for all \(x\in I\). Suppose that \({\cal P}_{2}\) is finer than \({\cal P}_{1}\), i.e., \({\cal P}_{1}\subseteq {\cal P}_{2}\). We have  \(L({\cal P}_{2},f)\geq L({\cal P}_{1},f)\), i.e.,

\[s_{{\cal P}_{1}}(x)\leq s_{{\cal P}_{2}}(x)\mbox{ for all }x\in I\]

and

\[\int_{I}s_{{\cal P}_{1}}(x)dx=L({\cal P}_{1},f)\leq L({\cal P}_{2},f)=\int_{I}s_{{\cal P}_{2}}(x)dx.\]

By referring to (\ref{ma28}) and (\ref{raeq241}), we have

\begin{equation}{\label{ma29}}\tag{7}
\underline{\int}_{a}^{b}f(x)dx=\sup_{\{s:s\leq f\}}\int_{a}^{b}s(x)dx,
\end{equation}

where the supremum is taken over all step functions \(s\) satisfying \(s(x)\leq f(x)\) for all \(x\in I\). Suppose that \(f\) is continuous on \([a,b]\). Then \(f\) is Riemann-integrable, i.e.,

\[\int_{a}^{b}f(x)dx=\underline{\int}_{a}^{b}f(x)dx=\overline{\int}_{a}^{b}f(x)dx.\]

In this case, we can take a sequence \(\{{\cal P}_{n}\}_{n=1}\) of partitions of \([a,b]\) such that the following conditions are satisfied.

• \({\cal P}_{n+1}\) is finer than \({\cal P}_{n}\) for \(n=1,2,\cdots\).
• The step functions \(s_{n}\equiv s_{{\cal P}_{n}}\) established from the corresponding partitions \({\cal P}_{n}\) satisfy \(s_{n}\leq f\), \(s_{n}\leq s_{n+1}\) and\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ for all }x\in I.\]

Then, by referring to (\ref{ma29}), we can obtain

\begin{align}
\int_{a}^{b}f(x)dx & =\underline{\int}_{a}^{b}f(x)dx=\sup_{n\geq 1}\int_{a}^{b}s_{n}(x)dx\nonumber\\ =\lim_{n\rightarrow\infty}\int_{a}^{b}s_{n}(x)dx.\label{ma31}\tag{8}\end{align}

Inspired by the above settings, we introduce the concept of Lebesgue function as follows.

\begin{equation}{\label{mad200}}\tag{9}\mbox{}\end{equation}

Definition \ref{mad200}. A real-valued function \(f\) defined on an interval \(I\) is called a Lebesgue function when there exists an increasing sequence of step functions \(\{s_{n}\}_{n=1}^{\infty}\) such that the following conditions are satisfied.

• The increasing sequence \(\{s_{n}\}_{n=1}^{\infty}\) converges to \(f\) a.e. on \(I\);
• The limit \[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx\] is finite.

In this case, we also say that the sequence \(\{s_{n}\}_{n=1}\) generates the function \(f\). The integral of \(f\) over \(I\) is defined by the limit

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.\]

We first note that \(s_{n}(x)\leq f(x)\) a.e. on \(I\). If there is another increasing sequence \(\{t_{n}\}_{n=1}^{\infty}\) that generates \(f\), the integral of \(f\) should be given by

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\]

from the definition. The question is that, when

\[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx\neq\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx,\]

Definition \ref{mad200} is not well-defined. The following proposition shows that the integral of Lebesgue function is well-defined.

Proposition. Let \(f\) be a Lebesgue function defined on \(I\). Suppose that the two increasing sequences \(\{s_{n}\}_{n=1}^{\infty}\) and \(\{t_{n}\}_{n=1}^{\infty}\) generate \(f\). Then, we have

\[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx=\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx.\]

Proof. Since the sequence \(\{s_{n}\}_{n=1}^{\infty}\) satisfies the conditions in Definition \ref{mad200}, it means that the sequence \(\{s_{n}\}_{n=1}^{\infty}\) also satisfies the assumptions in Proposition \ref{map306}. Since \(t_{n}(x)\leq f(x)\) a.e. on \(I\), we have

\[\int_{I}t_{n}(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx,\]

which also implies

\[\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.\]

If we interchange the roles of \(\{s_{n}\}_{n=1}^{\infty}\) and \(\{t_{n}\}_{n=1}^{\infty}\), we can similarly obtain the following reverse inequality

\[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx.\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{map201}}\tag{10}\mbox{}\end{equation}

Proposition \ref{map201}. Let \(f\) and \(g\) be two Lebesgue functions defined on an interval \(I\). Then, we have the following properties.

(i) The addition \(f+g\) is also a Lebesgue function and

\[\int_{I}\left [f(x)+g(x)\right ]dx=\int_{I}f(x)dx+\int_{I}g(x)dx.\]

(ii)The scalar multiplication \(cf\) is also a Lebesgue function for \(c\geq 0\) and

\[\int_{I}cf(x)dx=c\int_{I}f(x)dx.\]

(iii) Suppose that \(f(x)\leq g(x)\) a.e. on \(I\). Then, we have

\begin{equation}{\label{maeq316}}\tag{11}
\int_{I}f(x)dx\leq\int_{I}g(x)dx.
\end{equation}

Moreover, if \(f(x)=g(x)\) a.e. on \(I\), we have

\begin{equation}{\label{maeq309}}\tag{12}
\int_{I}f(x)dx=\int_{I}g(x)dx.
\end{equation}

Proof. By definition, there exist two increasing sequences \(\{s_{n}\}_{n=1}^{\infty}\) and \(\{t_{n}\}_{n=1}^{\infty}\) of step functions satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on }I\mbox{ and }\lim_{n\rightarrow\infty}t_{n}(x)=g(x)\mbox{ a.e. on }I\]

and the limits

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx\mbox{ and }\int_{I}g(x)dx=\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\]

are finite. To prove part (i), Let \(u_{n}=s_{n}+t_{n}\). Then, it is clear to see that the sequence \(\{u_{n}\}_{n=1}^{\infty}\) is increasing satisfying

\begin{align*} \lim_{n\rightarrow\infty}u_{n}(x) & =\lim_{n\rightarrow\infty}\left (s_{n}(x)+t_{n}(x)\right )
\\ & =\lim_{n\rightarrow\infty}s_{n}(x)+\lim_{n\rightarrow\infty}t_{n}(x)\\ & =f(x)+g(x)\mbox{ a.e. on }I\end{align*}

and

\begin{align*}
\int_{I}\left (f(x)+g(x)\right )dx & =\lim_{n\rightarrow\infty}\int_{I}u_{n}(x)dx\\ & =\lim_{n\rightarrow\infty}\int_{I}\left (s_{n}(x)+t_{n}(x)\right )dx\\ & =\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx+\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\\ & =\int_{I}f(x)dx+\int_{I}g(x)dx.
\end{align*}

To prove part (ii), let \(u_{n}=cs_{n}\). Since \(c\geq 0\), it follows that the sequence \(\{u_{n}\}_{n=1}^{\infty}\) is increasing satisfying

\begin{align*} \lim_{n\rightarrow\infty}u_{n}(x) & =\lim_{n\rightarrow\infty}cs_{n}(x)\\ & =c\lim_{n\rightarrow\infty}s_{n}(x)=cf(x)\mbox{ a.e. on }I\end{align*}

and

\begin{align*} \int_{I}cf(x)dx & =\lim_{n\rightarrow\infty}\int_{I}u_{n}(x)dx =\lim_{n\rightarrow\infty}\int_{I}cs_{n}(x)dx\\ & =c\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx=c\int_{I}f(x)dxdx.\end{align*}

To prove part (iii), since \(s_{n}(x)\leq f(x)\leq g(x)\) a.e. on \(I\), using Proposition \ref{map306}, we have

\[\int_{I}s_{n}(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx=\int_{I}g(x)dx.\]

By taking \(n\rightarrow\infty\) on both sides, we obtain the desired inequality (\ref{maeq316}). Finally, if \(f(x)=g(x)\) a.e. on \(I\), we have \(f(x)\leq g(x)\) a.e. on \(I\) and \(g(x)\leq f(x)\) a.e. on \(I\). Therefore, we can obtain the desired equality (\ref{maeq309}), and the proof is complete. \(\blacksquare\)

We note that the condition \(c\geq 0\) in part (ii) of Proposition \ref{map201} cannot be omitted, since there is an example such that \(f\) is a Lebesgue function and \(-f\) is not a Lebesgue function. However, if \(s\) is a step function, then \(f-s=f+(-s)\) is a Lebesgue function.

\begin{equation}{\label{map313}}\tag{13}\mbox{}\end{equation}

Proposition \ref{map313}. Suppose that \(f\) and \(g\) are Lebesgue functions. Then \(\max\{f,g\}\) and \(\min\{f,g\}\) are also Lebesgue functions.

Proof. By definition, there exist two increasing sequences \(\{s_{n}\}_{n=1}^{\infty}\) and \(\{t_{n}\}_{n=1}^{\infty}\) of step functions satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on }I\mbox{ and }\lim_{n\rightarrow\infty}t_{n}(x)=g(x)\mbox{ a.e. on }I\]

and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx<+\infty\]

and

\[\int_{I}g(x)dx=\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx<+\infty.\]

We define

\[u_{n}(x)=\max\left\{s_{n}(x),t_{n}(x)\right\}\mbox{ and }v_{n}(x)=\min\left\{s_{n}(x),t_{n}(x)\right\}.\]

It is clear to see that the sequences \(\{u_{n}\}_{n=1}^{\infty}\) and \(\{v_{n}\}_{n=1}^{\infty}\) are increasing and satisfy

\begin{align*} \lim_{n\rightarrow\infty}u_{n}(x) & =\lim_{n\rightarrow\infty}\max\left\{s_{n}(x),t_{n}(x)\right\}
\\ & =\max\left\{\lim_{n\rightarrow\infty}s_{n}(x),\lim_{n\rightarrow\infty}t_{n}(x)\right\}
\\ & =\max\left\{f,g\right\}\mbox{ a.e. on \(I\)}\end{align*}

and

\begin{align*} \lim_{n\rightarrow\infty}v_{n}(x) & =\lim_{n\rightarrow\infty}\min\left\{s_{n}(x),t_{n}(x)\right\}
\\ & =\min\left\{\lim_{n\rightarrow\infty}s_{n}(x),\lim_{n\rightarrow\infty}t_{n}(x)\right\}
\\ & =\min\left\{f,g\right\}\mbox{ a.e. on \(I\)}.\end{align*}

In order to prove that \(\min\{f,g\}\) is a Lebesgue function, we remain to claim that the limit of the sequence \(\{\int_{I}v_{n}(x)dx\}_{n=1}^{\infty}\) is finite. Since

\[v_{n}(x)=\min\left\{s_{n}(x),t_{n}(x)\right\}\leq s_{n}(x)\leq f(x)\mbox{ a.e. on }I\]

part (iii) of Proposition \ref{map201} says

\[\int_{I}v_{n}(x)dx\leq\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx<+\infty .\]

This shows that \(\min\{f,g\}\) is indeed a Lebesgue function with

\[\int_{I}\min\left\{f(x),g(x)\right\}dx=\lim_{n\rightarrow\infty}\int_{I}v_{n}(x)dx<+\infty.\]

On the other hand, since

\[u_{n}+v_{n}=\max\{s_{n},t_{n}\}+\min\{s_{n},t_{n}\}=s_{n}+t_{n},\]

we have \(u_{n}=s_{n}+t_{n}-v_{n}\). Therefore, we obtain

\[\int_{I}u_{n}(x)dx=\int_{I}s_{n}(x)dx+\int_{I}t_{n}(x)dx-\int_{I}v_{n}(x)dx\]

which implies

\begin{align*}
\lim_{n\rightarrow\infty}\int_{I}u_{n}(x)dx & =\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx
+\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx-\lim_{n\rightarrow\infty}\int_{I}v_{n}(x)dx\\
& =\int_{I}f(x)dx+\int_{I}g(x)dx-\int_{I}\min\left\{f(x),g(x)\right\}dx<+\infty .
\end{align*}

This shows that \(\max\{f,g\}\) is also a Lebesgue function, and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{map315}}\tag{14}\mbox{}\end{equation}

Proposition \ref{map315}. Let \(I\) be an interval with \(I=I_{1}\cup I_{2}\), where \(I_{1}\) and \(I_{2}\) are sub-intervals of \(I\) satisfying \(I_{1}\cap I_{2}=\emptyset\). Then, we have the following properties.

(i) Suppose that \(f\) is a Lebesgue function on \(I\) and \(f\geq 0\) a.e. on \(I\). Then \(f\) is a Lebesgue function on \(I_{1}\) and \(I_{2}\) satisfying

\[\int_{I}f(x)dx=\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.\]

(ii) Suppose that \(f_{1}\) is a Lebesgue function on \(I_{1}\) and \(f_{2}\) is a Lebesgue function on \(I_{2}\). We define \(f\) on \(I\) as follows

\[f(x)=\left\{\begin{array}{ll}
f_{1}(x) & \mbox{if }x\in I_{1}\\
f_{2}(x) & \mbox{if }x\in I\setminus I_{1}.
\end{array}\right .\]

Then \(f\) is a Lebesgue function on \(I\) and

\[\int_{I}f(x)dx=\int_{I_{1}}f_{1}(x)dx+\int_{I_{2}}f_{2}(x)dx.\]

Proof. Let \(\{s_{n}\}_{n=1}^{\infty}\) be an increasing sequence of step functions satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on }I\mbox{ and }\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx<+\infty.\]

We define

\[s_{n}^{+}(x)=\max\left\{s_{n}(x),0\right\}\mbox{ for each }x\in I.\]

It is clear to see that the sequence \(\{s_{n}^{+}\}_{n=1}^{\infty}\) is increasing. For any subinterval \(J\) of \(I\), since \(f\) is nonnegative a.e. on \(I\), it follows

\begin{align*} \lim_{n\rightarrow\infty}s_{n}^{+}(x) & =\lim_{n\rightarrow\infty}\max\left\{s_{n}(x),0\right\}
\\ & =\max\left\{\lim_{n\rightarrow\infty}s_{n}(x),0\right\}\\ & =\max\left\{f(x),0\right\}=f(x)\mbox{ a.e. on }J.\end{align*}

Since \(f\) is nonnegative a.e. on \(I\) and

\begin{align*} s_{n}^{+}(x) & =\max\left\{s_{n}(x),0\right\}\\ & \leq s_{n}(x)\leq f(x)\mbox{ a.e. on }I,\end{align*}

using part (iii) of Proposition~\ref{map201}, we also have

\begin{align*} \int_{J}s_{n}^{+}(x)dx & \leq\int_{J}f(x)dx\leq\int_{I}f(x)dx\\ & =\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx<+\infty .\end{align*}

This shows that the sequence \(\{s_{n}^{+}\}\) generates \(f\) on \(J\), i.e.,

\[\lim_{n\rightarrow\infty}\int_{J}s_{n}^{+}(x)dx=\int_{J}f(x)dx.\]

Since

\[\int_{I}s_{n}^{+}(x)dx=\int_{I_{1}}s_{n}^{+}(x)dx+\int_{I_{2}}s_{n}^{+}(x)dx,\]

it follows

\begin{align*}
\int_{I}f(x)dx & =\lim_{n\rightarrow\infty}\int_{I}s_{n}^{+}(x)dx\\
& =\lim_{n\rightarrow\infty}\int_{I_{1}}s_{n}^{+}(x)dx+\lim_{n\rightarrow\infty}\int_{I_{2}}s_{n}^{+}(x)dx\\
& =\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.
\end{align*}

Part (ii) is left as an exercise. \(\blacksquare\)

The next theorem shows that the class of Lebesgue function contains all the Riemann-integrable functions. Therefore, we may say that the concept of integrals of Lebesgue functions extends the concept of Riemann integrals.

\begin{equation}{\label{mat204}}\tag{15}\mbox{}\end{equation}

Theorem \ref{mat204}. Let \(f\) be a real-valued function defined on a compact interval \([a,b]\). We also assume that \(f\) is continuous a.e. on \([a,b]\). Then \(f\) is a Lebesgue function on \([a,b]\) and the integral of \(f\) is equal to the Riemann integral of \(f\) on \([a,b]\). We can write

\[\int_{[a,b]}f(x)dx=\int_{a}^{b}f(x)dx.\]

Proof. Let \({\cal P}_{n}=\{x_{0},x_{1},\cdots ,x_{2^{n}}\}\) be a partition of \([a,b]\) into \(2^{n}\) equal sub-intervals with length \((b-a)/2^{n}\). This also says that \({\cal P}_{n+1}\) is obtained by bisecting the subintervals obtained from \({\cal P}_{n}\). We define

\[m_{k}=\inf_{x\in [x_{k-1},x_{k}]}f(x)\]

for \(1\leq k\leq 2^{n}\) and define a step function \(s_{n}\) on \([a,b]\) as follows

\[s_{n}(a)=m_{1}\mbox{ and }s_{n}(x)=m_{k}\mbox{ for }x_{k-1}<x\leq x_{k}\mbox{ for }k\geq 1.\]

Then, we have \(s_{n}(x)\leq f(x)\) for all \(x\in [a,b]\). We also see that the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is increasing, which also says that the sequence \(\{\int_{a}^{b}s_{n}(x)dx\}_{n=1}^{\infty}\) is increasing. Next, we want to claim that \(s_{n}(x)\rightarrow f(x)\) at the point \(x\) of continuity of \(f\). Given any \(\epsilon >0\), there exists \(\delta>0\) such that \(|y-x|<\delta\) implies \(|f(y)-f(x)|<\epsilon\). We define

\[m(\delta )=\inf_{y\in (x-\delta ,x+\delta )}f(y).\]

Then, we have \(f(x)-\epsilon\leq m(\delta )\), i.e., \(f(x)\leq m(\delta )+\epsilon\). There exists an integer \(N>0\) such that the partition \({\cal P}_{N}\) has a sub-interval \([x_{k-1},x_{k}]\) satisfying

\[x\in [x_{k-1},x_{k}]\mbox{ and }[x_{k-1},x_{k}]\subset (x-\delta ,x+\delta ).\]

Therefore, we obtain \(m(\delta )\leq m_{k}\), which implies

\begin{equation}{\label{ma30}}\tag{16}
s_{N}(x)=m_{k}\leq f(x)\leq m(\delta )+\epsilon\leq m_{k}+\epsilon =s_{N}(x)+\epsilon .
\end{equation}

Since \(s_{n}(x)\leq f(x)\) for all \(n\) and \(s_{N}(x)\leq s_{n}(x)\) for all \(n\geq N\) by the increasing property of the sequence \(\{s_{n}\}_{n=1}^{\infty}\), using (\ref{ma30}), we also obtain

\[s_{n}(x)\leq f(x)\leq s_{N}(x)+\epsilon\leq s_{n}(x)+\epsilon\mbox{ for }n\geq N,\]

which shows that \(s_{n}(x)\rightarrow f(x)\) as \(n\rightarrow\infty\). Since the set of discontinuities of \(f\) on \([a,b]\) has measure zero, we conclude that

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e on }[a,b].\]

Since the sequence \(\{\int_{a}^{b}s_{n}(x)dx\}_{n=1}^{\infty}\) is increasing and bounded above by \(M/(b-a)\), where \(M=\sup_{x\in [a,b]}f(x)\), it is convergent given by

\[\lim_{n\rightarrow\infty}\int_{a}^{b}s_{n}(x)dx=\sup_{n}\int_{a}^{b}s_{n}(x)dx<+\infty.\]

This shows that \(f\) is a Lebesgue function on \([a,b]\) and

\begin{equation}{\label{maeq311}}\tag{17}
\int_{[a,b]}f(x)dx=\lim_{n\rightarrow\infty}\int_{a}^{b}s_{n}(x)dx.
\end{equation}

On the other hand, we also have

\[\int_{a}^{b}s_{n}(x)dx=\sum_{k=1}^{2^{n}}m_{k}(x_{k}-x_{k-1})=L({\cal P}_{n},f),\]

where \(L({\cal P}_{n},f)\) is a lower Riemann sum. Since the Riemann integral \(\int_{a}^{b}f(x)dx\) exists  using (\ref{ma31}), it follows

\begin{equation}{\label{maeq312}}\tag{18}
\lim_{n\rightarrow\infty}\int_{a}^{b}s_{n}(x)dx=\sup_{n}\int_{a}^{b}s_{n}(x)dx=\sup_{n}L({\cal P}_{n},f)=\int_{a}^{b}f(x)dx.
\end{equation}

From (\ref{maeq311}) and (\ref{maeq312}), we complete the proof. \(\blacksquare\)

\begin{equation}{\label{mat320}}\tag{19}\mbox{}\end{equation}

Theorem \ref{mat320}. (Monotone Convergence Theorem for Step Functions). Let \(\{s_{n}\}_{n=1}^{\infty}\) be a sequence of step functions such that the following conditions are satisfied.

• The sequence \(\{s_{n}\}_{n=1}^{\infty}\) is increasing on \(I\);
• The limit \[\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx\] is finite.

Then, the sequence \(\{s_{n}\}_{n=1}^{\infty}\) converges a.e. on \(I\) to a limit function \(f\) such that \(f\) is a Lebesgue function on \(I\) and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.\]

Proof. If we define \(r_{n}=s_{n}-s_{1}\), then \(r_{n}\) is an increasing sequence of nonnegative step functions. It is obvious that the sequence \(\{r_{n}\}\) is convergent a.e. on \(I\) if and only if the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is convergent a.e. on \(I\). Therefore, we can simply assume that \(\{s_{n}\}_{n=1}^{\infty}\) is an increasing sequence of nonnegative step functions on \(I\). Let \(D\) be the set of \(x\in I\) such that the sequence \(\{s_{n}(x)\}_{n=1}^{\infty}\) is divergent. We are going to show that \(D\) has measure zero. In other words, given any \(\epsilon >0\), the set \(D\) can be covered by a countable collection of intervals whose sum of lengths is less that \(\epsilon\). Since the sequence \(\{\int_{I}s_{n}(x)dx\}_{n=1}^{\infty}\) is convergent, it is bounded by some positive constant \(M\), i.e.,

\begin{equation}{\label{ma32}}\tag{20}
\int_{I}s_{n}(x)dx\leq M\mbox{ for all }n.
\end{equation}

In this case, we define
\[t_{n}(x)=\left [\frac{\epsilon}{2M}\cdot s_{n}(x)\right ]\mbox{ for }x\in I,\]

where \([y]\) denotes the greatest integer that is less than or equal to \(y\), i.e.,

\begin{equation}{\label{ma33}}\tag{21}
t_{n}(x)\leq\frac{\epsilon}{2M}\cdot s_{n}(x)\mbox{ for }x\in I.
\end{equation}

Therefore, \(\{t_{n}\}_{n=1}^{\infty}\) is an increasing sequence of step functions such that \(t_{n}(x)\) is a nonnegative integer. Suppose that the sequence \(\{s_{n}(x)\}_{n=1}^{\infty}\) is convergent. Then the sequence \(\{s_{n}(x)\}_{n=1}^{\infty}\) is bounded, which says that the sequence \(\{t_{n}(x)\}_{n=1}^{\infty}\) is also bounded. Since \(\{t_{n}\}_{n=1}^{\infty}\) is increasing and each \(t_{n}(x)\) is a nonnegative integer, there exists \(N>0\) such that \(n\geq N\) implies \(t_{n+1}(x)=t_{n}(x)\), which says that \(\{t_{n}(x)\}_{n=1}^{\infty}\) is convergent. Suppose that the sequence \(\{s_{n}(x)\}_{n=1}^{\infty}\) is divergent. It is clear to see that the sequence \(\{t_{n}(x)\}_{n=1}^{\infty}\) is also divergent. Since each \(t_{n}(x)\) is a nonnegative integer, it follows \(t_{n+1}(x)-t_{n}(x)\geq 1\) for infinitely many values of \(n\). We define

\[D_{n}=\left\{x\in I:t_{n+1}(x)-t_{n}(x)\geq 1\right\}.\]

Since \(\{t_{n}\}_{n=1}^{\infty}\) is an increasing and bounded sequence of step functions, we see that \(D_{n}\) is the union of a finite number of intervals, and the sum of whose lengths is denoted by \(|D_{n}|\). We also have

\[D\subseteq\bigcup_{n=1}^{\infty}D_{n}.\]

Therefore, if we can show that \(\sum_{n=1}^{\infty}|D_{n}|<\epsilon\), then \(D\) has measure zero. Now, we have

\begin{align*} \int_{I}(t_{n+1}(x)-t_{n}(x))dx & \geq\int_{D_{n}}(t_{n+1}(x)-t_{n}(x))dx\\ & \geq\int_{D_{n}}1dx=|D_{n}|.\end{align*}

Therefore, for every \(m\geq 1\), using (\ref{ma32}) and (\ref{ma33}), we also have

\begin{align*}
\sum_{n=1}^{m}|D_{n}| & \leq\sum_{n=1}^{m}\int_{I}(t_{n+1}(x)-t_{n}(x))dx\\ & =\int_{I}t_{m+1}(x)dx-\int_{I}t_{1}(x)dx\\
& \leq \int_{I}t_{m+1}(x)dx\leq\frac{\epsilon}{2M}\int_{I}s_{m+1}(x)dx\\ & \leq\frac{\epsilon}{2},
\end{align*}

which also says

\[\sum_{n=1}^{\infty}|D_{n}|\leq\frac{\epsilon}{2}<\epsilon\]

by taking \(m\rightarrow\infty\). This shows that \(D\) has indeed measure \(0\). In other words, the sequence \(\{s_{n}\}_{n=1}^{\infty}\) is convergent a.e. on \(I\). Now, we define

\[f(x)=\left\{\begin{array}{ll}
{\displaystyle \lim_{n\rightarrow\infty}s_{n}(x)} & \mbox{if \(x\in I\setminus D\)}\\
0 & \mbox{if \(x\in D\)}.
\end{array}\right .\]

Then \(f\) is defined everywhere on \(I\) and

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on }I.\]

This shows that \(f\) is a Lebesgue function and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx.\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{mat329}}\tag{22}\mbox{}\end{equation}

Theorem \ref{mat329}. (Monotone Convergence Theorem for Lebesgue Functions). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue functions such that the following conditions are satisfied.

• The sequence \(\{f_{n}\}_{n=1}^{\infty}\) is increasing a.e. on \(I\);
• The limit \[\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx\] is finite.

Then, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) converges a.e. on \(I\) to a limit function \(f\) such that \(f\) is a Lebesgue function on \(I\) and

\begin{equation}{\label{maeq328}}\tag{23}
\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx.
\end{equation}

Proof. Since each \(f_{k}\) is a Lebesgue function, there exists an increasing sequence of step functions \(\{s_{n,k}\}\) which generates \(f_{k}\), i.e.,

\begin{equation}{\label{ma34}}\tag{24}
\lim_{n\rightarrow\infty}s_{n,k}(x)=f_{k}(x)\mbox{ a.e. on }I
\end{equation}

and

\[\int_{I}f_{k}(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n,k}(x)dx.\]

We define a new step function on \(I\) as follows:

\[t_{n}(x)=\max\left\{s_{n,1}(x),s_{n,2}(x),\cdots ,s_{n,n}(x)\right\}.\]

Since \(\{s_{n,k}\}\) is increasing on \(I\) for any fixed \(k\), we have

\begin{align*}
t_{n+1}(x) & =\max\left\{s_{n+1,1}(x),\cdots ,s_{n+1,n+1}(x)\right\}\\ & \geq\max\left\{s_{n,1}(x),\cdots ,s_{n,n}(x),s_{n,n+1}(x)\right\}\\
& \geq\max\left\{s_{n,1}(x),s_{n,2}(x),\cdots ,s_{n,n}(x)\right\}=t_{n}(x).
\end{align*}

This says that \(\{t_{n}\}_{n=1}^{\infty}\) is increasing on \(I\). Since \(s_{n,k}(x)\leq f_{k}(x)\) and \(\{f_{n}\}_{n=1}^{\infty}\) is increasing a.e. on \(I\), we also have

\begin{equation}{\label{maeq325}}\tag{25}
t_{n}(x)\leq\max\left\{f_{1}(x),\cdots ,f_{n}(x)\right\}=f_{n}(x)\mbox{ a.e. on \(I\).}
\end{equation}

Therefore, using part (iii) of Proposition \ref{map201}, we obtain

\begin{equation}{\label{maeq326}}\tag{26}
\int_{I}t_{n}(x)dx\leq\int_{I}f_{n}(x)dx.
\end{equation}

Since the sequence \(\{\int_{I}f_{n}(x)dx\}_{n=1}^{\infty}\) is convergent, i.e., bounded above, we see that the sequence \(\{\int_{I}t_{n}(x)dx\}_{n=1}^{\infty}\) is also bounded above and is convergent. Theorem \ref{mat320} says

\begin{equation}{\label{ma35}}\tag{27}
\lim_{n\rightarrow\infty}t_{n}(x)=f(x)\mbox{ a.e. on }I,
\end{equation}

such that \(f\) is a Lebesgue function on \(I\) and

\begin{equation}{\label{maeq459}}\tag{28}
\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx.
\end{equation}

Next, we want to prove

\begin{equation}{\label{ma37}}\tag{29}
\lim_{n\rightarrow\infty}f_{n}(x)=f(x)\mbox{ a.e. on }I.
\end{equation}

The definition of \(t_{n}(x)\) says \(s_{n,k}(x)\leq t_{n}(x)\) for all \(k\leq n\) and all \(x\in I\). Therefore, using (\ref{ma34}) and (\ref{ma35}) we obtain

\begin{equation}{\label{ma36}}\tag{30}
f_{k}(x)=\lim_{n\rightarrow\infty}s_{n,k}(x)\leq\lim_{n\rightarrow\infty}t_{n}(x)=f(x)\mbox{ a.e. on }I,
\end{equation}

which also implies

\begin{equation}{\label{maeq327}}\tag{31}
\int_{I}f_{k}(x)dx\leq\int_{I}f(x)dx
\end{equation}

by part (iii) of Proposition \ref{map201}. From (\ref{ma36}), since the increasing sequence \(\{f_{k}\}_{k=1}^{\infty}\) is bounded above by \(f\) a.e. on \(I\), which means that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) is convergent a.e. on \(I\) to a limit function \(g\) satisfying \(g(x)\leq f(x)\) a.e. on \(I\). From (\ref{maeq325}) and (\ref{ma35}), we have

\[f(x)=\lim_{n\rightarrow\infty}t_{n}(x)\leq\lim_{n\rightarrow\infty}f_{n}(x)=g(x)\mbox{ a.e. on }I.\]

This shows that \(f(x)=g(x)\) a.e. on \(I\). Therefore, we obtain (\ref{ma37}). Finally, using (\ref{maeq459}), (\ref{maeq326}) and (\ref{maeq327}) (in order), we have

\begin{align*}
\int_{I}f(x)dx & =\lim_{n\rightarrow\infty}\int_{I}t_{n}(x)dx\\
& \leq\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx\leq\int_{I}f(x)dx.
\end{align*}

Therefore, we obtain (\ref{maeq328}), and the proof is complete. \(\blacksquare\)

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

The Lebesgue-Integrable Functions.

If \(u\) and \(v\) are Lebesgue functions on \(I\), then \(u-v\) is not necessary a Lebesgue function on \(I\). We denote by \({\cal L}(I)\) the set of all functions \(f\) which has the form of \(f=u-v\), where \(u\) and \(v\) are Lebesgue functions on \(I\). Now, we can enlarge the class of integrable functions.

\begin{equation}{\label{mad203}}\tag{32}\mbox{}\end{equation}

Definition \ref{mad203}. Each function \(f\) in \({\cal L}(I)\) is said to be Lebesgue-integrable on \(I\) and its integral is defined by

\[\int_{I}f(x)dx=\int_{I}u(x)dx-\int_{I}v(x)dx.\]

Every Lebesgue function \(f\) is Lebesgue-integrable, since we can write \(f=f-0\), where \(0\) is a Lebesgue function. Each function \(f\) in \({\cal L}(I)\) can be more than one difference of Lebesgue functions. For example, we can have

\[f=u_{1}-v_{1}=u_{2}-v_{2},\]

where \(u_{1},u_{2},v_{1},v_{2}\) are all Lebesgue functions on \(I\). Therefore, by definition, there are two Lebesgue integrals for \(f\). In other words, the Definition \ref{mad203} is well-defined when

\[\int_{I}u_{1}(x)dx-\int_{I}v_{1}(x)dx=\int_{I}u_{2}(x)dx-\int_{I}v_{2}(x)dx.\]

This can be shown below.

Proposition. Let \(u_{1},u_{2},v_{1},v_{2}\) be all Lebesgue functions on \(I\) satisfying \(u_{1}-v_{1}=u_{2}-v_{2}\). Then, we have

\[\int_{I}u_{1}(x)dx-\int_{I}v_{1}(x)dx=\int_{I}u_{2}(x)dx-\int_{I}v_{2}(x)dx.\]

Proof. We have \(u_{1}+v_{2}=u_{2}+v_{1}\). By part (i) of Proposition \ref{map201}, we have

\[\int_{I}u_{1}(x)dx+\int_{I}v_{2}(x)dx=\int_{I}u_{2}(x)dx+\int_{I}v_{1}(x)dx,\]

which implies the desired equality, and the proof is complete. \(\blacksquare\)

Any Riemann-integrable function on a compact interval \([a,b]\) is also a Lebesgue function on \([a,b]\) by Theorem \ref{mat204}, which is also a Lebesgue-integrable function on \([a,b]\). In other words, the concept of Lebesgue integral extends the concept of Riemann integral. Let \(f\) be a real-valued function defined on \(I\). The positive part of \(f\) is denoted by \(f^{+}\) and is defined by

\[f^{+}(x)=\max\{f(x),0\}.\]

The negative part of \(f\) is denoted by \(f^{-}\) and is defined by

\[f^{-}(x)=\max\{-f(x),0\}.\]

We see that \(f^{+}\geq 0\) and \(f^{-}\geq 0\) satisfy

\[f(x)=f^{+}(x)-f^{-}(x)\]

and

\[|f(x)|=f^{+}(x)+f^{-}(x).\]

If \(f\) is a Lebesgue function on \(I\), then \(f^{+}=\max\{f,0\}\) is also a Lebesgue function on \(I\) by Proposition \ref{map313}. Although, the negative part \(f^{-}\) is not necessary a Lebesgue function on \(I\), the following result shows that \(f^{-}\) is Lebesgue-integrable on \(I\).

\begin{equation}{\label{map318}}\tag{33}\mbox{}\end{equation}

Proposition \ref{map318}. Let \(f\) be a Lebesgue function on \(I\). Then \(f^{+}\) is a Lebesgue function on \(I\) and \(f^{-}\) is Lebesgue-integrable on \(I\). Moreover, we have

\begin{equation}{\label{maeq319}}\tag{34}
\int_{I}f(x)dx=\int_{I}f^{+}(x)dx-\int_{I}f^{-}(x)dx,
\end{equation}

Proof. Proposition \ref{map313} says that \(f^{+}=\max\{f,0\}\) is a Lebesgue function. This says that \(f^{-}=f^{+}-f\) is Lebesgue-integrable on \(I\) by definition, which also shows the following equality

\[\int_{I}f^{-}(x)dx=\int_{I}f^{+}(x)dx-\int_{I}f(x)dx.\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{map317}}\tag{35}\mbox{}\end{equation}

Proposition \ref{map317}. Suppose that \(f\) and \(g\) are Lebesgue-integrable functions on \(I\). Then, we have the following properties.

(i) The addition \(\alpha f+\beta g\) is also a Lebesgue-integrable function on \(I\), where \(\alpha ,\beta\in\mathbb{R}\), and we have

\[\int_{I}\left [\alpha f(x)+\beta g(x)\right ]dx=\alpha\int_{I}f(x)dx+\beta\int_{I}g(x)dx.\]

(ii) If \(f(x)\geq 0\) a.e. on \(I\), then

\[\int_{I}f(x)dx\geq 0.\]

(iii) If \(g(x)\leq f(x)\) a.e. on \(I\), then

\[\int_{I}g(x)dx\leq\int_{I}f(x)dx.\]

Moreover, if \(g(x)=f(x)\) a.e. on \(I\), then

\[\int_{I}g(x)dx=\int_{I}f(x)dx.\]

Proof. To prove part (i), suppose that \(f=u-v\) and \(g=s-t\) with

\[\int_{I}f(x)dx=\int_{I}u(x)dx-\int_{I}v(x)dx\]

and

\[\int_{I}g(x)dx=\int_{I}s(x)dx-\int_{I}t(x)dx,\]

where \(u,v,s,t\) are Lebesgue functions on \(I\). We consider the following cases.

• Suppose that \(\alpha\geq 0\) and \(\beta\geq 0\). Proposition \ref{map201} says that \(p(x)\equiv\alpha u(x)+\beta s(x)\) and \(q(x)\equiv\alpha v(x)+\beta t(x)\) are Lebesgue functions. Using Definition \ref{mad203} and Proposition \ref{map201}, we have
  \begin{align*}
  & \int_{I}\left [\alpha f(x)+\beta g(x)\right ]dx\\ & =\int_{I}\left [\alpha u(x)-\alpha v(x)+\beta s(x)-\beta t(x)\right ]dx\\
  & \quad=\int_{I}\left [\left (\alpha u(x)+\beta s(x)\right )-\left (\alpha v(x)+\beta t(x)\right )\right ]dx\\ & =\int_{I}\left [p(x)-q(x)\right ]dx\\
  & \quad=\int_{I}p(x)dx-\int_{I}q(x)dx\\
  & \quad=\int_{I}\left [\alpha u(x)+\beta s(x)\right ]-\int_{I}\left [\alpha v(x)+\beta t(x)\right ]dx\\
  & \quad=\alpha\int_{I}u(x)dx+\beta\int_{I}s(x)dx-\alpha\int_{I}v(x)dx-\beta\int_{I}t(x)dx\\
  & \quad=\alpha\left (\int_{I}u(x)dx-\int_{I}v(x)dx\right )+\beta\left (\int_{I}s(x)dx-\int_{I}t(x)dx\right )\\
  & \quad=\alpha\int_{I}f(x)dx+\beta\int_{I}g(x)dx.
  \end{align*}
• Suppose that \(\alpha\geq 0\) and \(\beta<0\). Proposition \ref{map201} says that \(p(x)\equiv\alpha u(x)-\beta t(x)\) and \(q(x)\equiv\alpha v(x)-\beta s(x)\) are Lebesgue functions. Using Definition \ref{mad203} and Proposition \ref{map201}. we have
  \begin{align*}
  & \int_{I}\left [\alpha f(x)+\beta g(x)\right ]dx\\ & =\int_{I}\left [\alpha u(x)-\alpha v(x)+\beta s(x)-\beta t(x)\right ]dx\\
  & \quad=\int_{I}\left [\left (\alpha u(x)-\beta t(x)\right )-\left (\alpha v(x)-\beta s(x)\right )\right ]dx\\ & =\int_{I}\left [p(x)-q(x)\right ]dx\\
  & \quad=\int_{I}p(x)dx-\int_{I}q(x)dx\\
  & \quad=\int_{I}\left [\alpha u(x)-\beta t(x)\right ]-\int_{I}\left [\alpha v(x)-\beta s(x)\right ]dx\\
  & \quad=\alpha\int_{I}u(x)dx-\beta\int_{I}t(x)dx-\alpha\int_{I}v(x)dx+\beta\int_{I}s(x)dx\\
  & \quad=\alpha\left (\int_{I}u(x)dx-\int_{I}v(x)dx\right )+\beta\left (\int_{I}s(x)dx-\int_{I}t(x)dx\right )\\
  & \quad=\alpha\int_{I}f(x)dx+\beta\int_{I}g(x)dx.
  \end{align*}
• Suppose that \(\alpha<0\) and \(\beta\geq 0\). Proposition \ref{map201} says that  \(p(x)\equiv-\alpha v(x)+\beta s(x)\) and \(q(x)\equiv-\alpha u(x)+\beta t(x)\) are Lebesgue functions. Using Definition \ref{mad203} and Proposition \ref{map201}, we have
  \begin{align*}
  & \int_{I}\left [\alpha f(x)+\beta g(x)\right ]dx\\ & =\int_{I}\left [\alpha u(x)-\alpha v(x)+\beta s(x)-\beta t(x)\right ]dx\\
  & \quad=\int_{I}\left [\left (-\alpha v(x)+\beta s(x)\right )-\left (-\alpha u(x)+\beta t(x)\right )\right ]dx\\ & =\int_{I}\left [p(x)-q(x)\right ]dx\\
  & \quad=\int_{I}p(x)dx-\int_{I}q(x)dx\\
  & \quad=\int_{I}\left [-\alpha v(x)+\beta s(x)\right ]-\int_{I}\left [-\alpha u(x)+\beta t(x)\right ]dx\\
  & \quad=-\alpha\int_{I}v(x)dx+\beta\int_{I}s(x)dx+\alpha\int_{I}u(x)dx-\beta\int_{I}t(x)dx\\
  & \quad=\alpha\left (\int_{I}u(x)dx-\int_{I}v(x)dx\right )+\beta\left (\int_{I}s(x)dx-\int_{I}t(x)dx\right )\\
  & \quad=\alpha\int_{I}f(x)dx+\beta\int_{I}g(x)dx.
  \end{align*}
• Suppose that \(\alpha<0\) and \(\beta<0\). Proposition \ref{map201} says \(p(x)\equiv-\alpha v(x)-\beta t(x)\) and \(q(x)\equiv-\alpha u(x)-\beta s(x)\) are Lebesgue functions. Using Definition \ref{mad203} and Proposition \ref{map201}, we have
  \begin{align*}
  & \int_{I}\left [\alpha f(x)+\beta g(x)\right ]dx\\ & =\int_{I}\left [\alpha u(x)-\alpha v(x)+\beta s(x)-\beta t(x)\right ]dx\\
  & \quad=\int_{I}\left [\left (-\alpha v(x)-\beta t(x)\right )-\left (-\alpha u(x)-\beta s(x)\right )\right ]dx\\ & =\int_{I}\left [p(x)-q(x)\right ]dx\\
  & \quad=\int_{I}p(x)dx-\int_{I}q(x)dx\\
  & \quad=\int_{I}\left [-\alpha v(x)-\beta t(x)\right ]-\int_{I}\left [-\alpha u(x)-\beta s(x)\right ]dx\\
  & \quad=-\alpha\int_{I}v(x)dx-\beta\int_{I}t(x)dx+\alpha\int_{I}u(x)dx+\beta\int_{I}s(x)dx\\
  & \quad=\alpha\left (\int_{I}u(x)dx-\int_{I}v(x)dx\right )+\beta\left (\int_{I}s(x)dx-\int_{I}t(x)dx\right )\\
  & \quad=\alpha\int_{I}f(x)dx+\beta\int_{I}g(x)dx.
  \end{align*}

To prove part (ii), we write \(f=u-v\), where \(u\) and \(v\) are Lebesgue functions. Since \(f(x)\geq 0\) a.e. on \(I\), it follows \(u(x)\geq v(x)\) a.e. on \(I\). By part (iii) of Proposition \ref{map201}, we have

\[\int_{I}f(x)dx=\int_{I}u(x)dx-\int_{I}v(x)dx\geq 0.\]

To prove part (iii), let \(h=f-g\). Then \(h(x)\geq 0\) a.e. on \(I\). Therefore, we have

\begin{align*}
\int_{I}f(x)dx-\int_{I}g(x)dx & =\int_{I}\left [f(x)-g(x)\right ]dx\mbox{ (using part (i))}\\
& =\int_{I}h(x)dx\geq 0\mbox{(using part (ii))}.
\end{align*}

Finally, since \(g(x)=f(x)\) a.e. on \(I\) if and only if \(g(x)\leq f(x)\) a.e. on \(I\) and \(g(x)\geq f(x)\) a.e. on \(I\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{map331}}\tag{36}\mbox{}\end{equation}

Proposition \ref{map331}. We have the following properties.

(i) Suppose that \(f\) is Lebesgue-integrable on \(I\). Then \(f^{+}\), \(f^{-}\) and \(|f|\) are also Lebesgue-integrable on \(I\). We also have

\begin{equation}{\label{maeq314}}\tag{37}
\left |\int_{I}f(x)dx\right |\leq\int_{I}|f(x)|dx.
\end{equation}

(ii) Suppose that \(f\) and \(g\) are Lebesgue-integrable functions on \(I\). Then \(\max\{f,g\}\) and \(\min\{f,g\}\) are also Lebesgue-integrable functions on \(I\).

Proof. We write \(f=u-v\), where \(u\) and \(v\) are Lebesgue functions on \(I\). we also have

\[f^{+}=\max\left\{u-v,0\right\}=\max\left\{u-v,v-v\right\}=\max\left\{u,v\right\}-v.\]

Since \(\max\{u,v\}\) is a Lebesgue functions by Proposition \ref{map313} and \(v\) is a Lebesgue function, it says that \(f^{+}\) is Lebesgue-integrable by definition. Since \(f^{-}=f^{+}-f\), it also says that \(f^{-}\) is Lebesgue-integrable by part (i) of Proposition \ref{map317}. Finally, since \(|f|=f^{+}+f^{-}\),
it follows that \(|f|\) is Lebesgue-integrable using part (i) of Proposition \ref{map317} again. Since \(-|f(x)|\leq f(x)\leq |f(x)|\) for all \(x\in I\), we have

\[-\int_{I}|f(x)|dx\leq\int_{I}f(x)dx\leq\int_{I}|f(x)|dx,\]

which implies (\ref{maeq314}). This proves part (i). On the other hand, part (ii) follows from the following relations

\[\max\left\{f,g\right\}=\frac{1}{2}\left (f+g+|f-g|\right )\]

and

\[\min\left\{f,g\right\}=\frac{1}{2}\left (f+g-|f-g|\right ).\]

This completes the proof. \(\blacksquare\)

For any \(c\in\mathbb{R}\), we define

\[I+c=\{x+c:x\in I\}\mbox{ and }cI=\{cx:x\in I\}.\]

Proposition. Suppose that \(f\) is Lebesgue-integrable on \(I\). Then, we have the following properties.

(i) Suppose that \(g(x)=f(x-c)\) for \(x\in I+c\). Then \(g\) is Lebesgue-integrable on \(I+c\) and

\[\int_{I+c}g(x)dx=\int_{I}f(x)dx.\]

(ii) Suppose that \(g(x)=f(x/c)\) for \(x\in cI\) and \(c>0\). Then \(g\) is Lebesgue-integrable on \(cI\) and

\[\int_{cI}g(x)dx=c\int_{I}f(x)dx.\]

(iii) Suppose that \(g(x)=f(-x)\) for \(x\in -I\). Then \(g\) is Lebesgue-integrable on \(-I\) and

\[\int_{-I}g(x)dx=\int_{I}f(x)dx.\]

Proof. The procedure for proving these results is as follows. We first verify the results for step functions, then for Lebesgue functions, and finally for Lebesgue-integrable functions. The detailed proof is left as an exercise. \(\blacksquare\)

\begin{equation}{\label{map324}}\tag{38}\mbox{}\end{equation}

Proposition \ref{map324}. Let \(I\) be an interval with \(I=I_{1}\cup I_{2}\), where \(I_{1}\) and \(I_{2}\) are sub-intervals of \(I\) satisfying \(I_{1}\cap I_{2}=\emptyset\). Suppose that \(f\) is a Lebesgue function on \(I\). Then, we have the following properties.

(i) \(f^{+}\) and \(-f^{-}\) are Lebesgue functions on \(I\).

(ii) \(f\) is a Lebesgue function on \(I_{1}\) and \(I_{2}\) satisfying

\[\int_{I}f(x)dx=\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.\]

Proof. Since \(f^{+}=\max\{f,0\}\) is a nonnegative Lebesgue function on \(I\) by Proposition \ref{map313}, we have

\begin{equation}{\label{maeq322}}\tag{39}
\int_{I}f^{+}(x)dx=\int_{I_{1}}f^{+}(x)dx+\int_{I_{2}}f^{+}(x)dx.
\end{equation}

by part (i) of Proposition\ref{map315}. Next, we are going to show

\begin{equation}{\label{maeq321}}\tag{40}
\int_{I}f^{-}(x)dx=\int_{I_{1}}f^{-}(x)dx+\int_{I_{2}}f^{-}(x)dx
\end{equation}

by monotone convergence theorem. It is obvious that \(f\) is also a Lebesgue function on any sub-interval \(J\) of \(I\). Let \(\{s_{n}\}_{n=1}^{\infty}\) be an increasing sequence of step functions that generates \(f\) on \(J\), i.e.,

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on }J\]

and

\[\int_{J}f(x)dx=\lim_{n\rightarrow\infty}\int_{J}s_{n}(x)dx.\]

We define \(s_{n}^{-}=\max\{-s_{n},0\}\). Then, we have

\begin{align*}
\lim_{n\rightarrow\infty}s_{n}^{-}(x) & =\lim_{n\rightarrow\infty}\max\left\{-s_{n}(x),0\right\}\\
& =\max\left\{-\lim_{n\rightarrow\infty}s_{n}(x),0\right\}\\ & =\max\left\{-f(x),0\right\}=f^{-}(x)\mbox{ a.e. on }J,
\end{align*}

which implies

\begin{equation}{\label{ma38}}\tag{41}
\lim_{n\rightarrow\infty}-s_{n}^{-}(x)=-f^{-}(x)\mbox{ a.e. on }J,
\end{equation}

From Proposition \ref{map318}, we also see that \(-f^{-}\) is Lebesgue-integrable on \(J\). Since \(\{-s_{n}^{-}\}_{n=1}^{\infty}\) is an increasing sequence of step functions on \(J\), using (\ref{ma38}), we have

\[\int_{J}(-s_{n}^{-}(x))dx\leq\int_{J}(-f^{-}(x))dx<+\infty ,\]

which also says that the limit of sequence \(\{\int_{J}(-s_{n}^{-}(x))dx\}_{n=1}^{\infty}\) exists. Using the monotone convergence Theorem \ref{mat320}, we conclude that \(-f^{-}\) is a Lebesgue function on \(J\) and

\[\int_{J}(-f^{-}(x))dx=\lim_{n\rightarrow\infty}\int_{J}(-s_{n}^{-}(x))dx.\]

Using part (i) of Proposition \ref{map317}, we obtain

\begin{equation}{\label{maeq323}}\tag{42}
\int_{J}f^{-}(x)dx=\lim_{n\rightarrow\infty}\int_{J}s_{n}^{-}(x)dx.
\end{equation}

Since \(s_{n}^{-}\) is a step function on \(I\), we have

\[\int_{I}s_{n}^{-}(x)dx=\int_{I_{1}}s_{n}^{-}(x)dx+\int_{I_{2}}s_{n}^{-}(x)dx,\]

Using (\ref{maeq323}), we obtain

\begin{align*}
\int_{I}f^{-}(x)dx & =\lim_{n\rightarrow\infty}\int_{I}s_{n}^{-}(x)dx
\\ & =\lim_{n\rightarrow\infty}\int_{I_{1}}s_{n}^{-}(x)dx+\lim_{n\rightarrow\infty}\int_{I_{2}}s_{n}^{-}(x)dx\\
& =\int_{I_{1}}f^{-}(x)dx+\int_{I_{2}}f^{-}(x)dx,
\end{align*}

which proves the equality (\ref{maeq321}). Finally, using using Proposition \ref{map318}, (\ref{maeq322}), (\ref{maeq321}) and Proposition \ref{map318} (in order), we have

\begin{align*}
\int_{I}f(x)dx & =\int_{I}f^{+}(x)dx-\int_{I}f^{-}(x)dx\\
& =\left (\int_{I_{1}}f^{+}(x)dx+\int_{I_{2}}f^{+}(x)dx\right )-\left (\int_{I_{1}}f^{-}(x)dx+\int_{I_{2}}f^{-}(x)dx\right )\\
& =\left (\int_{I_{1}}f^{+}(x)dx-\int_{I_{1}}f^{-}(x)dx\right )+\left (\int_{I_{2}}f^{+}(x)dx-\int_{I_{2}}f^{-}(x)dx\right )\\
& =\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.
\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{map236}}\tag{43}\mbox{}\end{equation}

Proposition \ref{map236}. Let \(I\) be an interval with \(I=I_{1}\cup I_{2}\), where \(I_{1}\) and \(I_{2}\) are subintervals of \(I\) satisfying \(I_{1}\cap I_{2}=\emptyset\). Then, we have the following properties.

(i) Suppose that \(f\) is a Lebesgue-integrable on \(I\). Then \(f\) is also Lebesgue-integrable on \(I_{1}\) and \(I_{2}\) satisfying

\[\int_{I}f(x)dx=\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.\]

(ii) Suppose that \(f_{1}\) is Lebesgue-integrable on \(I_{1}\) and \(f_{2}\) is Lebesgue-integrable on \(I_{2}\). We define \(f\) on \(I\) as follows

\[f(x)=\left\{\begin{array}{ll}
f_{1}(x) & \mbox{if }x\in I_{1} \\
f_{2}(x) & \mbox{if }x\in I\setminus I_{2} .
\end{array}\right .\]

Then \(f\) is Lebesgue-integrable on \(I\) and

\[\int_{I}f(x)dx=\int_{I_{1}}f_{1}(x)dx+\int_{I_{2}}f_{2}(x)dx.\]

Proof. We write \(f=u-v\), where \(u\) and \(v\) are Lebesgue functions on \(I\). Since \(u=u^{+}-u^{-}\) and \(v=v^{+}-v^{-}\), using Proposition \ref{map324}, we have

\begin{align*}
\int_{I}f(x)dx & =\int_{I}u(x)dx-\int_{I}v(x)dx\\
& =\left (\int_{I_{1}}u(x)dx+\int_{I_{2}}u(x)dx\right )-\left (\int_{I_{1}}v(x)dx+\int_{I_{2}}v(x)dx\right )\\
& =\left (\int_{I_{1}}u(x)dx-\int_{I_{1}}v(x)dx\right )+\left (\int_{I_{2}}u(x)dx-\int_{I_{2}}v(x)dx\right )\\
& =\int_{I_{1}}f(x)dx+\int_{I_{2}}f(x)dx.
\end{align*}

Part (ii) is left as an exercise by using part (ii) of Proposition \ref{map315}. This completes the proof. \(\blacksquare\)

\begin{equation}{\label{map330}}\tag{44}\mbox{}\end{equation}

Proposition \ref{map330}. Suppose that \(f\) is Lebesgue-integrable on \(I\). Given any \(\epsilon >0\), we have the following properties.

(i) There exist Lebesgue functions \(u\) and \(v\) on \(I\) satisfying

\[f=u-v\mbox{ with }v\geq 0\mbox{ and }\int_{I}v(x)dx<\epsilon.\]

(ii) There exists a step function \(s\) and a Lebesgue-integrable function \(g\) on \(I\) satisfying

\[f=s+g\mbox{ and }\int_{I}g(x)dx<\epsilon.\]

Proof. To prove part (i), since \(f\) is Lebesgue-integrable on \(I\), we can write \(f=\hat{u}-\hat{v}\), where \(\hat{u}\) and \(\hat{v}\) are Lebesgue functions on \(I\). Let \(\{t_{n}\}_{n=1}^{\infty}\) be an increasing sequence of step functions that generates \(\hat{v}\), i.e.,

\[\lim_{n\rightarrow\infty}t_{n}(x)=\hat{v}\mbox{ a.e. on }I\mbox{ and } \lim_{n\rightarrow\infty}\int_{I}t_{n}(dx)dx=\int_{I}\hat{v}(x)dx.\]

There exists an integer \(N\) satisfying

\[0\leq\int_{I}\left [\hat{v}(x)-t_{N}(x)\right ]dx<\epsilon .\]

Let \(v=\hat{v}-t_{N}\) and \(u=\hat{u}-t_{N}\). Then, both \(u\) and \(v\) are also Lebesgue functions on \(I\), since \(t_{N}\) is a step function. We also have

\[u-v=\hat{u}-\hat{v}=f\mbox{ with }v\geq 0\mbox{ and }0\leq\int_{I}v(x)dx<\epsilon .\]

To prove part (ii), using part (i), we can choose the Lebesgue functions \(u\) and \(v\) with \(v\geq 0\) satisfying

\[f=u-v\mbox{ and }0\leq\int_{I}v(x)dx<\frac{\epsilon}{2}.\]

Let \(\{s_{n}\}_{n=1}^{\infty}\) be an increasing sequence of step functions that generates \(u\), i.e.,

\[\lim_{n\rightarrow\infty}s_{n}(x)=u\mbox{ a.e. on }I\mbox{ and }\lim_{n\rightarrow\infty}\int_{I}s_{n}(dx)dx=\int_{I}u(x)dx.\].

Therefore, there exists an integer \(N\) satisfying

\[0\leq\int_{I}\left [u(x)-s_{N}(x)\right ]dx<\frac{\epsilon}{2}.\]

Then, we obtain

\[f=u-v=s_{N}+(u-s_{N})-v\equiv s_{N}+g,\]

where \(g=(u-s_{N})-v\). Since \(u-s_{N}\) is a Lebesgue function on \(I\), we see that \(g\) is Lebesgue-integrable on \(I\). We also have

\begin{align*}
\int_{I}|g(x)|dx & \leq\int_{I}|u(x)-s_{N}(x)|dx+\int_{I}|v(x)|dx\\
& =\int_{I}|u(x)-s_{N}(x)|dx+\int_{I}v(x)dx<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}

Since \(s_{N}\) is a step function on \(I\), this completes the proof. \(\blacksquare\)

\begin{equation}{\label{map338}}\tag{45}\mbox{}\end{equation}

Proposition \ref{map338}. We have the following properties.

(i) Suppose that \(f(x)=0\) a.e. on \(I\). Then \(f\) is a Lebesgue function and Lebesgue-integrable on \(I\) with

\[\int_{I}f(x)dx=0.\]

(ii) Suppose that \(f\) is Lebesgue-integrable on \(I\) and \(g(x)=f(x)\) a.e. on \(I\). Then \(g\) is also Lebesgue-integrable on \(I\) and

\[\int_{I}f(x)dx=\int_{I}g(x)dx.\]

Proof. To prove part (i), let \(s_{n}(x)=0\) for all \(x\in I\). Then \(\{s_{n}\}_{n=1}^{\infty}\) is an increasing sequence of step functions that converges to \(0\) everywhere on \(I\). Since \(f(x)=0\) a.e. on \(I\), we also see that \(\{s_{n}\}_{n=1}^{\infty}\) converges to \(f\) a.e. on \(I\). Since \(\int_{I}s_{n}(x)dx=0\) for all \(n\), the sequence \(\{\int_{I}s_{n}(x)dx\}_{n=1}^{\infty}\) is convergent. This says that \(f\) is a Lebesgue function, i.e., Lebesgue-integrable on \(I\) with

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}s_{n}(x)dx=0.\]

To prove part (ii), since \(f-g=0\) a.e. on \(I\), using part (i), we see that \(f-g\) is Lebesgue-integrable on \(I\) and

\[\int_{I}\left [f(x)-g(x)\right ]dx=0.\]

Using Proposition \ref{map317}, we conclude that \(g=f-(f-g)\) is Lebesgue-integrable on \(I\) and

\[\int_{I}g(x)dx=\int_{I}f(x)dx-\int_{I}\left [f(x)-g(x)\right ]dx=\int_{I}f(x)dx.\]

This completes the proof. \(\blacksquare\)

Example. We define a function \(f\) on \([0,1]\) as follows

\[f(x)=\left\{\begin{array}{ll}
1 & \mbox{if }x\in\mathbb{Q}\\
0 & \mbox{if }x\not\in\mathbb{Q}.
\end{array}\right .\]

Then \(f\) is not Riemann-integrable on \([0,1]\). Since \(f(x)=0\) a.e. on \([0,1]\), Proposition \ref{map338} says that \(f\) is Lebesgue-integrable on \([0,1]\) and
\[\int_{[0,1]}f(x)dx=\int_{[0,1]}0dx=0.\]

\begin{equation}{\label{mat232}}\tag{46}\mbox{}\end{equation}

Theorem \ref{mat232}. (Monotone Convergence Theorem for Series of Lebesgue-Integrable Functions).
Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue-integrable functions such that the following conditions are satisfied.

• Each \(f_{n}\) is nonnegative a.e. on \(I\).
• The series
  \[\sum_{k=1}^{\infty}\int_{I}f_{k}(x)dx\]
  is convergent.

Then, the series \(\sum_{k=1}^{\infty}f_{k}\) is convergent a.e. on \(I\) to a sum function \(f\) such that \(f\) is a Lebesgue-integrable function on \(I\) and

\[\int_{I}f(x)dx=\int_{I}\left [\sum_{k=1}^{\infty}f_{k}(x)\right ]dx=\sum_{k=1}^{\infty}\int_{I}f_{k}(x)dx.\]

Proof. Since each \(f_{n}\) is Lebesgue-integrable on \(I\), given \(\epsilon =1/2^{n}\), Proposition \ref{map330} says that we can write \(f_{n}=u_{n}-v_{n}\) such that \(u_{n}\) and \(v_{n}\) are Lebesgue functions on \(I\) with

\[v_{n}\geq 0\mbox{ and }\int_{I}v_{n}(x)dx<1/2^{n}.\]

This also says that the series \(\sum_{n=1}^{\infty}\int_{I}v_{n}(x)dx\) is convergent. Since \(u_{n}=f_{n}+v_{n}\) is nonnegative a.e. on \(I\), the partial sum

\[U_{n}(x)=\sum_{k=1}^{n}u_{k}(x)\]

form a sequence of Lebesgue functions \(\{U_{n}\}_{n=1}^{\infty}\) which is increasing a.e. on \(I\). Now, we have

\begin{align}
\int_{I}U_{n}(x)dx & =\int_{I}\left [\sum_{k=1}^{n}u_{k}(x)dx\right ]=\sum_{k=1}^{n}
\int_{I}u_{k}(x)dx\nonumber\\ &=\sum_{k=1}^{n}\int_{I}f_{k}(x)dx+\sum_{k=1}^{n}\int_{I}v_{k}(x)dx.\label{ma39}\tag{47}
\end{align}

Since the series \(\sum_{k=1}^{\infty}\int_{I}f_{k}(x)dx\) and \(\sum_{k=1}^{\infty}\int_{I}v_{k}(x)dx\) are convergent, it follows that the sequence \(\{\int_{I}U_{n}(x)dx\}_{n=1}^{\infty}\) is also convergent. Therefore, using the monotone convergence Theorem \ref{mat329}, the sequence \(\{U_{n}\}_{n=1}^{\infty}\) is convergent a.e. on \(I\) to a limit function \(U\) such that \(U\) is a Lebesgue function on \(I\) and

\begin{equation}{\label{ma41}}\tag{48}
\int_{I}U(x)dx=\lim_{n\rightarrow\infty}\int_{I}U_{n}(x)dx.
\end{equation}

Similarly, the sequence of partial sums \(\{V_{n}\}_{n=1}^{\infty}\) defined by

\[V_{n}(x)=\sum_{k=1}^{n}v_{k}(x)\]

is convergent a.e. on \(I\) to a limit function \(V\) such that \(V\) is a Lebesgue function on \(I\) and

\begin{equation}{\label{ma40}}\tag{49}
\int_{I}V_{n}(x)dx=\sum_{k=1}^{n}
\int_{I}v_{k}(x)dx\mbox{ and }\int_{I}V(x)dx=\lim_{n\rightarrow\infty}\int_{I}v_{n}(x)dx.
\end{equation}

Therefore, the sequence

\begin{align*} \left\{\sum_{k=1}^{n}f_{k}\right\}_{n=1}^{\infty} & =\left\{\sum_{k=1}^{n}u_{k}-
\sum_{k=1}^{n}v_{k}\right\}_{n=1}^{\infty}\\ & =\left\{U_{n}-V_{n}\right\}_{n=1}^{\infty}\end{align*}

is convergent a.e. on \(I\) to \(U-V=f\). Since \(U\) and \(V\) are Lebesgue functions, using (\ref{ma41}), (\ref{ma40}), (\ref{ma39}) and (\ref{ma40}) (in order), \(f=U-V\) is Lebesgue-integrable on \(I\) with

\begin{align*}
\int_{I}f(x) & =\int_{I}U(x)dx-\int_{I}V(x)dx\\
& =\lim_{n\rightarrow\infty}\int_{I}U_{n}(x)dx-\lim_{n\rightarrow\infty}\int_{I}V_{n}(x)dx\\
& =\lim_{n\rightarrow\infty}\left (\int_{I}U_{n}(x)dx-\int_{I}V_{n}(x)dx\right )\\
& =\lim_{n\rightarrow\infty}\left (\sum_{k=1}^{n}\int_{I}u_{k}(x)dx-\sum_{k=1}^{n}\int_{I}v_{k}(x)dx\right )\\
& =\lim_{n\rightarrow\infty}\left (\sum_{k=1}^{n}\int_{I}\left [u_{k}(x)-v_{k}(x)\right ]dx\right )\\ & =\lim_{n\rightarrow\infty}\left (\sum_{k=1}^{n}\int_{I}f_{k}(x)dx\right )=\sum_{n=1}^{\infty}\int_{I}f_{n}(x)dx.
\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{mat220}}\tag{50}\mbox{}\end{equation}

Theorem \ref{mat220} (Monotone Convergence Theorem for Lebesgue-Integrable Functions). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue-integrable functions such that the following conditions are satisfied.

• The sequence \(\{f_{n}\}_{n=1}^{\infty}\) is increasing a.e. on \(I\).
• The limit
  \[\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx\]
  exists.

Then, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) converges a.e. on \(I\) to a limit function \(f\) such that \(f\) is a Lebesgue-integrable function on \(I\) and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx.\]

Proof. We define \(g_{1}=f_{1}\) and \(g_{n}=f_{n}-f_{n-1}\geq 0\) for \(n\geq 2\). Then, we have

\[f_{n}=\sum_{k=1}^{n}g_{k}.\]

Using the second condition, the following series

\begin{align*} \sum_{k=1}^{\infty}\int_{I}g_{k}(x)dx & =\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\int_{I}g_{k}(x)dx\\ & =\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx\end{align*}

is convergent. Using Theorem~\ref{mat232} to the sequence \(\{g_{n}\}_{n=1}^{\infty}\), it follows that the series \(\sum_{k=1}^{\infty}g_{k}\) is convergent a.e. on \(I\), and we have

\begin{align*} \int_{I}f(x)dx & =\int_{I}\left [\sum_{k=1}^{\infty}g_{k}(x)\right ]dx\\ & =\sum_{k=1}^{\infty}\int_{I}g_{k}(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx.\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{maex226}}\tag{51}\mbox{}\end{equation}

Example \ref{maex226}. Let \(f(x)=x^{s}\) for \(x>0\) and \(f(0)=0\). We are going to claim that \(f\) is Lebesgue-integrable on \([0,1]\) when \(s>-1\), and its Lebesgue integral is \(1/(s+1)\). If \(s\geq 0\), then \(f\) is bounded on \([0,1]\), which says that \(f\) is Riemann-integrable on \([0,1]\) and its Riemann integral is \(1/(s+1)\). If \(s<0\), then \(f\) is not bounded on \([0,1]\). Therefore, \(f\) is not Riemann-integrable on \([0,1]\). We define a sequence of functions \(\{f_{n}\}_{n=1}^{\infty}\) as follows:

\[f_{n}(x)=\left\{\begin{array}{ll}
x^{s} & \mbox{if \(x\geq 1/n\)}\\
0 & \mbox{if \(0\leq x<1/n\).}
\end{array}\right .\]

Then, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) is increasing on \([0,1]\), and \(f_{n}\) converges to \(f\) everywhere on \([0,1]\). We also see that each \(f_{n}\) is Riemann-integrable, i.e., Lebesgue-integrable on \([0,1]\) with the integral

\begin{align*} \int_{[0,1]}f_{n}(x)dx & =\int_{0}^{1}f_{n}(x)dx=\int_{1/n}^{1}x^{s}dx\\ & =\frac{1}{s+1}\left (1-\frac{1}{n^{s+1}}\right ).\end{align*}

If \(s+1>0\), the sequence \(\{\int_{1}^{1}f_{n}(x)dx\}_{n=1}^{\infty}\) converges to \(1/(s+1)\). Therefore, Theorem \ref{mat220} says that \(f\) is Lebesgue-integrable on \([0,1]\) and

\[\int_{[0,1]}f(x)dx=\lim_{n\rightarrow\infty}\int_{[0,1]}f_{n}(x)dx=\frac{1}{s+1}.\]

\begin{equation}{\label{maex227}}\tag{52}\mbox{}\end{equation}

Example \ref{maex227}. The similar arguments in Example \ref{maex226} can be used to show that the Lebesgue integral

\[\int_{[0,1]}e^{-x}x^{y-1}dx=\int_{0}^{1}e^{-x}x^{y-1}dx\]

exists for every \(y>0\). \(\sharp\)

\begin{equation}{\label{mat334}}\tag{53}\mbox{}\end{equation}

Theorem \ref{mat334}. (Dominated Convergence Theorem). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue-integrable functions on \(I\).
Suppose that the following conditions are satisfied.

• The sequence \(\{f_{n}\}_{n=1}^{\infty}\) is convergent a.e. on \(I\) to a limit function \(f\).
• There is a nonnegative and Lebesgue-integrable function \(g\) on \(I\) satisfying \(|f_{n}(x)|\leq g(x)\) a.e. on \(I\) for all \(n\).

Then, the limit function \(f\) is Lebesgue-integrable on \(I\) and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx.\]

Proof. We define a sequence \(\{G_{n,1}\}_{n=1}^{\infty}\) as follows

\[G_{n,1}(x)=\max\left\{f_{1}(x),f_{2}(x),\cdots ,f_{n}(x)\right\}.\]

Since \(f_{k}\) are Lebesgue-integrable on \(I\) for \(k=1,\cdots ,n\), part (ii) of Proposition \ref{map331} says that the function \(G_{n,1}\) is also Lebesgue-integrable on \(I\). We also see that the sequence \(\{G_{n,1}\}_{n=1}^{\infty}\) is increasing on \(I\). It follows that the sequence \(\{\int_{I}G_{n,1}(x)dx\}_{n=1}^{\infty}\) is also increasing. Since \(|G_{n,1}(x)|\leq g(x)\) a.e. on \(I\), we have

\begin{equation}{\label{maeq332}}\tag{54}
\left |\int_{I}G_{n,1}(x)dx\right |\leq\int_{I}\left |G_{n,1}(x)\right |dx\leq\int_{I}g(x)dx.
\end{equation}

This says that the sequence \(\{\int_{I}G_{n,1}(x)dx\}_{n=1}^{\infty}\) is bounded above by \(\int_{I}g(x)dx\). Therefore, the sequence \(\{\int_{I}G_{n,1}(x)dx\}_{n=1}^{\infty}\) is convergent. By the monotone convergence Theorem \ref{mat220}, the sequence \(\{G_{n,1}\}_{n=1}^{\infty}\) is convergent a.e. on \(I\) to a limit function \(G_{1}\) such that \(G_{1}\) is Lebesgue-integrable on \(I\) and
\[\int_{I}G_{1}(x)dx=\lim_{n\rightarrow\infty}\int_{I}G_{n,1}(x)dx\leq\int_{I}g(x)dx.\]

From (\ref{maeq332}), we also have

\[-\int_{I}g(x)dx\leq\int_{I}G_{n,1}(x)dx\mbox{ for all }n,\]

which imply

\[-\int_{I}g(x)dx\leq\lim_{n\rightarrow\infty}\int_{I}G_{n,1}(x)dx=\int_{I}G_{1}(x)dx.\]

Therefore, we obtain

\[-\int_{I}g(x)dx\leq\int_{I}G_{1}(x)dx\leq\int_{I}g(x)dx.\]

Since the sequence \(\{G_{n,1}\}_{n=1}^{\infty}\) is increasing on \(I\) and

\[\lim_{n\rightarrow\infty}G_{n,1}(x)=G_{1}(x)\mbox{ a.e. on }I,\]

it follows that

\[G_{1}(x)=\sup\left\{f_{1}(x),f_{2}(x),\cdots\right\}\mbox{ a.e. on }I.\]

Using the same technique, for each fixed \(r\geq 1\), we define

\[G_{n,r}(x)=\max\left\{f_{r}(x),f_{r+1}(x),\cdots ,f_{n}(x)\right\}\]

for \(r\leq n\). Then, the sequence \(\{G_{n,r}\}_{n=1}^{\infty}\) is increasing and convergent a.e. on \(I\) to a limit function \(G_{r}\) such that \(G_{r}\) is Lebesgue-integrable on \(I\) and

\begin{equation}{\label{ma42}}\tag{55}
-\int_{I}g(x)dx\leq\int_{I}G_{r}(x)dx\leq\int_{I}g(x)dx.
\end{equation}

We also have

\[G_{r}(x)=\sup\left\{f_{r}(x),f_{r+1}(x),\cdots\right\}\mbox{ a.e. on }I,\]

which says that \(f_{r}(x)\leq G_{r}(x)\) a.e. on \(I\). For \(x\in I\) with \(f_{n}(x)\rightarrow f(x)\), given any \(\epsilon >0\), there exists an integer \(N\) satisfying

\[f(x)-\epsilon <f_{n}(x)<f(x)+\epsilon\mbox{ for all }n\geq N.\]

In other words, if \(r\geq N\), we have

\[f(x)-\epsilon\leq\sup\left\{f_{r}(x),f_{r+1}(x),\cdots\right\}\leq f(x)+\epsilon ;\]

that is,

\[f(x)-\epsilon\leq G_{r}(x)\leq f(x)+\epsilon\mbox{ for any }r\geq N.\]

This shows

\begin{equation}{\label{maeq333}}\tag{56}
\lim_{r\rightarrow\infty}G_{r}(x)=f(x)\mbox{ a.e. on }I.
\end{equation}

Since the sequence \(\{G_{r}(x)\}_{r=1}^{\infty}\) is decreasing a.e. on \(I\), from (\ref{ma42}), we see that the decreasing sequence \(\{\int_{I}G_{r}(x)dx\}_{r=1}^{\infty}\) is bounded below by \(-\int_{I}g(x)dx\), which says that the sequence \(\{\int_{I}G_{r}(x)dx\}_{r=1}^{\infty}\) is convergent. Using (\ref{maeq333}) and the monotone convergence Theorem \ref{mat220}, we see that \(f\) is Lebesgue-integrable and

\begin{equation}{\label{ma45}}\tag{57}
\lim_{r\rightarrow\infty}\int_{I}G_{r}(x)dx=\int_{I}f(x)dx.
\end{equation}

Now, we consider the following sequence

\[H_{n,r}(x)=\min\left\{f_{r}(x),f_{r+1}(x),\cdots\right\}\]

for \(r\leq n\), we can similarly show that the sequence \(\{H_{n,r}\}_{n=1}^{\infty}\) is decreasing and convergent a.e. on \(I\) to a limit function

\[H_{r}=\inf\left\{f_{r}(x),f_{r+1}(x),\cdots\right\}\]

such that \(H_{r}\) is Lebesgue-integrable on \(I\) and

\begin{equation}{\label{ma44}}\tag{58}
-\int_{I}g(x)dx\leq\int_{I}H_{r}(x)dx\leq\int_{I}g(x)dx.
\end{equation}

We also have \(H_{r}(x)\leq f_{r}(x)\) a.e. on \(I\). By referring to (\ref{maeq333}), we can similarly obtain

\begin{equation}{\label{ma43}}\tag{59}
\lim_{r\rightarrow\infty}H_{r}(x)=f(x)\mbox{ a.e. on }I.
\end{equation}

Since the sequence \(\{H_{r}\}_{r=1}^{\infty}\) is increasing a.e. on \(I\), from (\ref{ma44}), we see that the increasing sequence \(\{\int_{I}H_{r}(x)dx\}_{r=1}^{\infty}\) is bounded above by \(\int_{I}g(x)dx\), which says that the sequence \(\{\int_{I}H_{r}(x)dx\}_{r=1}^{\infty}\) is convergent. Using (\ref{ma43}) and the monotone convergence Theorem \ref{mat220}, we have

\begin{equation}{\label{ma46}}\tag{60}
\lim_{r\rightarrow\infty}\int_{I}H_{r}(x)dx=\int_{I}f(x)dx.
\end{equation}

Since

\[H_{r}(x)\leq f_{r}(x)\leq G_{r}(x)\mbox{ a.e. on }I,\]

we obtain

\[\int_{I}H_{r}(x)dx\leq\int_{I}f_{r}(x)dx\leq\int_{I}G_{r}(x)dx,\]

which imply

\begin{align*} \int_{I}f(x)dx & =\lim_{r\rightarrow\infty}\int_{I}H_{r}(x)dx\leq\lim_{r\rightarrow\infty}\int_{I}f_{r}(x)dx
\\ & \leq\lim_{r\rightarrow\infty}\int_{I}G_{r}(x)dx=\int_{I}f(x)dx\end{align*}

by using (\ref{ma45}) and (\ref{ma46}). Therefore, we obtain

\[\lim_{r\rightarrow\infty}\int_{I}f_{r}(x)dx=\int_{I}f(x)dx.\]

This completes the proof. \(\blacksquare\)

Theorem.  (Bounded Convergence Theorem). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue-integrable functions on a bounded interval \(I\). We also assume there exist a limit function \(f\) and a positive constant \(M\) such that \(|f_{n}(x)|\leq M\) for all \(n\) and \(\{f_{n}(x)\}_{n=1}^{\infty}\) converges to \(f(x)\) a.e. on \(I\). Then \(f\) is Lebesgue-integrable on \(I\) and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx.\]

Proof. We use Theorem \ref{mat334} by taking \(g(x)=M\) for all \(x\in I\). Since \(I\) is a bounded interval, the function \(g\) is Lebesgue-integrable on \(I\), and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{mat336}}\tag{61}\mbox{}\end{equation}

Theorem \ref{mat336}. Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of Lebesgue-integrable functions on \(I\) and converge a.e. on \(I\) to a limit function \(f\). If there is a nonnegative Lebesgue-integrable function \(g\) on \(I\) satisfying \(|f(x)|\leq g(x)\) a.e. on \(I\), then \(f\) is Lebesgue-integrable on \(I\).

Proof. We define a new sequence of functions \(\{g_{n}\}_{n=1}^{\infty}\) on \(I\) as

\[g_{n}=\max\left\{\min\left\{f_{n},g\right\},-g\right\}.\]

Then, we have \(|g_{n}(x)|\leq g(x)\) a.e. on \(I\). Since

\[-g(x)\leq f(x)\leq g(x)\mbox{ a.e. on }I,\]

we also have

\begin{align*} \lim_{n\rightarrow\infty}g_{n}(x) & =\lim_{n\rightarrow\infty}\max\left\{\min\left\{f_{n}(x),g(x)\right\},-g(x)\right\}\\ & =\max\left\{\min\left\{\lim_{n\rightarrow\infty}f_{n}(x),g(x)\right\},-g(x)\right\}\\ & =\max\left\{\min\left\{f(x),g(x)\right\},-g(x)\right\}\\ & =\max\left\{f(x),-g(x)\right\}=f(x)\mbox{ a.e. on }I\end{align*}

By Theorem \ref{mat334}, the function \(f\) is Lebesgue-integrable on \(I\) and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{mat221}}\tag{62}\mbox{}\end{equation}

Theorem \ref{mat221}. Let \(f\) be defined on the half-infinite interval \(I=[a,+\infty )\). Suppose that \(f\) is Lebesgue-integrable on the closed interval \([a,b]\). If there is a positive constant \(M\) satisfying

\[\int_{[a,b]}|f(x)|dx\leq M\mbox{ for all }b\geq a,\]

then \(f\) is Lebesgue-integrable on \(I\) and

\begin{equation}{\label{maeq222}}\tag{63}
\int_{I}f(x)dx=\int_{[a,+\infty )}f(x)dx=\lim_{b\rightarrow +\infty}\int_{[a,b]}f(x)dx.
\end{equation}

Proof. Let \(\{b_{n}\}_{n=1}^{\infty}\) be an increasing sequence of real numbers satisfying \(b_{n}\geq a\) and \(b_{n}\rightarrow +\infty\) as \(n\rightarrow\infty\). We define a sequence \(\{f_{n}\}_{n=1}^{\infty}\) as

\[f_{n}(x)=\left\{\begin{array}{ll}
f(x) & \mbox{if }a\leq x\leq b_{n}.\\
0 & \mbox{otherwise}.
\end{array}\right .\]

Then, we see that each \(f_{n}\) is Lebesgue-integrable on \(I\) and

\[\lim_{n\rightarrow\infty}f_{n}=f\mbox{ and }\lim_{n\rightarrow\infty}|f_{n}|=|f|\mbox{ on }I.\]

Since the sequence \(\{|f_{n}|\}_{n=1}^{\infty}\) is increasing and the increasing sequence \(\{\int_{I}|f_{n}(x)|dx\}_{n=1}^{\infty}\) is bounded above by \(M\). The sequence \(\{\int_{I}|f_{n}(x)|dx\}_{n=1}^{\infty}\) is convergent. By Theorem \ref{mat220}, the limit function \(|f|\) is Lebesgue-integrable on \(I\). Since \(|f_{n}|\leq |f|\) on \(I\) for all \(n\), the dominated convergence Theorem \ref{mat334} says that \(f\) is Lebesgue-integrable on \(I\) and

\[\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx=\int_{I}f(x)dx,\]

which says

\[\lim_{n\rightarrow\infty}\int_{[a,b_{n}]}f(x)dx=\int_{[a,+\infty )}f(x)dx.\]

This completes the proof. \(\blacksquare\)

Example. Let \(f(x)=1/(1+x^{2})\) for all \(x\in\mathbb{R}\). We shall prove that \(f\) is Lebesgue-integrable on \(\mathbb{R}\) with

\[\int_{\mathbb{R}}f(x)dx=\int_{\mathbb{R}}\frac{1}{1+x^{2}}dx=\pi.\]

Since \(f\) is nonnegative, for \(c\leq b\), we have

\[\int_{c}^{b}f(x)dx=\int_{c}^{b}\frac{1}{1+x^{2}}dx=\tan^{-1}b-\tan^{-1}c\leq\pi .\]

By Theorem \ref{mat221}, the function \(f\) is Lebesgue-integrable on \(\mathbb{R}\) and

\begin{align*}\int_{\mathbb{R}}f(x)dx & =\int_{-\infty}^{+\infty}\frac{1}{1+x^{2}}dx\\ & =
\lim_{c\rightarrow -\infty}\int_{c}^{0}\frac{1}{1+x^{2}}dx+\lim_{b\rightarrow +\infty}\int_{0}^{b}\frac{1}{1+x^{2}}dx\\ & =\frac{\pi}{2}+\frac{\pi}{2}=\pi .\end{align*}

\begin{equation}{\label{maex223}}\tag{64}\mbox{}\end{equation}

Example \ref{maex223}. Now, we are going to present an example such that the limit of (\ref{maeq222}) exists and it is not Lebesgue-integrable. Let \(I=[0,+\infty )\) and define \(f\) on \(I\) as

\[f(x)=\frac{(-1)^{n}}{n}\mbox{ if \(n-1\leq x<n\) for \(n=1,2,\cdots\).}\]

For \(b>0\), let \(m=[b]\) denote the greatest integer less than or equal to \(b\). Then, we have

\begin{align*} \int_{[0,b]}f(x)dx & =\int_{0}^{b}f(x)dx=\int_{0}^{m}f(x)dx+\int_{m}^{b}f(x)dx\\ & =
\sum_{k=1}^{m}\frac{(-1)^{k}}{k}+\frac{(b-m)(-1)^{m+1}}{m+1}.\end{align*}

We see that the last term converges to zero as \(b\rightarrow +\infty\). Therefore, we obtain

\[\lim_{b\rightarrow+\infty}\int_{0}^{b}f(x)dx=\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k}=-\ln 2.\]

Now, we assume that \(f\) is Lebesgue-integrable on \(I\) to obtain a contradiction. Let \(f_{n}\) be defined by

\[f_{n}(x)=\left\{\begin{array}{ll}
|f(x)| & \mbox{if \(0\leq x\leq n\)}\\
0 & \mbox{if \(x>n\)}.
\end{array}\right .\]

Then, we see that the sequence \(\{f_{n}\}_{n=1}^{\infty}\) increases and converges to \(|f|\) everywhere on \(I\). Since \(f\) is assumed to be Lebesgue-integrable on \(I\), it says that \(|f|\) is also Lebesgue-integrable on \(I\). Since \(|f_{n}(x)|\leq |f(x)|\) everywhere on \(I\), by the dominated convergence theorem, the sequence \(\{\int_{I}f_{n}(x)dx\}_{n=1}^{\infty}\) is convergent. This leads to a contradiction, since

\[\int_{I}f_{n}(x)dx=\int_{0}^{n}|f(x)|dx=\sum_{k=1}^{n}\frac{1}{k}\rightarrow +\infty\mbox{ as }n\rightarrow+\infty .\]

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

The Improper Riemann Integrals.

We are going to investigate the Riemann integrals on unbounded intervals.

Definition. Suppose that the real-valued function \(f\) is Riemann-integrable on \([a,b]\) for any \(b\). If the limit

\[\lim_{b\rightarrow +\infty}\int_{a}^{b}f(x)dx\]

exists, then \(f\) is said to be improper Riemann-integrable on \([a,+\infty )\), and the improper Riemann integral of \(f\) is denoted and defined by

\[\int_{a}^{+\infty}f(x)dx=\lim_{b\rightarrow +\infty}\int_{a}^{b}f(x)dx.\]

Example \ref{maex223} shows that the function \(f\) is improper Riemann-integrable on \([0,+\infty )\), but it is not Lebesgue-integrable on \([0,+\infty )\). The following result will present the connection.

\begin{equation}{\label{mat225}}\tag{65}\mbox{}\end{equation}

Theorem \ref{mat225}. Suppose that \(f\) is Riemann-integrable on \([a,b]\) for any \(b\). If there exists a positive constant \(M\) satisfying

\begin{equation}{\label{maeq335}}\tag{66}
\int_{a}^{b}|f(x)|dx\leq M
\end{equation}

for any \(b\), then both \(f\) and \(|f|\) are improper Riemann-integrable on \([a,+\infty )\). Moreover, \(f\) is Lebesgue-integrable on \([a,+\infty )\) and the Lebesgue integral of \(f\) on \([a,+\infty )\) is equal to the improper Riemann integral of \(f\) on \([a,+\infty )\), i.e.,

\[\int_{a}^{+\infty}f(x)dx=\int_{[a,+\infty )}f(x)dx.\]

Proof. We define the function

\[F(b)=\int_{a}^{b}|f(x)|dx.\]

Then \(F\) is an increasing function that is bounded above by \(M\). This says that the limit

\[\lim_{b\rightarrow +\infty}F(b)=\int_{a}^{\infty}|f(x)|dx\mbox{ exists},\]

which also says that \(|f|\) is improper Riemann-integrable on \([a,+\infty )\). Since

\[0\leq |f(x)|-f(x)\leq 2|f(x)|,\]

the limit

\begin{align*} \lim_{b\rightarrow +\infty}\left (\int_{a}^{b}|f(x)|dx-\int_{a}^{b}f(x)dx\right )
& =\lim_{b\rightarrow +\infty}\int_{a}^{b}\left [|f(x)|-f(x)\right ]dx\\ & \leq 2\int_{a}^{\infty}|f(x)|dx\mbox{ exists},\end{align*}

which implies that the limit

\[\lim_{b\rightarrow +\infty}\int_{a}^{b}f(x)dx\mbox{ exists}.\]

This proves that \(f\) is improper Riemann-integrable on \([a,+\infty )\). Using (\ref{maeq335}) and Theorem \ref{mat221}, we also see that \(f\) is Lebesgue-integrable on \([a,+\infty )\), and that the Lebesgue integral of \(f\) is equal to the improper Riemann integral of \(f\). This completes the proof. \(\blacksquare\)

We can similarly discuss the following forms of improper Riemann integrals:

\begin{align*}
\int_{-\infty}^{b}f(x)dx & =\lim_{a\rightarrow -\infty}\int_{a}^{b}f(x)dx\\
\int_{a}^{c-}f(x)dx & =\lim_{b\rightarrow c-}\int_{a}^{b}f(x)dx\\
\int_{c+}^{b}f(x)dx & =\lim_{a\rightarrow c+}\int_{a}^{b}f(x)dx.
\end{align*}

If both integrals

\[\int_{-\infty}^{a}f(x)dx\mbox{ and }\int_{a}^{+\infty}f(x)dx\]

exist, then we say that the integral \(\int_{-\infty}^{+\infty}f(x)dx\) exists and its integrals is defined by

\[\int_{-\infty}^{+\infty}f(x)dx=\int_{-\infty}^{a}f(x)dx+\int_{a}^{+\infty}f(x)dx.\]

On the other hand, if the integral \(\int_{-\infty}^{+\infty}f(x)dx\) exists, then its value is also equal to the symmetric limit

\begin{equation}{\label{maeq224}}\tag{67}
\lim_{b\rightarrow +\infty}\int_{-b}^{b}f(x)dx.
\end{equation}

However, if the symmetric limit (\ref{maeq224}) exists, we cannot always guarantee the integral \(\int_{-\infty}^{+\infty}f(x)dx\) exists, which can be realized by the function \(f(x)=x\) for all \(x\in\mathbb{R}\). In this case, the symmetric limit (\ref{maeq224}) is called the Cauchy principal value. Therefore, the Cauchy principal value of the integral is \(\int_{-\infty}^{+\infty}xdx=0\), but the integral does not exist.

\begin{equation}{\label{maex228}}\tag{68}\mbox{}\end{equation}

Example \ref{maex228}. We consider the function \(f(x)=e^{-x}x^{y-1}\), where \(y\) is a fixed real number. Since \(e^{-x/2}x^{y-1}\rightarrow 0\) as \(x\rightarrow +\infty\), there exists a constant \(M\) satisfying \(e^{-x/2}x^{y-1}\leq M\) for all \(x\geq 1\). Therefore, we obtain \(e^{-x}x^{y-1}\leq Me^{-x/2}\) and

\begin{align*} \int_{1}^{b}|f(x)|dx & \leq M\int_{1}^{b}e^{-x/2}dx\leq M\int_{0}^{b}e^{-x/2}dx\\ & =2M(1-e^{-b/2})<2M.\end{align*}

By Theorem \ref{mat225}, both the improper Riemann integral and Lebesgue integral

\[\int_{[1,+\infty )}e^{-x}x^{y-1}dx=\int_{1}^{+\infty}e^{-x}x^{y-1}dx\]

exists and the values are equal. From Example \ref{maex227}, the Lebesgue integral

\begin{align*} \Gamma (y) & =\int_{[0,+\infty )}e^{-x}x^{y-1}dx=\int_{0}^{+\infty}e^{-x}x^{y-1}dx
\\ & =\int_{0}^{1}e^{-x}x^{y-1}dx+\int_{1}^{+\infty}e^{-x}x^{y-1}dx\end{align*}

exists for each \(y>0\). This defines a Gamma function on \(\mathbb{R}_{+}\). \(\sharp\)

The following formula of integration by parts

\[\int_{a}^{b}f(x)g'(x)dx=f(b)g(b)-f(a)g(a)-\int_{a}^{b}g(x)f'(x)dx\]

can be converted into the formula concerning the improper Riemann integrals. Since \(b\) appears in three terms of the above formula, there are three limits to consider \(b\rightarrow +\infty\). If any two of these limits exist, the third limit also exists and we obtain the following formula

\[\int_{a}^{+\infty}f(x)g'(x)dx=\lim_{b\rightarrow +\infty}f(b)g(b)-f(a)g(a)-\int_{a}^{+\infty}g(x)f'(x)dx.\]

Many other theorems on Riemann integrals can be converted in much the same way to improper Riemann integrals.

Example. Continued from Example \ref{maex228}, if \(0<a<b\), the integration by parts gives

\[\int_{a}^{b}e^{-x}x^{y}dx=a^{y}e^{-a}-b^{y}e^{-b}+y\int_{a}^{b}e^{-x}x^{y-1}dx.\]

By taking \(a\rightarrow 0+\) and \(b\rightarrow +\infty\), we obtain the equality \(\Gamma (y+1)=y\Gamma (y)\). \(\sharp\)

Example. The Riemann zeta function \(\zeta\) is defined, for \(s>1\), by

\[\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}}.\]

Let \(\Gamma\) be the gamma function defined in Example \ref{maex228}. We are going to derive the following formula

\[\zeta (s)\Gamma (s)=\int_{0+}^{\infty}\frac{x^{s}-1}{e^{x}-1}dx,\]

which exists as a Lebesgue integral. For the integral \(\Gamma (s)\), we make the change of variable by setting \(t=nx\) for \(n>0\) to obtain

\[\Gamma (s)=\int_{0}^{\infty}e^{-t}t^{s-1}dt=n^{s}\int_{0}^{\infty}e^{-nx}x^{s-1}dx.\]

Therefore, if \(s>0\), we have

\begin{equation}{\label{maeq231}}\tag{69}
\frac{1}{n^{s}}\Gamma (s)=\int_{0}^{\infty}e^{-nx}x^{s-1}dx.
\end{equation}

If \(s>1\), the infinite series \(\sum_{n=1}^{\infty}n^{-s}\) converges. Therefore, from (\ref{maeq231}), we obtain

\begin{equation}{\label{maeq233}}\tag{70}
\zeta (s)\Gamma (s)=\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-nx}x^{s-1}dx.
\end{equation}

Since the integrands of (\ref{maeq233}) is nonnegative, Theorem~\ref{mat232} says that the infinite series \(\sum_{n=1}^{\infty}e^{-nx}x^{s-1}\) converges a.e. to a sum function which is Lebesgue integrable on \([0,+\infty )\), and that

\begin{equation}{\label{maeq234}}\tag{71}
\zeta (s)\Gamma (s)=\sum_{k=1}^{\infty}\int_{0}^{\infty}e^{-nx}x^{s-1}dx.=\int_{0}^{\infty}\left [\sum_{k=1}^{\infty}e^{-nx}x^{s-1}\right ]dx.
\end{equation}

Now, if \(x>0\), we have \(0<e^{-x}<1\). Therefore, we obtain

\[\sum_{n=1}^{\infty}e^{-nx}=\frac{e^{-x}}{1-e^{-x}}=\frac{1}{e^{x}-1},\]

which also implies

\[\sum_{n=1}^{\infty}e^{-nx}x^{s-1}=\frac{x^{s-1}}{e^{x}-1}\mbox{ a.e. on \([0,+\infty )\)}.\]

In fact, the above infinite series converges every where except at \(x=0\). Therefore, from (\ref{maeq234}), we obtain

\[\zeta (s)\Gamma (s)=\int_{0}^{\infty}\left [\sum_{k=1}^{\infty}e^{-nx}x^{s-1}\right ]dx=\int_{0+}^{\infty}\frac{x^{s-1}}{e^{x}-1}dx.\]

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Measurable Functions and Measurable Sets.

We have known that every function \(f\) that is Lebesgue-integrable on an interval \(I\) is the limit of a certain sequence of step functions in which the convergence is almost everywhere on \(I\). However, the converse is not true. For example, the constant function \(f(x)=1\) is a limit function of step functions on \(\mathbb{R}\), but this function is not Lebesgue-integrable on \(\mathbb{R}\). This says that the class of functions that are limits of step functions is larger than the class of Lebesgue-integrable functions. The functions in this larger class are called measurable functions and are defined below.

Definition. A real-valued function \(f\) defined on \(I\) is called  measurable when there exists a sequence of step functions \(\{s_{n}\}_{n=1}^{\infty}\) defined on \(I\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on \(I\)}.\]

It is obvious that if the function \(f\) is measurable on \(I\), then it is also measurable on every sub-interval of \(I\).

\begin{equation}{\label{map337}}\tag{72}\mbox{}\end{equation}

Proposition \ref{map337}. We have the following properties.

(i) Suppose that \(f\) is measurable on \(I\) and \(|f(x)|\leq g(x)\) a.e. on \(I\) for some nonnegative function \(g\) such that \(g\) is Lebesgue-integrable on \(I\). Then \(f\) is also Lebesgue-integrable on \(I\).

(ii) Suppose that \(f\) is measurable on \(I\) and \(|f|\) is Lebesgue-integrable on \(I\). Then \(f\) is Lebesgue-integrable on \(I\).

(iii) Suppose that \(f\) is measurable and bounded on a bounded interval \(I\). Then \(f\) is Lebesgue-integrable on \(I\).

Proof. There exists a sequence of step function \(\{s_{n}\}_{n=1}^{\infty}\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ a.e. on \(I\)}.\]

Using Theorem \ref{mat336}, we see that \(f\) is Lebesgue-integrable on \(I\). This proves part (i). Parts (ii) and (iii) are left as exercise. \(\blacksquare\)

Proposition. Let \(\phi\) be a continuous real-valued function defined on \(\mathbb{R}^{2}\). Suppose that \(f\) and \(g\) are measurable functions defined on \(I\). Then, the function \(h(x)=\phi (f(x),g(x))\) defined on \(I\) is measurable on \(I\). In particular, the functions \(f+g\), \(f-g\), \(f\cdot g\), \(|f|\), \(\max\{f,g\}\) and \(\min\{f,g\}\) are measurable on \(I\). If \(f(x)\neq 0\) a.e. on \(I\), then \(1/f\) is also measurable on \(I\).

Proof. Let \(\{s_{n}\}_{n=1}^{\infty}\) and \(\{t_{n}\}_{n=1}^{\infty}\) denote the sequences of step functions satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x)=f(x)\mbox{ and }\lim_{n\rightarrow\infty}t_{n}(x)=g(x)\mbox{ a.e. on \(I\)}.\]

Then, the function \(u_{n}=\phi (s_{n},t_{n})\) is also a step function satisfying

\[\lim_{n\rightarrow\infty}u_{n}(x)=h(x)\mbox{ a.e. on \(I\)}.\]

This shows that \(h\) is measurable on \(I\), and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{mat339}}\tag{73}\mbox{}\end{equation}

Theorem \ref{mat339}. Let \(f\) be defined on \(I\). Assume that \(\{f_{n}\}_{n=1}^{\infty}\) is a sequence of measurable functions defined on \(I\) satisfying

\[\lim_{n\rightarrow\infty}f_{n}(x)=f(x)\mbox{ a.e. on \(I\)}.\]

Then \(f\) is measurable on \(I\).

Proof. We define \(g(x)=1/(1+x^{2})\) on \(I\). Then \(g\) is positive and Lebesgue-integrable on \(I\), which also says that \(g\) is measurable on \(I\). We define

\[F_{n}(x)=g(x)\cdot\frac{f_{n}(x)}{1+|f_{n}(x)|}\mbox{ for }x\in I.\]

Then

\[\lim_{n\rightarrow\infty}F_{n}(x)=\frac{g(x)f(x)}{1+|f(x)|}\equiv F(x)\mbox{ a.e. on \(I\)}.\]

Since each \(F_{n}\) is measurable on \(I\) and \(|F_{n}(x)|<g(x)\) for all \(x\in I\), part (i) of Proposition \ref{map337} says that each \(F_{n}\) is Lebesgue-integrable on \(I\). Since \(|F(x)|<g(x)\) for all \(x\in I\), Theorem \ref{mat336} says that \(F\) is Lebesgue-integrable on \(I\), which also implies that \(F\) is measurable on \(I\). Now, we have

\begin{align*} f(x)\left [g(x)-|F(x)|\right ] & =f(x)g(x)\left [1-\frac{f(x)}{1+|f(x)|}\right ]\\ & =\frac{f(x)g(x)}{1+|f(x)|}=F(x)\mbox{ for all \(x\in I\),}\end{align*}

which implies

\[f(x)=\frac{F(x)}{g(x)-|F(x)|}.\]

Since \(F,g,|F|\) are measurable on \(I\) and \(g(x)-|F(x)|>0\) for all \(x\in I\), we conclude that \(f\) is measurable on \(I\), and the proof is complete. \(\blacksquare\)

Definition. Given nonempty subset \(S\) of \(\mathbb{R}\). The function \(\chi_{S}\) defined by

\[\chi_{S}=\left\{\begin{array}{ll}
1 & \mbox{if \(x\in S\)}\\
0 & \mbox{if \(x\not\in S\)}
\end{array}\right .\]

is called the characteristic function of \(S\). When \(S=\emptyset\), we define \(\chi_{S}(x)=0\) for all \(x\in\mathbb{R}\). \(\sharp\)

\begin{equation}{\label{map238}}\tag{74}\mbox{}\end{equation}

Proposition \ref{map238}. We have the following properties.

(i) Suppose that \(S\) has measure zero. Then \(\chi_{S}\) is Lebesgue-integrable on \(\mathbb{R}\) and

\[\int_{\mathbb{R}}\chi_{S}(x)dx=0.\]

(ii) Suppose that \(\chi_{S}\) is Lebesgue-integrable on \(\mathbb{R}\) and

\[\int_{\mathbb{R}}\chi_{S}(x)dx=0.\]

Then \(S\) has measure zero.

Proof. To prove part (i), since \(S\) has measure zero, it says that \(\chi_{S}=0\) a.e. on \(S\). Therefore, the desired result follows from part (i) of Proposition \ref{map338}. To prove part (ii), we define \(f_{n}=\chi_{S}\) for all \(n\). Then \(|f_{n}|=\chi_{S}\). Therefore, we have

\[\sum_{n=1}^{\infty}\int_{\mathbb{R}}|f_{n}(x)|dx=\sum_{n=1}^{\infty}\int_{\mathbb{R}}\chi_{S}(x)dx=0.\]

By Theorem \ref{mat232}, the series \(\sum_{k=1}^{\infty}f_{k}(x)\) is convergent a.e. on \(\mathbb{R}\); that is, there exists a subset \(T\) of \(\mathbb{R}\) such that \(T\) has measure zero and the series \(\sum_{k=1}^{\infty}f_{k}(x)\) is divergent on \(T\). For \(x\in S\), the series \(\sum_{k=1}^{\infty}f_{k}(x)\) cannot be convergent, since each term is \(1\). For \(x\not\in S\), the series \(\sum_{k=1}^{\infty}f_{k}(x)\) is convergent, since each term is \(0\). This shows that \(S=T\) and has measure zero. This completes the proof. \(\blacksquare\)

Definition. A subset \(S\) of \(\mathbb{R}\) is said to be {\bf measurable} when its characteristic function \(\chi_{S}\) is measurable. If the characteristic function \(\chi_{S}\) is Lebesgue-integrable on \(\mathbb{R}\), then the measure $\mu (S)$ of \(S\) is defined by

\[\mu (S)=\int_{\mathbb{R}}\chi_{S}(x)dx.\]

If \(\chi_{S}\) is measurable and is not Lebesgue-integrable on \(\mathbb{R}\), we define \(\mu (S)=+\infty\). The set function \(\mu\) is called a Lebesgue measure. \(\sharp\)

We have the following observations.

• Proposition \ref{map238} says that a set \(S\) of measure zero is measurable and \(\mu (S)=0\).
• If \(A\) and \(B\) are measurable satisfying \(A\subseteq B\), then \(\mu (A)\leq\mu (B)\).
• Every interval \(I\) (bounded or unbounded) is measurable. If \(I\) is a bounded interval with endpoints \(a\) and \(b\) satisfying \(a\leq b\), then \(\mu (I)=b-a\). If \(I\) is an unbounded interval, then \(\mu (I)=+\infty\).

Proposition. We have the following properties.

(i) Suppose that \(S\) and \(T\) are measurable. Then \(S\setminus T\) is also measurable.

(ii) Suppose that \(S_{1},S_{2},\cdots\) are measurable. Then

\[\bigcup_{i=1}^{\infty}S_{i}\mbox{ and }\bigcap_{i=1}^{\infty}S_{i}\]

are also measurable.

Proof. Since the characteristic function of \(S\setminus T\) is given by \(\chi_{S}-\chi_{S}\chi_{T}\) and is measurable, it says that \(S\setminus T\) is measurable, which proves part (i). To prove part (ii), we define

\[\begin{array}{cccc}
{\displaystyle U_{n}=\bigcup_{i=1}^{n}S_{i}}, & {\displaystyle V_{n}=\bigcap_{i=1}^{n}S_{i}}, &
{\displaystyle U=\bigcup_{i=1}^{\infty}S_{i}}, & {\displaystyle V=\bigcap_{i=1}^{\infty}S_{i}}.
\end{array}\]

Then, we have

\[\chi_{U_{n}}=\max\left\{\chi_{S_{1}},\cdots ,\chi_{S_{n}}\right\}\]

and

\[\chi_{V_{n}}=\min\left\{\chi_{S_{1}},\cdots ,\chi_{S_{n}}\right\}.\]

This shows that each of \(U_{n}\) and \(V_{n}\) is measurable. Since

\[\chi_{U}=\lim_{n\rightarrow\infty}\chi_{U_{n}}\]

and

\[\chi_{V}=\lim_{n\rightarrow\infty}\chi_{V_{n}},\]

using Theorem \ref{mat339}, we conclude that \(U\) and \(V\) are measurable, and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{map342}}\tag{75}\mbox{}\end{equation}

Proposition \ref{map342}. Suppose that \(A\) and \(B\) are disjoint measurable sets. Then, we have

\begin{equation}{\label{maeq340}}\tag{76}
\mu (A\cup B)=\mu (A)+\mu (B).
\end{equation}

More general, when \(\{A_{1},\cdots ,A_{n}\}\) is a finite disjoint collection of measurable sets, we have

\begin{equation}{\label{maeq341}}\tag{77}
\mu\left (\bigcup_{i=1}^{n}A_{i}\right )=\sum_{i=1}^{n}\mu (A_{i}).
\end{equation}

Proof. Let \(S=A\cup B\). Since \(A\) and \(B\) are disjoint, we have

\begin{equation}{\label{ma47}}\tag{78}
\chi_{S}=\chi_{A}+\chi_{B}.
\end{equation}

Suppose that \(\chi_{S}\) is Lebesgue-integrable. Since both \(\chi_{A}\) and \(\chi_{B}\) are measurable satisfying

\[0\leq\chi_{A}(x)\leq\chi_{S}(x)\mbox{ and }0\leq\chi_{B}(x)\leq\chi_{S}(x)\]

for all \(x\), part (i) of Proposition~\ref{map337} says that both \(\chi_{A}\) and \(\chi_{B}\) are Lebesgue-integrable. Therefore, we also have

\begin{align*} +\infty & >\mu (S)=\int_{\mathbb{R}}\chi_{S}(x)dx\\ & =\int_{\mathbb{R}}\chi_{A}(x)dx+
\int_{\mathbb{R}}\chi_{B}(x)dx\\ & =\mu (A)+\mu (B).\end{align*}

In this case, the equality (\ref{maeq340}) holds true and both terms are finite. Now, we assume that \(\chi_{S}\) is not Lebesgue-integrable. Since \(S\) is measurable, the definition says \(\mu (S)=+\infty\). The equality (\ref{ma47}) also says that at least one of \(\chi_{A}\) or \(\chi_{B}\) is not Lebesgue-integrable. Otherwise, \(\chi_{S}\) is Lebesgue-integrable. In this case, the equality (\ref{maeq340}) holds true for both terms to be infinite. The equality (\ref{maeq341}) can be obtained by induction. This completes the proof. \(\blacksquare\)

Theorem. Suppose that \(\{A_{1},A_{2},\cdots\}\) is a disjoint collection of measurable sets. Then, we have

\begin{equation}{\label{maeq343}}\tag{79}
\mu\left (\bigcup_{i=1}^{\infty}A_{i}\right )=\sum_{i=1}^{\infty}\mu (A_{i}).
\end{equation}

Proof. We define

\[T_{n}=\bigcup_{i=1}^{n}A_{i}\mbox{ and }T=\bigcup_{i=1}^{\infty}A_{i}.\]

Then, we have

\[\lim_{n\rightarrow\infty}\chi_{T_{n}}=\chi_{T}.\]

Proposition \ref{map342} says

\[\mu (T_{n})=\sum_{i=1}^{n}\mu (A_{i})\]

for each \(n\). We are going to prove

\[\lim_{n\rightarrow\infty}\mu (T_{n})=\mu (T).\]

Since \(\mu (T_{n})\leq\mu (T_{n+1})\) for all \(n\), the sequence \(\{\mu (T_{n})\}_{n=1}^{\infty}\) is increasing. Suppose that \(\mu (T)\) is finite. Then \(\chi_{T}\) and each \(\chi_{T_{n}}\) is Lebesgue-integrable. Since the sequence \(\{\mu (T_{n})\}_{n=1}^{\infty}\) is bounded above by \(\mu (T)\), it is convergent. By the dominated convergence Theorem \ref{mat334}, we obtain

\begin{align*} \lim_{n\rightarrow\infty}\mu (T_{n}) & =\lim_{n\rightarrow\infty}\int_{\mathbb{R}}\chi_{T_{n}}(x)dx
\\ & =\int_{\mathbb{R}}\chi_{T}(x)dx=\mu (T).\end{align*}

Now, we assume that \(\mu (T)=+\infty\). This says that \(\chi_{T}\) is not Lebesgue-integrable. Theorem \ref{mat334} also says that either some \(\chi_{T_{n}}\) is not Lebesgue-integrable or every \(\chi_{T_{n}}\) is Lebesgue-integrable with \(\mu (T_{n})\rightarrow +\infty\) as \(n\rightarrow\infty\). In either case, the equality (\ref{maeq343}) holds true for both terms to be infinite. This completes the proof. \(\blacksquare\)

\begin{equation}{\label{f}}\tag{F}\mbox{}\end{equation}

Properties of Lebesgue Integrals.

In what follows, we are going to present some interesting and important properties and results regarding Lebesgue integrals. We first consider the continuity of functions defined by Lebesgue integral. Let \(f\) be a real-valued function of two variables defined on a subset \(X\times Y\) of \(\mathbb{R}^{2}\), where \(X\) and \(Y\) are intervals in \(\mathbb{R}\). Therefore, we can consider the function like

\[F(y)=\int_{X}f(x,y)dx.\]

We shall discuss the continuity, differentiability and integrability of function \(F\).

\begin{equation}{\label{mat235}}\tag{80}\mbox{}\end{equation}

Theorem \ref{mat235}. Let \(X\) and \(Y\) be two intervals in \(\mathbb{R}\), and let \(f\) be a function defined on \(X\times Y\) satisfying the following conditions.

• For each fixed \(y\in Y\), the function \(f_{y}\) defined on \(X\) by \(f_{y}=f(x,y)\) is measurable on \(X\).
• There exists a nonnegative and Lebesgue-integrable function on \(X\) satisfying, for each \(y\in Y\), \(|f(x,y)|\leq g(x)\) a.e. on \(X\).
•  For each fixed \(y\in Y\),
  \begin{equation}{\label{maeq344}}\tag{81}
  \lim_{t\rightarrow y}f(x,t)=f(x,y)\mbox{ a.e. in \(X\)}.
  \end{equation}

Then, the Lebesgue integral \(\int_{X}f(x,y)dx\) exists for each \(y\in Y\), and the function \(F\) defined on \(Y\) by

\[F(y)=\int_{X}f(x,y)dx\]

is continuous on \(Y\). In other words, for \(y\in Y\), we have

\[\lim_{t\rightarrow y}\int_{X}f(x,t)dx=\int_{X}\left [\lim_{t\rightarrow y}f(x,t)\right ]dx.\]

Proof.  Since \(f_{y}\) is measurable on \(X\) and dominated a.e. on \(X\) by a nonnegative and Lebesgue-integrable function \(g\) on \(X\), part (i) of Proposition \ref{map337} says that \(f_{y}\) is Lebesgue-integrable on \(X\). In other words, the Lebesgue integral \(\int_{X}f(x,y)dx\) exists for each \(y\in Y\). Now, given any fixed \(y\in Y\), let \(\{y_{n}\}_{n=1}^{\infty}\) be any sequence of points in \(Y\) satisfying \(y_{n}\rightarrow y\) as \(n\rightarrow\infty\). We are going to show that \(F(y_{n})\rightarrow F(y)\) as \(n\rightarrow\infty\). We define \(G_{n}(x)=f(x,y_{n})\). Then each \(G_{n}\) is Lebesgue-integrable on \(X\). Condition (\ref{maeq344}) also says that \(G_{n}(x)\rightarrow f(x,y)\) a.e. on \(X\). Since \(|f(x,y)|\leq g(x)\) a.e. on \(X\) by the second condition and

\[F(y_{n})=\int_{X}G_{n}(x)dx,\]

Theorem \ref{mat334} says that the sequence \(\{F(y_{n})\}_{n=1}^{\infty}\) is convergent and

\[\lim_{n\rightarrow\infty}F(y_{n})=\int_{X}f(x,y)dx=F(y).\]

This completes the proof. \(\blacksquare\)

Example. Continued from Example \ref{maex228}, we are going to claim that the gamma function

\begin{align*} \Gamma (y) & =\int_{[0,+\infty )}e^{-x}x^{y-1}dx=\int_{0}^{+\infty}e^{-x}x^{y-1}dx
\\ & =\int_{0}^{1}e^{-x}x^{y-1}dx+\int_{1}^{+\infty}e^{-x}x^{y-1}dx\end{align*}

is continuous by applying Theorem \ref{mat235} and considering \(X=[0,+\infty )\) and \(Y=(0,+\infty )\). We need to check the three conditions of Theorem \ref{mat235}.

• For each \(y>0\), the integrand, as a function of \(x\), is continuous a.e. on \(X\), i.e., measurable a.e. on \(X\). This says that the first condition is satisfied.
• For each fixed \(x>0\), the integrand, as a function of \(y\), is continuous on \(Y\). This says that the third condition is satisfied.
• For any fixed \(a>0\), we consider the interval \(I_{a}=[a,+\infty )\). For each \(y\in I_{a}\), the integrand is dominated by the following function
  \[g(x)=\left\{\begin{array}{ll}
  x^{a-1} & \mbox{if \(0<x\leq 1\)}\\
  Me^{-x/2} & \mbox{if \(x\geq 1\)},
  \end{array}\right .\]
  where \(M\) is some positive constant. This function \(g\) is Lebesgue-integrable on \(X\) by Proposition \ref{map236}. This says that the second condition is satisfied.

Therefore, Theorem \ref{mat235} says that \(\Gamma\) is continuous on \(I_{a}\). Since \(a\) can be any positive number, we conclude that the gamma function \(\Gamma\) is continuous on \((0,+\infty )\). \(\sharp\)

Example. We consider the following function

\[F(y)=\int_{0+}^{+\infty}e^{-xy}\frac{\sin x}{x}dx\]

for \(y>0\). Let \(X=[0,+\infty )\) and \(Y=(0,+\infty )\). The first and third conditions of Theorem \ref{mat235} can be easily checked. For the second condition, we also consider the interval \(I_{a}=[a,+\infty )\) for any fixed \(a>0\). Since \(|\sin x/x|\leq 1\), the integrand is dominated on \(I_{a}\) by the function \(g(x)=e^{-ax}\) for \(x\geq 0\). It is obvious that \(g\) is Lebesgue-integrable on \(X\). This shows that \(F(y)\) is continuous on \(I_{a}\). Therefore, we conclude that the function \(F\) is continuous on \(Y=(0,+\infty )\). \(\sharp\)

\begin{equation}{\label{mat237}}\tag{82}\mbox{}\end{equation}

Theorem \ref{mat237}. Let \(X\) and \(Y\) be two intervals in \(\mathbb{R}\), and let \(f\) be a function defined on \(X\times Y\) satisfying the following conditions.

• For each fixed \(y\in Y\), the function \(f_{y}\) defined on \(X\) by \(f_{y}(x)=f(x,y)\) is measurable on \(X\), and \(f_{a}\) is Lebesgue-integrable on \(X\) for some \(a\in Y\).
• The partial derivative \(\partial f(x,y)/\partial y\) exists for each interior point \((x,y)\in X\times Y\).
• There is a nonnegative and Lebesgue-integrable function \(g\) on \(X\) satisfying
  \begin{equation}{\label{maeq345}}\tag{83}
  \left |\frac{\partial}{\partial y}f(x,y)\right |\leq g(x)
  \end{equation}
  for all interior points of \(X\times Y\).

Then, the Lebesgue integral \(\int_{X}f(x,y)dx\) exists for every \(y\in Y\), and the function \(F\) defined by

\[F(y)=\int_{X}f(x,y)dx\]

is differentiable at each interior point of \(Y\). Moreover, the derivative is given by

\[F'(y)=\int_{X}\left [\frac{\partial}{\partial y}f(x,y)\right ]dx.\]

Proof. Given any interior point \((x,y)\) of \(X\times Y\), the mean-value theorem says

\[f(x,y)-f(x,a)=(y-a)\frac{\partial}{\partial y}f(x,c),\]

where \(c\) lies between \(a\) and \(y\). According to (\ref{maeq345}), we also have

\begin{align*} |f(x,y)| & \leq |f(x,a)|+|y-a|\left |\frac{\partial}{\partial y}f(x,c)\right |\\ & \leq |f(x,a)|+|y-a|g(x),\end{align*}

which also says

\[|f_{y}(x)|\leq |f_{a}(x)|+|y-a|g(x).\]

Since \(f_{y}\) is measurable on \(X\), and is dominated by a nonnegative and Lebesgue-integrable function \(|f_{a}(x)|+|y-a|g(x)\) on \(X\), part (i) of Proposition \ref{map337} says that \(f_{y}\) is Lebesgue-integrable on \(X\), i.e., the Lebesgue integral \(\int_{X}f(x,y)dx\) exists for each \(y\in Y\). Now, we choose any sequence \(\{y_{n}\}_{n=1}^{\infty}\) in \(Y\) satisfying \(y_{n}\neq y\) and \(y_{n}\rightarrow y\) as \(n\rightarrow\infty\). We define a sequence of functions \(\{q_{n}\}_{n=1}^{\infty}\) on \(X\) by

\[q_{n}(x)=\frac{f(x,y_{n})-f(x,y)}{y_{n}-y}.\]

Then, we see that each \(q_{n}\) is Lebesgue-integrable on \(X\) satisfying

\begin{align*} \lim_{n\rightarrow\infty}q_{n}(x) & =\lim_{n\rightarrow\infty}\frac{f(x,y_{n})-f(x,y)}{y_{n}-y}\\ & =\frac{\partial}{\partial y}f(x,y)\end{align*}

at each interior point of \(X\). Using the mean-value theorem again, we obtain \(q_{n}(x)=\partial f(x,c_{n})/\partial y\), where \(c_{n}\) lies between \(y_{n}\) and \(y\). By (\ref{maeq345}), we also have \(|q_{n}(x)|\leq g(x)\) a.e. on \(X\). Therefore, the dominated convergence Theorem \ref{mat334} says

\begin{align}\lim_{n\rightarrow\infty}\int_{X}q_{n}(x)dx & =\int_{X}\left [\lim_{n\rightarrow\infty}q_{n}(x)\right ]dx\nonumber\\ & =\int_{X}\frac{\partial}{\partial y}f(x,y)dx.\label{maeq346}\tag{84}\end{align}

We also have

\begin{align} \int_{X}q_{n}(x)dx & =\frac{1}{y_{n}-y}\int_{X}\left [f(x,y_{n})-f(x,y)\right ]dx\nonumber\\ & =\frac{F(y_{n})-F(y)}{y_{n}-y}.\label{maeq347}\tag{85}\end{align}

Therefore, from (\ref{maeq346}) and (\ref{maeq347}), we have

\begin{align*} F'(y) & =\lim_{n\rightarrow\infty}\frac{F(y_{n})-F(y)}{y_{n}-y}\\ & =
\lim_{n\rightarrow\infty}\int_{X}q_{n}(x)dx\\ & =\int_{X}\frac{\partial}{\partial y}f(x,y)dx,\end{align*}

and the proof is complete. \(\blacksquare\)

Example. Continued from Example \ref{maex228}, we are going discuss the derivative of gamma function by applying Theorem \ref{mat237}. The first and third conditions of Theorem \ref{mat237} can be easily checked. For the second condition, we consider the interval \(I_{a}=[a,+\infty )\) for any fixed \(a>0\). Then, we have

\[\frac{\partial}{\partial y}(e^{-x}x^{y-1})=e^{-x}x^{y-1}\ln x\]

for \(x>0\), which is dominated by the following function

\[g(x)=\left\{\begin{array}{ll}
x^{a-1}|\ln x| & \mbox{if \(0<x\leq 1\)}\\
Me^{-x/2} & \mbox{if \(x\geq 1\)}\\
0 & \mbox{if \(x=0\)},
\end{array}\right .\]

where \(M\) is some positive constant. It is not difficult to show that \(g\) is Lebesgue-integrable on \([0,+\infty )\). Therefore, we obtain

\begin{align*} \Gamma^{\prime}(y) & =\int_{0+}^{+\infty}\left [\frac{\partial}{\partial y}(e^{-x}x^{y-1})
\right ]dx\\ & =\int_{0+}^{+\infty}e^{-x}x^{y-1}\ln xdx.\end{align*}

Example. We are going to apply Theorem \ref{mat237} to evaluate the following improper Riemann integral

\[F(y)=\int_{0+}^{+\infty}e^{-xy}\frac{\sin x}{x}dx.\]

The three conditions of Theorem\ref{mat237} can be easily checked by considering the interval \(I_{a}=[a,+\infty )\) for any fixed \(a>0\). Therefore, we obtain

\begin{align*} F'(y) & =\int_{0+}^{+\infty}\left [\frac{\partial}{\partial y}
\left (e^{-xy}\frac{\sin x}{x}\right )\right ]dx\\ & =-\int_{0+}^{+\infty}e^{-xy}\sin xdx\end{align*}

for \(y>0\). Using integration by parts, we have

\[\int_{0}^{b}e^{-xy}\sin xdx=\frac{e^{-by}\left (-y\sin b-\cos b\right )}{1+y^{2}}+\frac{1}{1+y^{2}}\]

for all \(y\in\mathbb{R}\). By taking \(b\rightarrow +\infty\), we find

\[F'(y)=-\int_{0+}^{+\infty}e^{-xy}\sin xdx=-\frac{1}{1+y^{2}}\]

for \(y>0\). By taking integration, we also obtain

\begin{align*} F(y)-F(b) & =\int_{b}^{y}F'(t)dt=-\int_{b}^{y}\frac{dt}{1+t^{2}}\\ & =\tan^{-1}b-\tan^{-1}y\end{align*}

for \(y>0\) and \(b>0\). Since

\[\lim_{b\rightarrow\infty}\tan^{-1}b=\frac{\pi}{2}\mbox{ and }\lim_{b\rightarrow\infty}F(b)=0,\]

we obtain

\[F(y)=\int_{0+}^{+\infty}e^{-xy}\frac{\sin x}{x}dx=\frac{\pi}{2}-\tan^{-1}y\]

for \(y>0\). \(\sharp\)

Example. The function \(\sin x/x\) does not have indefinite integral. However, we are going to find

\[\int_{0+}^{+\infty}\frac{\sin x}{x}dx=\lim_{b\rightarrow +\infty}\int_{0+}^{b}\frac{\sin x}{x}dx=\frac{\pi}{2}.\]

Let \(\{g_{n}\}_{n=1}^{\infty}\) be a sequence of functions defined on \(\mathbb{R}\) by

\[g_{n}(y)=\int_{0}^{n}e^{-xy}\frac{\sin x}{x}dx.\]

Since

\begin{align*} |g_{n}(n)| & \leq\int_{0}^{n}\left |e^{-nx}\frac{\sin x}{x}\right |dx\\ & \leq\int_{0}^{n}e^{-nx}dx=\frac{1}{n}\int_{0}^{n^{2}}e^{-t}dx<\frac{1}{n},\end{align*}

it follows

\begin{equation}{\label{ma48}}\tag{86}
\lim_{n\rightarrow\infty}g_{n}(n)=0.
\end{equation}

Now, we have

\begin{align*} g’_{n}(y) & =\int_{0}^{n}\left [\frac{\partial}{\partial y}e^{-xy}\frac{\sin x}{x}\right ]dx
\\ & =-\int_{0}^{n}e^{-xy}\sin xdx=-\frac{e^{-ny}(-y\sin n-\cos n)+1}{1+y^{2}}\end{align*}

for all \(y\in\mathbb{R}\), which implies

\[\lim_{n\rightarrow\infty}g’_{n}(y)=-\frac{1}{1+y^{2}}\]

for all \(y\in\mathbb{R}\). We also have

\[\left |g’_{n}(y)\right |\leq\frac{e^{-y}(y+1)+1}{1+y^{2}}\mbox{ for }y\geq 0,\]

which says hat the function \(f_{n}\) defined by

\[f_{n}(y)=\left\{\begin{array}{ll}
g’_{n}(y) & \mbox{if \(0\leq y\leq n\)}\\
0 & \mbox{if \(y>n\)},
\end{array}\right .\]

is Lebesgue-integrable on \([0,+\infty )\) and is dominated by the nonnegative and Lebesgue-integrable function

\[g(y)=\frac{e^{-y}(y+1)+1}{1+y^{2}}\]

on \([0,+\infty )\). Since

\[\lim_{n\rightarrow\infty}f_{n}(y)=-\frac{1}{1+y^{2}}\mbox{ for }y\geq 0,\]

the dominated convergence Theorem \ref{mat334} says
\begin{equation}{\label{ma49}}\tag{87}
\lim_{n\rightarrow\infty}\int_{0}^{+\infty}f_{n}(x)dx=-\int_{0}^{+\infty}\frac{dy}{1+y^{2}}=-\frac{\pi}{2},
\end{equation}

On the other hand, we have

\begin{align*} \int_{0}^{+\infty}f_{n}(x)dx=\int_{0}^{n}g’_{n}(y)dy\\ & =g_{n}(n)-g_{n}(0).\end{align*}

Using (\ref{ma48}) and (\ref{ma49}), we obtain

\begin{equation}{\label{ma50}}\tag{88}
\lim_{n\rightarrow\infty}g_{n}(0)=\lim_{n\rightarrow\infty}g_{n}(n)-\lim_{n\rightarrow\infty}\int_{0}^{+\infty}f_{n}(x)dx=\frac{\pi}{2}.
\end{equation}

For \(b>0\), let \(n=[b]\). Then, we have

\begin{align*} \int_{0+}^{b}\frac{\sin x}{x}dx & =\int_{0+}^{n}\frac{\sin x}{x}dx+\int_{n}^{b}\frac{\sin x}{x}dx\\ & =g_{n}(0)+\int_{n}^{b}\frac{\sin x}{x}dx.\end{align*}

Since

\begin{align*} 0 & \leq\left |\int_{n}^{b}\frac{\sin x}{x}dx\right |\leq\int_{n}^{b}\frac{1}{n}dx=\frac{b-n}{n}\\ & \leq\frac{1}{n}\rightarrow 0\mbox{ as }b\rightarrow +\infty ,\end{align*}

using (\ref{ma50}), we obtain

\begin{align*} \int_{0+}^{+\infty}\frac{\sin x}{x}dx & =\lim_{b\rightarrow\infty}\int_{0+}^{b}\frac{\sin x}{x}dx\\ & =\lim_{n\rightarrow\infty}g_{n}(0)=\frac{\pi}{2}.\end{align*}

Theorem.  (Fubini’s Theorem). Let \(X\) and \(Y\) be two intervals in \(\mathbb{R}\), and let \(k\) be a function which is continuous and bounded on \(X\times Y\) with \(|k(x,y)|\leq M\) for all \((x,y)\in X\times Y\). We also assume that \(f\) is Lebesgue-integrable on \(X\) and \(g\) is Lebesgue-integrable on \(Y\). Then, we have the following properties.

(i) For each \(y\in Y\), the Lebesgue integral \(\int_{X}f(x)k(x,y)dx\) exists, and the function \(F\) defined on \(Y\) by

\[F(y)=\int_{X}f(x)k(x,y)dx\]

is continuous on \(Y\).

(ii) For each \(x\in X\), the Lebesgue integral \(\int_{Y}g(y)k(x,y)dy\) exists, and the function \(G\) defined on \(X\) by

\[G(x)=\int_{Y}g(y)k(x,y)dy\]

is continuous on \(X\).

(iii) The two Lebesgue integrals \(\int_{X}f(x)k(x,y)dx\) and \(\int_{Y}g(y)k(x,y)dy\) exist and are equal; that is ,

\begin{equation}{\label{maeq348}}\tag{89}
\int_{X}f(x)\left [\int_{Y}g(y)k(x,y)dy\right ]dx=\int_{Y}g(y)\left [\int_{X}f(x)k(x,y)dx\right ]dy.
\end{equation}

Proof. Given any fixed \(y\in Y\), we define \(f_{y}(x)=f(x)k(x,y)\). Then \(f_{y}\) is measurable on \(X\) and satisfies the following inequality

\[\left |f_{y}(x)\right |=\left |f(x)k(x,y)\right |\leq M\left |f(x)\right |\mbox{ for all }x\in X.\]

Since \(k\) is continuous on \(X\times Y\), we have

\[\lim_{t\rightarrow y}f(x)k(x,t)=f(x)k(x,y)\mbox{ for all }x\in X.\]

Therefore, part (i) follows from Theorem \ref{mat235}. The similar arguments can also prove part (ii).

To prove part (iii), the product \(f\cdot G\) is measurable on \(X\) and satisfies the following inequality

\begin{align*} \left |f(x)G(x)\right | & \leq\left |f(x)\right |\int_{Y}\left |g(y)\right |
\cdot\left |k(x,y)\right |dy\\ & \leq\left |f(x)\right |\int_{Y}\left |g(y)\right |\cdot Mdy\equiv M’\left |f(x)\right |,\end{align*}

where

\[M’=M\int_{Y}\left |g(y)\right |dy.\]

By part (i) of Proposition \ref{map337}, we see that \(f\cdot G\) is Lebesgue-integrable on \(X\). The similar argument can also show that \(g\cdot F\) is Lebesgue-integrable on \(Y\). We remain to prove (\ref{maeq348}). Suppose that \(f\) and \(g\) are taken to be step functions \(s\) and \(t\), respectively. Then, the equality (\ref{maeq348}) holds true by Corollary \ref{mac349} and is given below

\begin{equation}{\label{ma52}}\tag{90}
\int_{X}s(x)\left [\int_{Y}t(y)k(x,y)dy\right ]dx=\int_{Y}t(y)\left [\int_{X}s(x)k(x,y)dx\right ]dy.
\end{equation}

Given \(\epsilon >0\), part (ii) of Proposition \ref{map330} says that there exist step functions \(s\) and \(t\) satisfying

\[\int_{X}\left |f(x)-s(x)\right |dx<\epsilon\mbox{ and }\int_{Y}\left |g(y)-t(y)\right |dy<\epsilon .\]

Let

\begin{equation}{\label{maeq350}}\tag{91}
\int_{X}f(x)G(x)dx=\int_{X}s(x)G(x)dx+A_{1}.
\end{equation}

Then, we have

\begin{align*}
\left |A_{1}\right |& =\left |\int_{X}\left [f(x)-s(x)\right ]G(x)dx\right |\\ &  \leq\int_{X}\left [
\left |f(x)-s(x)\right |\int_{Y}\left |g(y)\right|\cdot\left |k(x,y)\right |dy\right ]dx\\
& <\epsilon\int_{Y}\left |g(y)\right|\cdot\left |k(x,y)\right |dy\leq\epsilon M\int_{Y}\left |g(y)\right |dy .
\end{align*}

Let

\[G(x)=\int_{Y}g(y)k(x,y)dy=\int_{Y}t(y)k(x,y)dy+A_{2}.\]

Then, we have

\begin{align*} \left |A_{2}\right | & =\left |\int_{Y}\left [g(y)-t(y)\right ]k(x,y)dy\right |\\ & \leq M\int_{Y}\left |g(y)-t(y)\right |dy<\epsilon M.\end{align*}

Let

\begin{equation}{\label{ma51}}\tag{92}
\int_{X}s(x)G(x)dx=\int_{X}s(x)\left [\int_{Y}t(y)k(x,y)dy\right ]dx+A_{3}.
\end{equation}

Then, we have

\begin{align*}
\left |A_{3}\right | & =\int_{X}s(x)\left [G(x)-\int_{Y}t(y)k(x,y)dy\right ]dx
\\ & =\left |A_{2}\int_{X}s(x)dx\right |\leq\epsilon M\int_{X}\left |s(x)\right |dx\\
& \leq\epsilon M\int_{X}\left (\left |s(x)-f(x)\right |+\left |f(x)\right |\right )dx\\
& <\epsilon^{2}M+\epsilon M\int_{X}\left |f(x)\right |dx.
\end{align*}

Using (\ref{maeq350}) and (\ref{ma51}), we obtain

\begin{equation}{\label{maeq351}}\tag{93}
\int_{X}f(x)G(x)dx=\int_{X}s(x)\left [\int_{Y}t(y)k(x,y)dy\right ]dx+A_{1}+A_{3}.
\end{equation}

The same argument can obtain

\begin{equation}{\label{maeq352}}\tag{94}
\int_{Y}g(y)F(y)dy=\int_{Y}t(y)\left [\int_{X}s(x)k(x,y)dx\right ]dy+B_{1}+B_{3},
\end{equation}

where

\[\left |B_{1}\right |<\epsilon M\int_{X}\left |f(x)\right |dx\]

and

\[\left |B_{3}\right |\leq\epsilon M\int_{Y}\left |t(y)\right |dy<\epsilon^{2}M+\epsilon M\int_{Y}\left |g(y)\right |dy.\]

By applying the equality (\ref{ma52}) to (\ref{maeq351}) and (\ref{maeq352}), we have

\begin{align*}
\left |\int_{X}f(x)G(x)dx-\int_{Y}g(y)F(y)dy\right | & \leq\left |A_{1}\right |+\left |A_{3}\right |+\left |B_{1}\right |+\left |B_{3}\right |\\
& <2\epsilon^{2}M+2\epsilon M\left [\int_{X}\left |f(x)\right |dx+\int_{Y}\left |g(y)\right |dy\right ].
\end{align*}

Since \(\epsilon\) can be any positive number, we conclude that the equality (\ref{maeq348}) holds true, and the proof is complete. \(\blacksquare\)

Let \(f\) and \(g\) be two Lebesgue-integrable function on \(I\) such that the product function \(f\cdot g\) is also Lebesgue-integrable on \(I\). The inner product of \(f\) and \(g\) is denoted and defined by

\[\langle f,g\rangle =\int_{I}f(x)g(x)dx.\]

If \(f^{2}\) is Lebesgue-integrable on \(I\), then the \(L^{2}\)-norm of \(f\) is denoted and defined by

\[\parallel f\parallel =\langle f,f\rangle^{1/2}=\left (\int_{I}f^{2}(x)dx\right )^{1/2}.\]

We denote by \({\cal L}^{2}(I)\) the collection of all measurable and Lebesgue-integrable functions on \(I\) satisfying \(f^{2}\in {\cal L}(I)\).

\begin{equation}{\label{map353}}\tag{95}\mbox{}\end{equation}

Proposition \ref{map353}. Suppose that \(f\) is Lebesgue-integrable on \(I\) and is bounded a.e. on \(I\). Then \(f^{2}\) is Lebesgue-integrable on \(I\).

Proof. Since \(f\) is Lebesgue-integrable on \(I\), it is measurable on \(I\). This also says that \(f^{2}\) is measurable on \(I\) and satisfies the inequality

\[\left |f(x)\right |^{2}\leq M\left |f(x)\right |\mbox{ a.e. on \(I\)},\]

where \(M\) is an upper bound for \(|f|\). By part (i) of Proposition \ref{map337}, we see that \(f^{2}\) is Lebesgue-integrable on \(I\). This completes the proof. \(\blacksquare\)

Proposition. If \(f,g\in {\cal L}^{2}(I)\), then \(fg\) is Lebesgue-integrable on \(I\) and \(af+bg\in {\cal L}^{2}(I)\).

Proof. Since \(f\) and \(g\) are measurable on \(I\), the product \(fg\) is also measurable on \(I\). Since

\[\left |f(x)g(x)\right |\leq\frac{1}{2}\left [f^{2}(x)+g^{2}(x)\right ],\]

and \(f^{2}\) and \(g^{2}\) are Lebesgue-integrable on \(I\) by Proposition \ref{map353}, part (i) of Proposition \ref{map337} says that \(fg\) is Lebegsue-integrable on \(I\). On the other hand, since \(af+bg\) is measurable on \(I\) and

\[(af+bg)^{2}=a^{2}f^{2}+2abfg+b^{2}g^{2}\]

is Lebesgue-integrabe on \(I\), we have \(af+bg\in {\cal L}^{2}(I)\). This completes the proof. \(\blacksquare\)

Proposition. Suppose that \(f,g,h\in {\cal L}^{2}(I)\) and \(c\in\mathbb{R}\). Then, we have the following properties:

(i) \(\langle f,g\rangle =\langle g,f\rangle\);

(ii) \(\langle f+g,h\rangle =\langle f,h\rangle +\langle g,h\rangle\);

(iii) \(\langle cf,g\rangle =c\langle f,g\rangle\);

(iv) \(\parallel cf\parallel =|c|\cdot\parallel f\parallel\);

(v) \(|\langle f,g\rangle |\leq\parallel f\parallel\cdot\parallel g\parallel\);

(vi) \(\parallel f+g\parallel\leq\parallel f\parallel +\parallel g\parallel\).

Proof. Parts (i) through (iv) can be obtained by the definition. Part (v) can be obtained by the following inequality

\[\int_{I}\left [\int_{I}\left |f(x)g(y)-g(x)f(y)\right |^{2}dy\right ]dx\geq 0.\]

Finally, using part (v), we have

\begin{align*}\parallel f+g\parallel & =\langle f+g,f+g\rangle\\ & =\langle f,f\rangle +2\langle f,g\rangle +\langle g,g\rangle
\\ & =\parallel f\parallel^{2}+\parallel g\parallel^{2}+2\langle f,g\rangle ,\end{align*}

which proves part (vi). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{mat358}}\tag{96}\mbox{}\end{equation}

Theorem \ref{mat358}. Let \(\{g_{n}\}_{n=1}^{\infty}\) be a sequence of functions in \({\cal L}^{2}(I)\) such that the infinite series \(\sum_{k=1}^{\infty}\parallel g_{k}\parallel\) is convergent. Then, the series of functions \(\sum_{k=1}^{\infty}g_{k}\) converges a.e. on \(I\) to a sum function \(g\in {\cal L}^{2}(I)\). Moreover, we have

\begin{equation}{\label{maeq356}}\tag{97}
\parallel g\parallel =\lim_{n\rightarrow\infty}\left |\!\left |\sum_{k=1}^{n}g_{k}\right |\!\right |\leq\sum_{k=1}^{\infty}\parallel g_{k}\parallel .
\end{equation}

Proof. We define

\[M=\sum_{k=1}^{\infty}\parallel g_{k}\parallel .\]

The triangle inequality says

\[\left |\!\left |\sum_{k=1}^{n}|g_{k}|\right |\!\right |\leq\sum_{k=1}^{n}\parallel g_{k}\parallel\leq M,\]

which implies

\begin{equation}{\label{maeq354}}\tag{98}
\int_{I}\left (\sum_{k=1}^{n}\left |g_{k}(x)\right |\right )^{2}dx=\left |\!\left |\sum_{k=1}^{n}|g_{k}|\right |\!\right |^{2}\leq M^{2}.
\end{equation}

For \(x\in I\), we define

\[f_{n}(x)=\left (\sum_{k=1}^{n}\left |g_{k}(x)\right |\right )^{2}.\]

Then, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) is increasing. Since each \(g_{k}\in {\cal L}^{2}(I)\), from (\ref{maeq354}), we have \(\int_{I}f_{n}(x)dx\leq M^{2}\) for all \(n\), which says that the sequence \(\{\int_{I}f_{n}(x)dx\}_{n=1}^{\infty}\) is convergent. By the monotone Theorem \ref{mat220}, there exists a Lebesgue-integrable function \(f\) on \(I\) satisfying

\[\lim_{n\rightarrow\infty}f_{n}=f\mbox{ a.e. on }I\]

and

\[\int_{I}f(x)dx=\lim_{n\rightarrow\infty}\int_{I}f_{n}(x)dx\leq M^{2}.\]

This shows that the series \(\sum_{k=1}^{\infty}g_{k}(x)\) is absolutely convergent a.e. on \(I\). We define the function

\[g(x)=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}g_{k}(x)\]

at those points in which the limit exists. We also define

\[G_{n}(x)=\left |\sum_{k=1}^{n}g_{k}(x)\right |^{2}.\]

Then, we see that each \(G_{n}\) is Lebesgue-integrable on \(I\) and

\[\lim_{n\rightarrow\infty}G_{n}(x)=|g(x)|^{2}\mbox{ a.e. on }I.\]

We also have

\[G_{n}(x)\leq f_{n}(x)\leq f(x)\mbox{ a.e. on \(I\)}.\]

Therefore, using the dominated convergence Theorem \ref{mat334}, the function \(|g|^{2}\) is Lebesgue-integrable on \(I\) with

\begin{equation}{\label{maeq355}}\tag{99}
\int_{I}\left |g(x)\right |^{2}dx=\lim_{n\rightarrow\infty}\int_{I}G_{n}(x)dx.
\end{equation}

Since \(g\) is measurable on \(I\), it says that \(g\in {\cal L}^{2}(I)\). We obtain

\[\int_{I}G_{n}(x)dx=\int_{I}\left |\sum_{k=1}^{n}g_{k}(x)\right |^{2}dx=\left |\!\left |\sum_{k=1}^{n}g_{k}\right |\!\right |^{2}\]

and

\[\int_{I}G_{n}(x)dx\leq\int_{I}f_{n}(x)dx\leq M^{2}.\]

From (\ref{maeq355}), we have

\[\parallel g\parallel^{2}=\lim_{n\rightarrow\infty}\left |\!\left |\sum_{k=1}^{n}g_{k}\right |\!\right |^{2}\leq M^{2},\]

which proves (\ref{maeq356}), and the proof is complete. \(\blacksquare\)

Theorem. (Riesz-Fischer Theorem). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a Cauchy sequence in \({\cal L}^{2}(I)\); that is, given any \(\epsilon >0\), there is an integer \(N\) satisfying

\begin{equation}{\label{maeq357}}\tag{100}
\parallel f_{m}-f_{n}\parallel <\epsilon\mbox{ whenever \(m,n>N\)}.
\end{equation}

Then, there exists a function \(f\in {\cal L}^{2}(I)\) satisfying

\[\lim_{n\rightarrow\infty}\parallel f_{n}-f\parallel =0.\]

Proof. By applying (\ref{maeq357}) repeatedly, we can find an increasing sequence of integers \(n_{1}<n_{2}<\cdots\) satisfying

\begin{equation}{\label{maeq359}}\tag{101}
\parallel f_{m}-f_{n_{k}}\parallel <\frac{1}{2^{k}}\mbox{ whenever }m\geq n_{k}.
\end{equation}

We define \(g_{1}=f_{n_{1}}\) and \(g_{k}=f_{n_{k}}-f_{n_{k-1}}\) for \(k\geq 2\). Since

\begin{align*} \sum_{k=1}^{\infty}\parallel g_{k}\parallel & \leq\parallel f_{n_{1}}\parallel +
\sum_{k=2}^{\infty}\parallel f_{n_{k}}-f_{n_{k-1}}\parallel\\  & <\parallel f_{n_{1}}\parallel
+\sum_{k=1}^{\infty}\frac{1}{2^{k}}=\parallel f_{n_{1}}\parallel +1,\end{align*}

it says that the series \(\sum_{k=1}^{\infty}\parallel g_{k}\parallel\) is convergent. Since each \(g_{n}\in {\cal L}^{2}(I)\), by Theorem \ref{mat358}, the series \(\sum_{k=1}^{\infty}g_{k}\) is convergent a.e. on \(I\) to a sum function \(f\in {\cal L}^{2}(I)\). Since
\[f-f_{n_{k}}=\sum_{r=k+1}^{\infty}\left (f_{n_{r}}-f_{n_{r-1}}\right )\]

and the series

\[\sum_{r=k+1}^{\infty}\parallel g_{r}\parallel=\sum_{r=k+1}^{\infty}\parallel f_{n_{r}}-f_{n_{r-1}}\parallel\]

is convergent, using (\ref{maeq359}), we have

\begin{equation}{\label{maeq360}}\tag{102}
\parallel f-f_{n_{k}}\parallel\leq\sum_{r=k+1}^{\infty}\parallel f_{n_{r}}-f_{n_{r-1}}\parallel <\sum_{r=k+1}^{\infty}\frac{1}{2^{r-1}}=\frac{1}{2^{k-1}}.
\end{equation}

Using (\ref{maeq359}) and (\ref{maeq360}), we also have

\begin{align*} \parallel f_{m}-f\parallel & \leq\parallel f_{m}-f_{n_{k}}\parallel +
\parallel f_{n_{k}}-f\parallel\\ & \leq\frac{1}{2^{k}}+\frac{1}{2^{k-1}}\\ & =\frac{3}{2^{k}}\mbox{ for }m\geq n_{k}.\end{align*}

Since \(n_{k}\rightarrow\infty\) as \(k\rightarrow\infty\), we conclude that \(\parallel f_{m}-f\parallel\rightarrow 0\) as \(m\rightarrow\infty\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{g}}\tag{G}\mbox{}\end{equation}

Multiple Lebesgue Integrals.

We are going to study the multiple Lebesgue integrals. In this case, we need to consider the \(n\)-dimensional intervals. Let \({\bf I}\) be a compact interval in \(\mathbb{R}^{n}\) given by

\[{\bf I}=I_{1}\times\cdots\times I_{n},\]

where each \(I_{k}\) is a compact interval in \(\mathbb{R}\). Let \({\cal P}_{k}\) be a partition of \(I_{k}\). The Cartesian product

\[{\cal P}={\cal P}_{1}\times\cdots\times {\cal P}_{n}\]

is called a partition of \({\bf I}\). Suppose that \({\cal P}_{k}\) decomposes \(I_{k}\) into \(m_{k}\) subintervals in \(\mathbb{R}\). Then \({\cal P}\) decomposes \({\cal I}\) into \(m=\prod_{k=1}^{n}m_{k}\) sub-intervals in \(\mathbb{R}^{n}\), which are named to be \({\bf J}_{1},\cdots ,{\bf J}_{m}\).

A function \(s\) defined on \({\bf I}\) is called a step function when a partition \({\cal P}\) of \({\bf I}\) exists such that the function \(s\) is constant on the interior of each sub-interval \({\bf J}_{k}\). In this case, we say \(s({\bf x})=c_{k}\) for \({\bf x}\in\mbox{int}({\bf J}_{k})\). The integral of \(s\) over \({\bf I}\) is defined by

\[\int_{I}s({\bf x})d\mu =\sum_{k=1}^{m}c_{k}\mu ({\bf J}_{k}).\]

Now, let \(\hat{\bf I}\) be a general \(n\)-dimensional interval, i.e., an interval in \(\mathbb{R}^{n}\) which need not be compact. A real-valued function \(s\) defined on \(\hat{\bf I}\) is called a {\bf step function} when there is a compact \(n\)-dimensional subinterval \({\bf I}\) of \(\hat{\bf I}\) such that \(s\) is step function on \({\bf I}\) and \(s({\bf x})=0\) for \({\bf x}\in\hat{\bf I}\setminus {\bf I}\). The integral of \(s\) over \(\hat{\bf I}\) is defined by

\[\int_{\hat{\bf I}}s({\bf x})d\mu =\int_{\bf I}s({\bf x})d\mu .\]

The integral is independent of the choice of \({\bf I}\).

Definition. A real-valued function \(f\) defined on an interval \({\bf I}\) in \(\mathbb{R}^{n}\) is called a Lebesggue function on \({\bf I}\) when there exists an increasing sequence of step functions \(s_{n}\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}({\bf x})f({\bf x})\mbox{ a.e. on }{\bf I}\]

and the limit

\[\lim_{n\rightarrow\infty}\int_{\bf I}s_{n}({\bf x})d\mu\]

exists. The sequence \(\{s_{n}\}_{n=1}^{\infty}\) is said to generate \(f\). The integral of \(f\) over \({\bf I}\) is defined by

\[\int_{\bf I}f({\bf x})d\mu =\lim_{n\rightarrow\infty}\int_{\bf I}s_{n}({\bf x})d{\bf x}.\]

The real-valued function \(f\) defined on \({\bf I}\) is said to be Lebesgue-integrable on \({\bf I}\) when \(f\) can be expressed as the form \(f=u-v\) such that \(u\) and \(v\) are Lebesgue functions on \({\bf I}\). In this case, the integral of \(f\) on \({\bf I}\) is defined by

\[\int_{\bf I}f({\bf x})d\mu =\int_{\bf I}u({\bf x})d\mu-\int_{\bf I}v({\bf x})d\mu .\]

Since these definitions are completely analogous to the one-dimensional case, many of the theorems in the one-dimensional case are also valid for the multiple integrals. For example, the dominated convergence theorem is also valid for the multiple integrals.

A real-valued function \(f\) defined on an interval \({\bf I}\) in \(\mathbb{R}^{n}\) is called measurable on \({\bf I}\) when there exists a sequence \(\{s_{n}\}_{n=1}^{\infty}\) on \({\bf I}\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}({\bf x})=f({\bf x})\mbox{ a.e. on }{\bf I}.\]

The properties of measurable function in the one-dimensional case are also valid for this more general setting. A subset \(S\) of \(\mathbb{R}^{n}\) is called measurable when its characteristic function \(\chi_{S}\) is measurable. If \(\chi_{S}\) is also Lebegsue-integrable on \(\mathbb{R}^{n}\), then the measure of \(S\) is denoted and defined by

\[\mu (S)=\int_{\mathbb{R}^{n}}\chi_{S}({\bf x})d\mu .\]

If \(\chi_{S}\) is measurable, but not Lebesgue-integrable on \(\mathbb{R}^{n}\), then we define \(\mu (S)=+\infty\). This set function \(\mu\) is called the Lebesgue measure. The properties of Lebesgue measure in one-dimensional case are also valid for this more general setting. For example, if \(\{A_{n}\}_{n=1}^{\infty}\) is a countable disjoint collection of measurable sets in \(\mathbb{R}^{n}\), then the union \(\bigcup_{k=1}^{\infty} A_{k}\) is measurable and

\[\mu\left (\bigcup_{k=1}^{\infty}A_{k}\right )=\sum_{k=1}^{\infty}\mu (A_{k}).\]

Theorem. Every open set \(S\) in \(\mathbb{R}^{n}\) can be expressed as the countable disjoint union of bounded cubes whose closure is contained in \(S\). Therefore \(S\) is measurable. Moreover, if \(S\) is bounded, the Lebesgue measure \(\mu (S)\) is finite.

Proof. Fix an integer \(m\geq 1\), we consider all half-open intervals in \(\mathbb{R}\) of the form

\[\left (\frac{k}{2^{m}},\frac{k+1}{2^{m}}\right ]\mbox{ for }k=0,\pm 1,\pm 2,\cdots .\]

All the intervals are of length \(2^{-m}\), and they form a countable disjoint collection whose union is \(\mathbb{R}\). The Cartesian product of \(n\) such intervals is an \(n\)-dimensional cube of edge length \(2^{-m}\). Let \(F_{m}\) denote the collection of all these cubes. Then \(F_{m}\) is a countable disjoint collection whose union is \(\mathbb{R}^{m}\). Note that the cubes in \(F_{m+1}\) are obtained by bisecting the edges of those in \(F_{m}\). Therefore, if \(Q_{m}\) is a cube in \(F_{m}\) and if \(Q_{m+1}\) is a cube in \(F_{m+1}\), then either \(Q_{m+1}\subseteq Q_{m}\) or \(Q_{m+1}\) and \(Q_{m}\) are disjoint. Now, we extract a subcollection \(G_{m}\) from \(F_{m}\) as follows. If \(m=1\), \(G_{1}\) consists of all cubes in \(F_{1}\) whose closure lies in \(S\). If \(m=2\), \(G_{2}\) consists of all cubes in \(F_{2}\) whose closure lies in \(S\) but not in any of the cubes in \(G_{1}\). If \(m=3\), \(G_{3}\) consists of all cubes in \(F_{3}\) whose closure lies in \(S\) but not in any of the cubes in \(G_{1}\) or \(G_{2}\), and so on. We define

\[T=\bigcup_{m=1}^{\infty}\bigcup_{Q\in G_{m}}Q,\]

i.e., \(T\) is the union of all cubes in \(G_{1},G_{2},\cdots\). We shall prove that \(S=T\) and this will prove the theorem, since \(T\) is a countable disjoint collection of cubes whose closure lies in \(S\). Since each \(Q\) in \(G_{m}\) is a subset of \(S\), we immediately have \(T\subseteq T\). Therefore, we remain to show \(S\subseteq T\). Let \({\bf p}=(p_{1},\cdots ,p_{n})\) be a point in \(S\). Since \(S\) is open, there is a cube with center \({\bf p}\) and edge-length \(\delta >0\), which lies in \(S\). We choose \(m\) satisfying \(2^{-m}<\delta /2\). Then, for each \(i\), we have

\[p_{i}-\frac{\delta}{2}<p_{i}-\frac{1}{2^{m}}<p_{i}<p_{i}+\frac{1}{2^{m}}<p_{i}+\frac{\delta}{2}.\]

Now, we choose \(k_{i}\) satisfying

\[\frac{k_{i}}{2^{m}}<p_{i}\leq\frac{k_{i}+1}{2^{m}}.\]

Let \(Q\) be the Cartesian product of the intervals

\[\left (\frac{k_{i}}{2^{m}},\frac{k_{i}+1}{2^{m}}\right ]\mbox{ for }i=1,2,\cdots ,n.\]

Then \({\bf p}\in Q\) for some cube \(Q\) in \(F_{m}\). If \(m\) is the smallest integer with this property, then \(Q\in G_{m}\), which says that \({\bf p}\in T\). Therefore, we obtain \(S\subseteq T\). The statements about the measurability of \(S\) follow at once from the countably additive property of Lebesgue measure. This completes the proof. \(\blacksquare\)

Next, we are going to consider the Fubini’s theorem in \(\mathbb{R}^{2}\). If \({\bf I}=[a,b]\times [c,d]\) is a compact interval in \(\mathbb{R}^{2}\) and if \(f\) is Riemann-integrable on \({\bf I}\), then we have the following formula

\begin{equation}{\label{maeq397}}\tag{103}
\int\!\!\!\!\!\int_{\bf I}f(x,y)d(x,y)=\int_{c}^{d}\left [\underline{\int}_{a}^{b}f(x,y)dx\right ]dy.
\end{equation}

There is a companion formula with the lower integral replaced by the upper integral. The upper and lower integrals are needed, since the hypothesis of Riemann-integrability on \({\bf I}\) is not enough to ensure the existence of the one-dimensional Riemann integral \(\int_{a}^{b}f(x,y)dx\). Fortunately, this difficulty does not arise in the Lebesgue integrals. The Fubini’s theorem for double Lebesgue integrals gives us the following formula

\begin{align*} \int\!\!\!\!\!\int_{\bf I}f(x,y)d\mu (x,y) & =\int_{c}^{d}\left [\int_{a}^{b}f(x,y)d\mu (x)\right ]d\mu (y)\\ & =\int_{a}^{b}\left [\int_{c}^{d}f(x,y)d\mu (y)\right ]d\mu (x)\end{align*}

under the hypothesis that \(f\) is Lebesgue-integrable on \({\bf I}\). We shall show that the inner inetgarls always exist as Lebesgue integrals.

\begin{equation}{\label{mat405}}\tag{104}\mbox{}\end{equation}

Theorem \ref{mat405}. (Fubini’s Theorem for Step Functions). Let \(s\) be a step function on \(\mathbb{R}^{2}\). Then, for any fixed \(y\in\mathbb{R}\), the integral

\[\int_{\mathbb{R}}s(x,y)d\mu (x)\]

exists and, as a function of \(y\), is also Lebesgue-integrable on \(\mathbb{R}\). Moreover, we have

\begin{equation}{\label{maeq398}}\tag{105}
\int\!\!\!\!\!\int_{\mathbb{R}^{2}}s(x,y)d\mu (x,y)=\int_{\mathbb{R}}\left [\int_{\mathbb{R}}s(x,y)d\mu (x)\right ]d\mu (y).
\end{equation}

Similarly, for any fixed \(x\in\mathbb{R}\), the integral

\[\int_{\mathbb{R}}s(x,y)d\mu (y)\]

exists and, as a function of \(x\), is also Lebesgue-integrable on \(\mathbb{R}\). Moreover, we have

\begin{equation}{\label{maeq399}}\tag{106}
\int\!\!\!\!\!\int_{\mathbb{R}^{2}}s(x,y)d\mu (x,y)=\int_{\mathbb{R}}\left [\int_{\mathbb{R}}s(x,y)d\mu (y)\right ]d\mu (x).
\end{equation}

In other words, we have

\begin{align*} \int\!\!\!\!\!\int_{\mathbb{R}^{2}}s(x,y)d\mu (x,y) & =\int_{\mathbb{R}}
\left [\int_{\mathbb{R}}s(x,y)d\mu (x)\right ]d\mu (y)\\ & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}s(x,y)d\mu (y)\right ]d\mu (x).\end{align*}

Proof. This theorem can be derive from the reduction formula (\ref{maeq397}) for Riemann integrals. Here, we give a direct proof independent of the Riemann integration. There is a compact interval \(I=[a,b]\times [c,d]\) such that \(s\) is a step function on \(I\) and \(s(x,y)=0\) for \((x,y)\in\mathbb{R}^{2}\setminus I\). There is a partition of \(I\) into \(mn\) sub-intervals

\[I_{ij}=[x_{i-1},x_{i}]\times [y_{j-1},y_{j}]\]

such that \(s\) is constant on the interior of \(I_{ij}\), which is defined to be \(s(x,y)=c_{ij}\) for \((x,y)\in\mbox{int}I_{ij}\). Then, we have
\begin{align*} \int\!\!\!\!\!\int_{I_{ij}}s(x,y)d\mu (x,y) & =c_{ij}(x_{i}-x_{i-1})(y_{j}-y_{j-1})\\ & =\int_{y_{j-1}}^{y_{j}}\left [\int_{x_{i-1}}^{x_{i}}s(x,y)dx\right ]dy.\end{align*}

Summing on \(i\) and \(j\), we obtain

\[\int\!\!\!\!\!\int_{I}s(x,y)d\mu (x,y)=\int_{c}^{d}\left [\int_{a}^{b}s(x,y)dx\right ]dy.\]

Since \(s\) vanishes outside \(I\), this proves (\ref{maeq398}). The similar argument can also prove (\ref{maeq399}) and the proof is complete. \(\blacksquare\)

For extending Fubini’s theorem to Lebesgue-integrable functions, we need some further results concerning sets of measure zero.

\begin{equation}{\label{mat400}}\tag{107}\mbox{}\end{equation}

Theorem \ref{mat400}. Let \(S\) be a subset of \(\mathbb{R}^{n}\). Then \(S\) has measure zero if and only if there exists a countable collection of \(n\)-dimensional intervals \(\{{\bf J}_{k}\}_{k=1}^{\infty}\) such that the sum of their measures is finite and each point in \(S\) belongs to \({\bf J}_{k}\) for
infinitely many \(k\).

Proof. We first assume that \(S\) has \(n\) measure zero. Then for every \(m\geq 1\), \(S\) can be covered by a countable collection of \(n\)-dimensional intervals \(\{{\bf I}_{mr}\}_{r=1}^{\infty}\) such that the sum of their measure is less than \(2^{-m}\). The set \(A\) consisting of all intervals \(\{{\bf I}_{mr}\}_{m,r=1}^{\infty}\) is a countable collection which covers \(S\), and the sum of their measures is less than

\[\sum_{m=1}^{\infty}2^{-m}=1.\]

Moreover, if \({\bf a}\in S\), then, for each \(m\), we have \({\bf a}\in {\bf I}_{mr}\) for some \(r\). Therefore, if we rename \({\bf I}_{mr}\) to be \({\bf J}_{k}\), we see that \({\bf a}\) belongs to \({\bf J}_{k}\) for infinitely many \(k\). Conversely, we assume that there is a countable collection of \(n\)-dimensional intervals \(\{{\bf J}_{k}\}_{k=1}^{\infty}\) such that the series \(\sum_{k=1}^{\infty}\mu ({\bf J}_{k})\) is convergent, and such that each point in \(S\) belongs to \({\bf J}_{k}\) for infinitely many \(k\). The convergence of the series \(\sum_{k=1}^{\infty}\mu ({\bf J}_{k})\) says that, given any \(\epsilon >0\), there is an integer \(N\) satisfying

\[\sum_{k=N}^{\infty}\mu ({\bf J}_{k})<\epsilon .\]

Each point of \(S\) lies in the set \(\bigcup_{k=N}^{\infty}{\bf J}_{k}\). Therefore, we have

\[S\subseteq\bigcup_{k=N}^{\infty}{\bf J}_{k}.\]

This shows that \(S\) is covered by a countable collection of intervals such that the sum of their measures is less that \(\epsilon\). This says that \(S\) has measure zero, and the proof is complete. \(\blacksquare\)

Let \(S\) be a subset of \(\mathbb{R}^{2}\). Given any \((x,y)\in\mathbb{R}^{2}\), we define

\[S_{y}=\left\{x\in\mathbb{R}:(x,y)\in S\right\}\mbox{ and }S_{x}=\left\{y\in\mathbb{R}:(x,y)\in S\right\}.\]

Proposition. Suppose that \(S\) is a subset of \(\mathbb{R}^{2}\) with measure zero. Then \(S_{y}\) has measure zero for almost all \(y\in\mathbb{R}\),
and \(S_{x}\) has measure zero for almost all \(x\in\mathbb{R}\).

Proof. We are going to use Theorem \ref{mat400} to prove that \(S_{y}\) has measure zero for almost all \(x\in\mathbb{R}\). Since \(S\) has measure zero, Theorem \ref{mat400} says that there is a countable collection of rectangles \(\{{\bf I}_{k}\}_{k=1}^{\infty}\) such that the series

\begin{equation}{\label{maeq401}}\tag{108}
\sum_{k=1}^{\infty}\mu ({\bf I}_{k})
\end{equation}

is convergent and such that every point \((x,y)\) of \(S\) belongs to \({\bf I}_{k}\) for infinitely many \(k\). We write \({\bf I}_{k}=X_{k}\times Y_{k}\), where \(X_{k}\) and \(Y_{k}\) are intervals in \(\mathbb{R}\). Then, we have

\begin{align*} \mu ({\bf I}_{k}) & =\mu (X_{k})\mu (Y_{k})=\mu (X_{k})\int_{\mathbb{R}}\chi_{Y_{k}}(y)d\mu
\\ & =\int_{\mathbb{R}}\mu (X_{k})\chi_{Y_{k}}(y)d\mu ,\end{align*}

where \(\chi_{Y_{k}}\) is the characteristic function of the interval \(Y_{k}\). Let \(g_{k}(y)=\mu (X_{k})\chi_{Y_{k}}(y)\). Then, the series (\ref{maeq401}) implies that the series

\[\sum_{k=1}^{\infty}\int_{\mathbb{R}}g_{k}(y)d\mu\]

is convergent. Now \(\{g_{k}\}_{k=1}^{\infty}\) is a sequence of nonnegative and Lebesgue-integrable functions on \(\mathbb{R}\) such that the series (\ref{maeq401}) is convergent. By the monotone convergence Theorem \ref{mat232}, the series \(\sum_{k=1}^{\infty}g_{k}\) is convergent a.e. on \(\mathbb{R}\). In other words, there is a subset \(T\) of \(\mathbb{R}\) with measure zero such that the series

\begin{equation}{\label{maeq404}}\tag{109}
\sum_{k=1}^{\infty}\mu (X_{k})\chi_{Y_{k}}(y)
\end{equation}

is convergent for all \(\mathbb{R}\setminus T\). Given any \(y\in\mathbb{R}\setminus T\), we shall prove that the set \(S_{y}\) has measure zero. If \(S_{y}\) is empty, then the result is obvious. We assume that \(S_{y}\) is not empty. Let

\[A(y)=\left\{X_{k}:y\in Y_{k}\mbox{ for }k=1,2,\cdots\right\}.\]

Then \(A(y)\) is a countable collection of one-dimensional intervals which we re-label as \(\{J_{p}\}_{p=1}^{\infty}\). The sum of the lengths of all the intervals \(J_{p}\) is convergent because of (\ref{maeq404}). If \(x\in S_{y}\), then \((x,y)\in S\), which says

\[(x,y)\in {\bf I}_{k}=X_{k}\times Y_{k}\]

for infinitely many \(k\). Therefor, we have \(x\in J_{p}\) for infinitely many \(p\). By the one-dimensional version of Theorem \ref{mat400}, it follows that \(S_{y}\) has measure zero. This shows that \(S_{y}\) has measure zero for almost all \(y\in\mathbb{R}\). The similar argument can prove that \(S_{x}\) has measure zero for almost all \(x\in\mathbb{R}\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{mat262}}\tag{110}\mbox{}\end{equation}

Theorem \ref{mat262}. ( Fubini’s Theorem for Double Integrals). Suppose that the real-valued function \(f\) is Lebesgue-integrable on
$\mathbb{R}^{2}$. Then, we have the following properties.

(i) There is subset \(T\) of \(\mathbb{R}\) with measure zero such that the Lebesgue integral

\[\int_{\mathbb{R}}f(x,y)d\mu (x)\]

exists for all \(y\in\mathbb{R}\setminus T\).

(ii) The following function \(G\) defined on \(\mathbb{R}\) by

\[G(y)=\left\{\begin{array}{ll}
{\displaystyle \int_{\mathbb{R}}f(x,y)d\mu (x)} &
\mbox{if \(y\in\mathbb{R}\setminus T\)}\\
0 & \mbox{if \(y\in T\)}.
\end{array}\right .\]

is Lebesgue-integrable on \(\mathbb{R}\).

(iii) We have

\begin{align*}
\int_{\mathbb{R}}G(y)d\mu (y) & =\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y)\\
& =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f(x,y)d\mu (x)\right ]d\mu (y)
\\ & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f(x,y)d\mu (y)\right ]d\mu (x).
\end{align*}

Proof. The case of step functions has been proved in Theorem \ref{mat405}. Next, we first prove it for Lebesgue functions. For \(f\in {\cal L}(\mathbb{R}^{2})\), there is an increasing sequence of step functions \(\{s_{n}\}_{n=1}^{\infty}\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x,y)=f(x,y)\mbox{ for all }(x,y)\in\mathbb{R}^{2}\setminus S,\]

where \(S\) is a set of measure zero. We also have

\[\lim_{n\rightarrow\infty}\int\!\!\!\!\!\int_{\mathbb{R}^{2}}s_{n}(x,y)d\mu (x,y)=\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y).\]

Since \((x,y)\in\mathbb{R}^{2}\setminus S\) if and only if \(x\in\mathbb{R}\setminus S_{y}\), for any fixed \(y\), we have

\begin{equation}{\label{maeq407}}\tag{111}
\lim_{n\rightarrow\infty}s_{n}(x,y)=f(x,y)\mbox{ for }x\in\mathbb{R}\setminus S_{y}.
\end{equation}

We define

\[t_{n}(y)=\int_{\mathbb{R}}s_{n}(x,y)d\mu (x).\]

This integral exists for each \(y\) and is an integrable function of \(y\). Therefore, using Theorem \ref{mat405}, we have

\begin{align*} \int_{\mathbb{R}}t_{n}(y)d\mu (y) & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}
s_{n}(x,y)d\mu (x)\right ]d\mu (y)\\ & =\int\!\!\!\!\!\int_{\mathbb{R}^{2}}s_{n}(x,y)d\mu (x,y)
\\ & \leq\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y),\end{align*}

which says that the sequence \(\{\int_{\mathbb{R}}t_{n}(y)d\mu (y)\}_{n=1}^{\infty}\) is bounded above. Since the sequence \(\{t_{n}\}_{n=1}^{\infty}\) is increasing, it follows that the limit

\[\lim_{n\rightarrow\infty}\int_{\mathbb{R}}t_{n}(y)d\mu (y)\]

exists. Therefore, using the monotone convergence Theorem \ref{mat220}, there is a Lebesgue-integrable function \(t\) on \(\mathbb{R}\) satisfying

\[\lim_{n\rightarrow\infty}t_{n}(y)=t(y)\mbox{ a.e. on }\mathbb{R}\]

and

\[\int_{\mathbb{R}}t(y)d\mu (y)=\lim_{n\rightarrow\infty}\int_{\mathbb{R}}t_{n}(y)d\mu (y).\]

In other words, there exists a set \(T_{1}\) of measure zero satisfying

\[\lim_{n\rightarrow\infty}t_{n}(y)=t(y)\mbox{ for all }y\in\mathbb{R}\setminus T_{1}.\]

Since \(\{t_{n}\}_{n=1}^{\infty}\) is increasing, we also have

\[\int_{\mathbb{R}}s_{n}(x,y)d\mu (x)=t_{n}(y)\leq t(y)\mbox{ for }t\in\mathbb{R}\setminus T_{1}.\]

Applying the monotone convergence Theorem \ref{mat220} again to the sequence \(\{s_{n}\}_{n=1}^{\infty}\), for any fixed \(y\in\mathbb{R}\setminus T_{1}\), there exists a Lebesgue-integrable function \(g\) on \(\mathbb{R}\) satisfying

\[\lim_{n\rightarrow\infty}s_{n}(x,y)=g(x,y)\mbox{ for }x\in\mathbb{R}\setminus A_{y},\]

where \(A_{y}\) is a set of measure zero and depends on \(y\). Comparing this to (\ref{maeq407}), for any fixed \(y\in\mathbb{R}\setminus T_{1}\), we have

\begin{equation}{\label{maeq408}}\tag{112}
g(x,y)=f(x,y)\mbox{ for }x\in\mathbb{R}\setminus (A_{y}\cup S_{y}).
\end{equation}

Since \(S_{y}\) has measure zero for almost all \(y\), we can say that \(S_{y}\) has measure zero for all \(y\in\mathbb{R}\setminus T_{2}\), where \(T_{2}\) has measure zero. Let \(T=T_{1}\cup T_{2}\). Then \(T\) has measure zero. Since \(A_{y}\) has measure zero, the set \(A_{y}\cup S_{y}\) has measure zero for \(y\in\mathbb{R}\setminus T\). From (\ref{maeq408}), we have

\[g(x,y)=f(x,y)\mbox{ a.e. on }\mathbb{R}.\]

Since the integral \(\int_{\mathbb{R}}g(x,y)d\mu (x)\) exists for \(y\in\mathbb{R}\setminus T\), it follows that the integral \(\int_{\mathbb{R}}f(x,y)d\mu (x)\) also exists for \(y\in\mathbb{R}\setminus T\). This proves part (i).

For \(y\in\mathbb{R}\setminus T\), we also have

\begin{align}
\int_{\mathbb{R}}f(x,y)d\mu (x) & =\int_{\mathbb{R}}g(x,y)d\mu (x)\nonumber\\ =\lim_{n\rightarrow\infty}\int_{\mathbb{R}}s_{n}(x,y)d\mu (x)=t(y).\label{maeq409}\tag{113}\end{align}

Since \(t\) is Lebesgue-integrable on \(\mathbb{R}\), this proves part (ii). Now, we have

\begin{align*}
\int_{\mathbb{R}}t(y)d\mu (y) & =\lim_{n\rightarrow\infty}\int_{\mathbb{R}}t_{n}(y)d\mu (y)\\ & =\lim_{n\rightarrow\infty}
\int_{\mathbb{R}}\left [\int_{\mathbb{R}}s_{n}(x,y)d\mu (x)\right ]d\mu (y)\\
& =\lim_{n\rightarrow\infty}\int\!\!\!\!\!\int_{\mathbb{R}^{2}}s_{n}(x,y)d\mu (x,y)\\ & =\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y).
\end{align*}

Comparing this with (\ref{maeq409}), we obtain part (iii). This proves Fubini’s theorem for Lebesgue functions. To prove it for Lebesgue-integrable functions, we write \(f=u-v\), where \(u\) and \(v\) are Lebesgue functions on \(\mathbb{R}^{2}\). Then, we have

\begin{align*}
\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y)& =\int\!\!\!\!\!\int_{\mathbb{R}^{2}}u(x,y)d\mu (x,y)\int\!\!\!\!\!\int_{\mathbb{R}^{2}}v(x,y)d\mu (x,y)\\
& =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}u(x,y)d\mu (x)\right ]d\mu (y)-\int_{\mathbb{R}}\left [\int_{\mathbb{R}}v(x,y)d\mu (x)\right ]d\mu (y)\\
& =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}\left (u(x,y)-v(x,y)\right )d\mu (x)\right ]d\mu (y)\\ & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f(x,y)
d\mu (x)\right ]d\mu (y).
\end{align*}

This completes the proof. \(\blacksquare\)

Theorem.  (Fubini’s Theorem for Double Integrals). Suppose that the real-valued function is defined on a compact rectangle \({\bf I}=[a,b]\times [c,d]\), and that \(f\) is continuous a.e. on \({\bf I}\). Then \(f\) is Lebesgue-integrable on \({\bf I}\), and we have

\begin{align*} \int\!\!\!\!\!\int_{\bf I}f(x,y)d\mu (x,y) & =\int_{c}^{d}\left [\int_{a}^{b}f(x,y)d\mu (x)\right ]d\mu (y)
\\ & =\int_{a}^{b}\left [\int_{c}^{d}f(x,y)d\mu (y)\right ]d\mu (x).\end{align*}

Proof. The results follow immediately from Theorem \ref{mat262} and the two-dimensional analogue of Theorem \ref{mat204}. \(\blacksquare\)

We can extend the Fubini’s theorem to the higher-dimensional integrals. If \(f\) is Lebesgue-integrable on \(\mathbb{R}^{m+k}\), the analog of Theorem \ref{mat262} shows that

\begin{align*} \int_{\mathbb{R}^{m+k}}f({\bf x},{\bf y})d\mu ({\bf x},{\bf y})
& =\int_{\mathbb{R}^{k}}\left [\int_{\mathbb{R}^{m}}f({\bf x},{\bf y})d\mu ({\bf x})\right ]
d\mu ({\bf y})\\ & =\int_{\mathbb{R}^{m}}\left [\int_{\mathbb{R}^{k}}f({\bf x},{\bf y})d\mu ({\bf y})
\right ]d\mu ({\bf x}).\end{align*}

The proof can follow the similar arguments of Theorem \ref{mat262}. \(\blacksquare\)

Theorem.  (Tonelli-Hobson Test for Integrability). Suppose that \(f\) is measurable on \(\mathbb{R}^{2}\). We also assume that at least one of the following two iterated integrals

\[\int_{\mathbb{R}}\left [\int_{\mathbb{R}}|f(x,y)|d\mu (x)\right ]d\mu (y)
\mbox{ or }\int_{\mathbb{R}}\left [\int_{\mathbb{R}}|f(x,y)|d\mu (y)\right ]d\mu (x)\]

exist. Then, we have the following properties.

(i) The function \(f\) is Lebesgue-integrable on \(\mathbb{R}^{2}\).

(ii) We have

\begin{align*} \int\!\!\!\!\!\int_{\mathbb{R}^{2}}f(x,y)d\mu (x,y) & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f(x,y)d\mu (x)\right ]d\mu (y)
\\ & =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f(x,y)d\mu (y)\right ]d\mu (x).\end{align*}

Proof. We shall use Fubini’s theorem to prove part (i). Suppose that the iterated integral

\[\int_{\mathbb{R}}\left [\int_{\mathbb{R}}|f(x,y)|d\mu (x)\right ]d\mu (y)\]

exists. Let \(\{s_{n}\}_{n=1}^{\infty}\) denote the increasing sequence of nonnegative step functions defined as follows

\[s_{n}(x,y)=\left\{\begin{array}{ll}
n & \mbox{if \(|x|\leq n\) and \(|y|\leq n\)}\\
0 & \mbox{otherwise}.
\end{array}\right .\]

We define

\[f_{n}(x,y)=\min\left\{s_{n}(x,y),|f(x,y)|\right\}.\]

Both \(s_{n}\) and \(|f|\) are measurable, which says that \(f_{n}\) is measurable. We also have

\[0\leq f_{n}(x,y)\leq s_{n}(x,y).\]

Therefore \(f_{n}\) is dominated by a Lebesgue-integrable function, which says that \(f_{n}\) is Lebesgue-integrable on \(\mathbb{R}^{2}\). Now, we can apply Fubini’s theorem to \(f_{n}\) along with the inequality

\[0\leq f_{n}(x,y)\leq |f(x,y)|\]

to obtain

\begin{align*} \int\!\!\!\!\!\int_{\mathbb{R}^{2}}f_{n}(x,y)d\mu (x,y)
& =\int_{\mathbb{R}}\left [\int_{\mathbb{R}}f_{n}(x,y)d\mu (x)\right ]d\mu (y)
\\ & \leq\int_{\mathbb{R}}\left [\int_{\mathbb{R}}|f(x,y)|d\mu (x)\right ]d\mu (y),\end{align*}

which says that the sequence \(\{\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f_{n}(x,y)d\mu (x,y)\}_{n=1}^{\infty}\) is bounded above. Since \(\{f_{n}\}_{n=1}^{\infty}\) is increasing, it says that the limit

\[\lim_{n\rightarrow\infty}\int\!\!\!\!\!\int_{\mathbb{R}^{2}}f_{n}(x,y)d\mu (x,y)\]

exists. By the monotone convergence Theorem \ref{mat220}, the sequence \(\{f_{n}\}_{n=1}^{\infty}\) is convergent a.e. on \(\mathbb{R}^{2}\) to a limit function that is Lebesgue-integrable on \(\mathbb{R}^{2}\). Since

\[\lim_{n\rightarrow\infty}f_{n}(x,y)=|f(x,y)|,\]

it follows that \(|f|\) is Lebesgue-integrable on \(\mathbb{R}^{2}\). Since \(f\) is measurable, we conclude that \(f\) is Lebesgue-integrable on \(\mathbb{R}^{2}\). The proof is similar when the other iterated integral exists. Part (ii) follows from part (i) by using the Fubini’s theorem. This completes the proof. \(\blacksquare\)