The Lebesgue Integral Using Lebesgue Measure

Bartholomeus van Hove (1790-1880) was a Dutch painter.

We have sections

• The Lebesgue Integral of Simple Functions 
• The Lebesgue Integral of Bounded Measurable Functions
• The Lebesgue Integral of Nonnegative Functions
• The General Lebesgue Integrals
• Iterated Integrations
• Inequalities
• Differentiation and Integration
• Functions of Bounded Variation
• The Real-Valued Indefinite Integrals
• The Set-Valued Indefinite Integrals
• The Lebesgue-Stieltjes Integrals

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

The Lebesgue Integral of Simple Functions.

In the elementary calculus, the Riemann integral is defined based on the upper sum and lower sum. In this case, the following function

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

is not Riemann-integrable. Therefore, the Lebesgue integral tries to extend the Riemann integral such that the above function \(f\) can have the Lebesgue integral. Here, we are going to present the Lebesgue integral using Lebesgue measure. Let \(\nu\) be a Lebesgue measure on \(\bar{\mathbb{R}}^{n}\). Suppose that the simple function \(\phi :\bar{\mathbb{R}}^{n}\rightarrow \bar{\mathbb{R}}\) is Lebesgue-measurable and has the form of

\[\phi =\sum_{i=1}^{n}\alpha_{i}\chi_{E_{i}},\]

where \(\alpha_{1},\cdots ,\alpha_{n}\) are the distinct values of \(\phi\) with

\[E_{i}=\{{\bf x}:\phi ({\bf x})=\alpha_{i}\}.\]

Given any Lebesgue-measurable set \(E\) with \(\nu (E)<+\infty\), we define the Lebesgue integral as follows

\[\int_{E}\phi d\nu =\sum_{i=1}^{n}\alpha_{i}\nu\left (E_{i}\cap E\right ).\]

The convention \(0\cdot\infty=0\) is used here. It may happen that \(\alpha_{i}=0\) and \(\nu (E_{i}\cap E)=\infty\) for some \(i\). Suppose that \(\phi\) vanishes outside a set of finite measure. Then

\[\int_{\bar{\mathbb{R}}^{n}}\phi d\nu=\sum_{i=1}^{n}\alpha_{i}\nu\left (E_{i}\right ).\]

We also have

\[\int_{E}\phi d\nu =\int_{\bar{\mathbb{R}}^{n}}\phi\cdot\chi_{E}d\nu ,\]

where \(\chi_{E}\) is the characteristic function of set \(E\).

\begin{equation}{\label{rap76}}\tag{1}\mbox{}\end{equation}

Proposition \ref{rap76}. We have the following properties.

(i) Suppose that the simple function \(\phi\) satisfies \(\phi\geq 0\) a.e. Then, for any Lebesgue-measurable subset \(E\) of \(\bar{\mathbb{R}}^{n}\), we have

\[\int_{E}\phi d\nu\geq 0.\]

(ii) Given any Lebesgue-measurable set \(E\) with \(\nu (E)=0\), suppose that the simple function \(\phi\) satisfies \(\phi\geq 0\). Then

\[\int_{E}\phi d\nu =0.\]

(iii) Let \(F\) and \(E\) be Lebesgue measurable sets satisfying \(F\subseteq E\). Suppose that \(\phi\) is a simple function defined on \(E\) satisfying \(\phi\geq 0\) a.e. Then

\[\int_{F}\phi d\nu\leq\int_{E}\phi d\nu .\]

(iv) Given a function

\[\phi =\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}}\mbox{ with }A_{i}\cap A_{j}=\emptyset\mbox{ for }i\neq j.\]

Suppose that each \(A_{i}\) is Lebesgue measurable with \(\nu (A_{i})<+\infty\). Then, we have

\[\int_{E}\phi d\nu =\sum_{i=1}^{n}\alpha_{i}\nu (E\cap A_{i}).\]

Proof. Parts (i) and (ii) are obvious by definition. Part (iii) followsimmediately from Proposition \ref{rap62}. To prove part (iv), we first re-write \(\phi\) as a simple function. Let \(\{\beta_{1},\cdots,\beta_{m}\}\) be a subset of \(\{\alpha_{1},\cdots,\alpha_{n}\}\) with$\beta_{j_{1}}\neq\beta_{j_{2}}$ for \(j_{1}\neq j_{2}\). We have

\[E_{j}=\{{\bf x}:\phi ({\bf x})=\beta_{j}\}=\bigcup_{\{i:\alpha_{i}=\beta_{j}\}}A_{i}.\]

In this case, \(\phi\) is a simple function represented by

\[\phi=\sum_{j=1}^{m}\beta_{j}\chi_{E_{j}}.\]

Using the additivity of \(\nu\), we have

\[\beta_{j}\nu (E\cap E_{j})=\sum_{\{i:\alpha_{i}=\beta_{j}\}}\alpha_{i}\nu (E\cap A_{i}).\]

Therefore, by definition, we have

\[\int_{E}\phi d\nu=\sum_{j=1}^{m}\beta_{j}\nu (E\cap E_{j})=\sum_{i=1}^{n}\alpha_{i}\nu (E\cap A_{i}).\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap78}}\tag{2}\mbox{}\end{equation}

Proposition \ref{rap78}. Let \(\phi_{1}\) and \(\phi_{2}\) be two simple functions, and let \(E\) be a Lebesgue measurable set. We have the following properties.

(i) Given any two constants \(a\) and \(b\), we have

\[\int_{E}\left (a\phi_{1}+b\phi_{2}\right )d\nu =a\int_{E}\phi_{1}d\nu +b\int_{E}\phi_{2}d\nu .\]

(ii) Suppose that \(\phi_{1}\leq\phi_{2}\) a.e. Then, we have

\[\int_{E}\phi_{1}d\nu\leq\int_{E}\phi_{2}d\nu .\]

Proof. Suppose that the simple functions \(\phi_{1}\) and \(\phi_{2}\) can be represented by

\[\phi_{1}=\sum_{i=1}^{n}a_{i}\chi_{A_{i}}\mbox{ and }\phi_{2}=\sum_{j=1}^{m}b_{j}\chi_{B_{j}}.\]

Let \(\{E_{k}\}_{k=1}^{N}\) be obtained by taking all the intersections \(A_{i}\cap B_{j}\). Then \(\{E_{k}\}_{k=1}^{N}\) form a finite disjoint collection of Lebesgue-measurable sets. In this case, we can also write

\[\phi_{1}=\sum_{k=1}^{N}a_{k}\chi_{E_{k}}\mbox{ and }\phi_{2}=\sum_{k=1}^{N}b_{k}\chi_{E_{k}}.\]

Therefore, we obtain

\begin{align*} a\phi_{1}+b\phi_{2} & =a\sum_{k=1}^{N}a_{k}\chi_{E_{k}}+b\sum_{k=1}^{N}b_{k}\chi_{E_{k}}\\ & =\sum_{k=1}^{N}(aa_{k}+bb_{k})\chi_{E_{k}}.\end{align*}

Using part (ii) of Proposition \ref{rap76}, we also have

\begin{align*}
\int_{E}\left (a\phi_{1}+b\phi_{2}\right )d\nu & =\sum_{k=1}^{N}(aa_{k}+bb_{k})\nu (E\cap E_{k})\\
& =a\sum_{k=1}^{N}a_{k}\nu (E\cap E_{k})+b\sum_{k=1}^{N}b_{k}\nu (E\cap E_{k})=a\int_{E}\phi_{1}d\nu +b\int_{E}\phi_{2}d\nu .
\end{align*}

This proves part (i). To prove part (ii), we have \(\phi_{1}-\phi_{2}\geq 0\) a.e. Therefore, using part (i) of Proposition \ref{rap76}, we obtain

\[\int_{E}\phi_{1}d\nu -\int_{E}\phi_{2}d\nu =\int_{E}(\phi_{1}-\phi_{2})d\nu\geq 0.\]

This completes the proof. \(\blacksquare\)

Let \(f\) be a bounded real-valued function, and let \(E\) be a Lebesgue measurable set with finite measure. By referring to the form of upper and lower Riemann integral, we consider the numbers

\[\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu\mbox{ and }\sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu .\]

for all simple functions \(\psi\) and \(\phi\). Then, we have the following interesting result.

\begin{equation}{\label{rat84}}\tag{3}\mbox{}\end{equation}

Theorem \ref{rat84}. Let \(f\) be defined and bounded on a Lebesgue measurable set \(E\) with \(\nu (E)<\infty\). The function \(f\) is Lebesgue-measurable if and only if

\begin{equation}{\label{raeq79}}\tag{4}
\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu=\sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu ,
\end{equation}

where \(\psi\) and \(\phi\) are simple functions.

Proof. Suppose that \(f\) is bounded by \(M\) and is Lebesgue-measurable on \(E\). Then \(-M\leq f(x)\leq M\) and the sets

\[E_{k}=\left\{{\bf x}:\frac{kM}{n}\geq f({\bf x})>\frac{(k-1)M}{n}\right\}\mbox{ for }-n\leq k\leq n\]

are disjoint Lebesgue-measurable sets with \(E=\bigcup_{k=-n}^{n}E_{k}\). Therefore, we have \(\nu (E)=\sum_{k=-n}^{n}\nu (E_{k})\). Now, we define the simple functions

\[\psi_{n}=\frac{M}{n}\sum_{k=-n}^{n}k\chi_{E_{k}}\mbox{ and }\phi_{n}=\frac{M}{n}\sum_{k=-n}^{n}(k-1)\chi_{E_{k}}.\]

Then, we have \(\phi_{n}\leq f\leq\psi_{n}\). Therefore, we obtain

\begin{align*} \inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu & \leq\int_{E}\psi_{n}d\nu\\ & =\frac{M}{n}\sum_{k=-n}^{n}k\nu (E\cap E_{k})=\frac{M}{n}\sum_{k=-n}^{n}k\nu (E_{k})\end{align*}

and

\begin{align*} \sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu & \geq\int_{E}\phi_{n}d\nu\\ & =\frac{M}{n}\sum_{k=-n}^{n}(k-1)\nu (E\cap E_{k})=
\frac{M}{n}\sum_{k=-n}^{n}(k-1)\nu (E_{k}),\end{align*}

which implies

\[0\leq\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu-\sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu\leq
\frac{M}{n}\sum_{k=-n}^{n}\nu (E_{k})=\frac{M}{n}\nu (E).\]

For \(n\) sufficiently large, we obtain the desired equality (\ref{raeq79}). On the other hand, suppose that the equality (\ref{raeq79}) holds true. Then, given any \(n\), there exist simples functions \(\phi_{n}\) and \(\psi_{n}\) satisfying \(\phi_{n}\leq f\leq\psi_{n}\) and

\begin{equation}{\label{raeq82}}\tag{5}
\int_{E}\psi_{n}d\nu -\int_{E}\phi_{n}d\nu<\frac{1}{n}.
\end{equation}

Using Proposition~\ref{rap80}, we see that

\[\psi^{*}=\inf_{n\geq 1}\psi_{n}\mbox{ and }\phi^{*}=\sup_{n\geq 1}\phi_{n}\]

are Lebesgue measurable. We also have \(\phi^{*}\leq f\leq\psi^{*}\). If we can show \(\phi^{*}=\psi^{*}\) a.e., then \(f=\psi^{*}\) a.e. and \(f\) is Lebesgue-measurable by Proposition \ref{rap81}. Now, we define the set

\[F_{r}=\left\{{\bf x}:\phi^{*}({\bf x})<\psi^{*}({\bf x})-\frac{1}{r}\right\}.\]

Then, we have

\[F=\left\{{\bf x}:\phi^{*}({\bf x})<\psi^{*}({\bf x})\right\}=\bigcup_{r=1}^{\infty}F_{r}.\]

Since

\[\phi_{n}\leq\phi^{*}<\psi^{*}-\frac{1}{r}\leq\psi_{n}-\frac{1}{r},\]

we have

\begin{align*} F_{r} & \subseteq\left\{{\bf x}:\phi_{n}({\bf x})<\psi_{n}({\bf x})-\frac{1}{r}\right\}\\ & =\left\{{\bf x}:\psi_{n}({\bf x})-\phi_{n}({\bf x})
>\frac{1}{r}\right\}\equiv A.\end{align*}

From (\ref{raeq82}), we have

\begin{align*} \frac{1}{n} & >\int_{E}(\psi_{n}-\phi_{n})d\nu\\ & =\int_{A}(\psi_{n}-\phi_{n})d\nu
+\int_{E\setminus A}(\psi_{n}-\phi_{n})d\nu ,\end{align*}

which implies

\[\frac{\nu (A)}{r}=\int_{A}\frac{1}{r}d\nu\leq\int_{A}(\psi_{n}-\phi_{n})d\nu <\frac{1}{n}.\]

Therefore, we obtain

\[\nu (F_{r})\leq\nu (A)<r/n,\]

which implies \(\nu (F_{r})=0\) by taking \(n\rightarrow\infty\). This shows that \(\nu (F)=0\). It also means \(\phi^{*}=\psi^{*}\) a.e., since \(\phi^{*}\leq\psi^{*}\). This completes the proof. \(\blacksquare\)

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

The Lebesgue Integral of Bounded Measurable Functions.

Recall the upper Riemann integral can be written as

\begin{equation}{\label{raeq240}}\tag{6}
\overline{\int}_{a}^{b}f(x)dx=\inf_{\{\psi:\psi\geq f\}}\int_{a}^{b}\psi (x)dx
\end{equation}

where \(\psi\) is a step function, and the lower Riemann integral can be written as

\begin{equation}{\label{raeq241}}\tag{7}
\underline{\int}_{a}^{b}f(x)dx=\sup_{\{\phi:\phi\leq f\}}\int_{a}^{b}\phi (x)dx.
\end{equation}

where \(\phi\) is a step function. In order to distinguish the Riemann integral and Lebesgue integral on the closed interval, we write

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

to mean the Riemann integral, and write

\[\int_{[a,b]}fd\nu\]

to mean the Lebesgue integral.

\begin{equation}{\label{ma66}}\tag{8}\mbox{}\end{equation}

Definition \ref{ma66}. Let \(f\) be a bounded and Lebesgue mesaurable function defined on a Lebesgue measurable set \(E\) with \(\nu (E)<\infty\). The Lebesgue integral of \(f\) on \(E\) is defined by

\[\int_{E}fd\nu =\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu\]

for all simple functions \(\psi\). \(\sharp\)

According to Theorem \ref{rat84}, we also have

\[\int_{E}fd\nu =\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu=\sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu.\]

\begin{equation}{\label{rap7}}\tag{9}\mbox{}\end{equation}

Proposition \ref{rap7}. Let \(f\) be a bounded function defined on a compact interval \([a,b]\). Suppose that \(f\) is
Riemann-integrable on \([a,b]\). Then, it is Lebesgue-measurable satisfying

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

Proof. For all simple functions \(\psi\) and \(\phi\) satisfying \(\psi\geq f\geq\phi\), we have

\[\int_{[a,b]}\psi d\nu\geq\int_{[a,b]}\phi d\nu .\]

It follows

\[\inf_{\{\psi :\psi\geq f\}}\int_{[a,b]}\psi d\nu\geq\int_{[a,b]}\phi d\nu\]

for all simples function \(\phi\) satisfying \(f\geq\phi\), which implies

\begin{equation}{\label{ma64}}\tag{10}
\inf_{\{\psi :\psi\geq f\}}\int_{[a,b]}\psi d\nu\geq\sup_{\{\phi :\phi\leq f\}}\int_{[a,b]}\phi d\nu .
\end{equation}

Since each step function is also a simple function, using (\ref{raeq240}), (\ref{raeq241}) and (\ref{ma64}), we have

\begin{align}
\underline{\int_{a}^{b}}f(x)dx & =\sup_{\{\phi^{*}:\phi^{*}\leq f\}}\int_{a}^{b}\phi^{*}(x)dx.
\mbox{ (for all step functions \(\phi^{*}\))}\nonumber\\
& =\sup_{\{\phi^{*}:\phi^{*}\leq f\}}\int_{[a,b]}\phi^{*}d\nu\leq\sup_{\{\phi :\phi\leq f\}}\int_{[a,b]}\phi d\nu
\mbox{ (for all simple functions \(\phi\))}\nonumber\\
& \leq\inf_{\{\psi :\psi\geq f\}}\int_{[a,b]}\psi d\nu\mbox{ (for all simple functions \(\psi\))}\nonumber\\
& \leq\inf_{\{\psi^{*}:\psi^{*}\geq f\}}\int_{[a,b]}\psi^{*}d\nu\mbox{ (for all step functions \(\psi^{*}\))}\nonumber\\
& =\inf_{\{\psi^{*}:\psi^{*}\geq f\}}\int_{a}^{b}\psi^{*}(x)dx=\overline{\int}_{a}^{b}f(x)dx.\label{raeq86}\tag{11}
\end{align}

Since \(f\) is Riemann-integrable, the inequalities in (\ref{raeq86}) become all equalities; that is, we have

\begin{align*} \underline{\int_{a}^{b}}f(x)dx & =\inf_{\{\psi :\psi\geq f\}}\int_{[a,b]}\psi d\nu
\\ & =\sup_{\{\phi :\phi\leq f\}}\int_{[a,b]}\phi d\nu\\ & =\overline{\int}_{a}^{b}f(x)dx.\end{align*}

Using Theorem \ref{rat84}, we also conclude that \(f\) is Lebesgue-measurable. This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap143}}\tag{12}\mbox{}\end{equation}

Proposition \ref{rap143}. Suppose that \(f\) is bounded and Lebesgue-measurable defined on a Lebesgue-measurable set \(E\) with \(\nu (E)<\infty\). Then, for any constant \(c\), we have

\[\int_{E}(cf)d\nu =c\int_{E}fd\nu .\]

Proof. For \(c=0\), the result is obvious. For \(c>0\), we have

\begin{align*}
\int_{E}(cf)d\nu & =\inf_{\{\psi^{\prime}:\psi^{\prime}\geq cf\}}\int_{E}\psi^{\prime}d\nu =\inf_{\{c\psi :c\psi\geq cf\}}\int_{E}(c\psi )d\nu\\
& =\inf_{\{\psi :\psi\geq f\}}\int_{E}(c\psi )d\nu=c\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu =c\int_{E}fd\nu .
\end{align*}

For \(c<0\), we also have

\begin{align*}
\int_{E}(cf)d\nu & =\inf_{\{c\psi :c\psi\geq cf\}}\int_{E}(c\psi )d\nu=\inf_{\{\psi :\psi\leq f\}}\int_{E}(c\psi )d\nu =\inf_{\{\phi :\phi\leq f\}}\int_{E}(c\phi )d\nu\\
& =c\sup_{\{\phi :\phi\leq f\}}\int_{E}\phi d\nu=c\inf_{\{\psi :\psi\geq f\}}\int_{E}\psi d\nu\mbox{ (by Theorem~\ref{rat84})}\\
& =c\int_{E}fd\nu .
\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap144}}\tag{13}\mbox{}\end{equation}

Proposition \ref{rap144}. Suppose that \(f\) and \(g\) are bounded and Lebesgue-measurable defined on a Lebesgue-measurable set \(E\) with \(\nu (E)<\infty\). Then, we have

\[\int_{E}\left (f+g\right )d\nu =\int_{E}fd\nu +\int_{E}gd\nu .\]

Proof. Given any simple functions \(\psi_{1}\) and \(\psi_{2}\) satisfying \(f\leq\psi_{1}\) and \(g\leq\psi_{2}\), we see that \(\psi_{1}+\psi_{2}\) is a simple function and \(f+g\leq\psi_{1}+\psi_{2}\). Therefore, using Proposition \ref{rap78}, we obtain

\begin{align*} \int_{E}(f+g)d\nu & =\inf_{\{\psi :\psi\geq f+g\}}\int_{E}\psi d\nu\\ & \leq\int_{E}
\left (\psi_{1}+\psi_{2}\right )d\nu\\ & =\int_{E}\psi_{1}d\nu +\int_{E}\psi_{2}d\nu ,\end{align*}

which also implies, by taking infimum,

\begin{align*}
\int_{E}(f+g)d\nu & \leq\inf_{\{\psi_{1}:\psi_{1}\geq f\}}\left (\int_{E}\psi_{1}d\nu+\int_{E}\psi_{2}d\nu\right )\\
& =\int_{E}\psi_{2}d\nu +\inf_{\{\psi_{1}:\psi_{1}\geq f\}}\int_{E}\psi_{1}d\nu=\int_{E}\psi_{2}d\nu +\int_{E}fd\nu .
\end{align*}

By taking infimum again, we obtain

\begin{align*} \int_{E}(f+g)d\nu & \leq\int_{E}fd\nu +\inf_{\{\psi_{2}:\psi_{2}\geq g\}}
\int_{E}\psi_{1}d\nu\\ & =\int_{E}fd\nu +\int_{E}gd\nu .\end{align*}

On the other hand, given any simple functions \(\phi_{1}\) and \(\phi_{2}\) satisfying \(f\geq\phi_{1}\) and \(g\geq\phi_{2}\), we have that \(\phi_{1}+\phi_{2}\) is a simple function and \(f+g\geq\phi_{1}+\phi_{2}\). Therefore, using Theorem \ref{rat84} and Proposition \ref{rap78}, we obtain

\begin{align*}
\int_{E}(f+g)d\nu & =\inf_{\{\psi :\psi\geq f+g\}}\int_{E}\psi d\nu =\sup_{\{\phi :\phi\leq f+g\}}\int_{E}\phi d\nu\\
& \geq\int_{E}\left (\phi_{1}+\phi_{2}\right )d\nu =\int_{E}\phi_{1}d\nu +\int_{E}\phi_{2}d\nu .
\end{align*}

By taking the supremum, we obtain

\[\int_{E}(f+g)d\nu\geq\int_{E}fd\nu +\int_{E}gd\nu .\]

This completes the proof. \(\blaxksquare\)

\begin{equation}{\label{rap145}}\tag{14}\mbox{}\end{equation}

Proposition \ref{rap145}. Suppose that \(f\) and \(g\) are bounded and Lebesgue measurable defined on a Lebesgue measurable set \(E\) with \(\nu (E)<\infty\). If \(f\leq g\) a.e., then

\begin{equation}{\label{ma65}}\tag{15}
\int_{E}fd\nu\leq\int_{E}gd\nu .
\end{equation}

Therefore, we also have

\begin{equation}{\label{raeq146}}\tag{16}
\left |\int_{E}fd\nu\right |\leq\int_{E}|f|d\nu .
\end{equation}

In particular, If \(f=g\) a.e., then

\begin{equation}{\label{raeq147}}\tag{17}
\int_{E}fd\nu =\int_{E}gd\nu .
\end{equation}

Moreover, if \(a\leq f(x)\leq b\) on \(E\), then

\[a\nu (E)\leq\int_{E}fd\nu\leq b\nu (E).\]

Proof. Since \(g-f\geq 0\) a.e., given \(\psi\geq g-f\), we have \(\psi\geq 0\) a.e. From Propositions \ref{rap143}, \ref{rap144} and \ref{rap78}, we obtain

\[\int_{E}gd\nu -\int_{E}fd\nu =\int_{E}(g-f)d\nu=\inf_{\psi\geq g-f}\int_{E}\psi d\nu\geq 0,\]

which proves (\ref{ma65}). Since \(-|f|\leq f\leq |f|\), we have

\[-\int_{E}|f|d\nu\leq\int_{E}fd\nu\leq\int_{E}|f|d\nu ,\]

which proves (\ref{raeq146}). We also have \(f=g\) a.e. if and only if \(f\leq g\) a.e. and \(g\leq f\) a.e., which proves (\ref{raeq147}) and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{rap110}}\tag{18}\mbox{}\end{equation}

Proposition \ref{rap110}. Suppose that \(f\) is bounded and Lebesgue measurable defined on a Lebesgue measurable set \(E\) with \(\nu (E)<\infty\). Then, we have the following properties.

(i) Suppose that \(\nu (E)=0\). Then

\[\int_{E}fd\nu =0.\]

(ii) Suppose that \(F\) is a Lebesgue measurable set with \(F\subseteq E\), and that \(f\) is nonnegative. Then

\[\int_{F}fd\nu\leq\int_{E}fd\nu .\]

(iii) Suppose that \(A\) and \(B\) are disjoint Lebesgue measurable sets with finite measures. Then

\[\int_{A\cup B}fd\nu =\int_{A}fd\nu +\int_{B}fd\nu .\]

Proof. Part (i) follows from part (ii) of Proposition \ref{rap76} by considering the definition. Part (ii) follows from part (ii) of Proposition \ref{rap76}. To prove part (iii), since

\[\chi_{A\cup B}=\chi_{A}+\chi_{B},\]

using Proposition \ref{rap144}, we have

\begin{align*} \int_{A\cup B}fd\nu & =\int_{E}f\cdot\chi_{A\cup B}d\nu \\ & =
\int_{E}f\cdot\chi_{A}d\nu +\int_{E}f\cdot\chi_{B}d\nu \\ & =\int_{A}fd\nu +\int_{B}fd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rat88}}\tag{19}\mbox{}\end{equation}

Theorem \ref{rat88}. (Egorov’s Theorem) Suppose that \(\{f_{k}\}_{k=1}^{\infty}\) is a sequence of Lebesgue-measurable functions which converges a.e. on \(E\) with \(\nu (E)<\infty\) to a finite limit function \(f\). Then, given any \(\epsilon >0\), there exists a closed subset \(F\) of \(E\) such that \(\nu (E\setminus F)<\epsilon\) and \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(F\). \(\sharp\)

\begin{equation}{\label{rat96}}\tag{20}\mbox{}\end{equation}

Theorem \ref{rat96}. (Bounded Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue measurable functions defined on a Lebesgue measurable set \(E\) with \(\nu (E)<+\infty\). Suppose that there exists a real number \(M\) satisfying \(|f_{k}({\bf x})|\leq M\) for all \(k\) and all \({\bf x}\in E\). If

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E,\]

then

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. It is clear to see \(|f({\bf x})|\leq M\) for all \({\bf x}\in E\). Then, we have

\begin{align*} \left |f_{n}({\bf x})-f({\bf x})\right | & \leq\left |f_{n}({\bf x})\right |
+\left |f({\bf x})\right |\\ & \leq 2M\mbox{ for all }{\bf x}\in E.\end{align*}

Using the Egorov’s Theorem \ref{rat88}, given any \(\epsilon >0\), there is a closed subset \(F\) of \(E\) satisfying \(\nu (E\setminus F)<\epsilon /4M\) and \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(F\). The uniform convergence says that there exists an integer \(N\) satisfying \(|f_{n}({\bf x})-f({\bf x})|<\epsilon /2\nu (E)\) for all \(x\in F\). Therefore, using part (ii) of Proposition \ref{rap110}, we obtain

\begin{align*}
\left |\int_{E}f_{n}d\nu -\int_{E}fd\nu\right | & =\left |\int_{E}(f_{n}-f)d\nu\right |\leq\int_{E}\left |f_{n}-f\right |d\nu\\
& =\int_{F}\left |f_{n}-f\right |d\nu +\int_{E\setminus F}\left |f_{n}-f\right |d\nu \\
& \leq\int_{F}\frac{\epsilon}{2\nu (E)}d\nu +\int_{E\setminus F}2Md\nu\\
& =\nu (F)\cdot\frac{\epsilon}{2\nu (E)}+2M\cdot\nu (E\setminus F)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}

This completes the proof. \(\blacksquare\)

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

The Lebesgue Integral of Nonnegative Functions.

Two approaches will be presented. The first one is based on the concept of supremum by mimicing the lower Riemann integral. The second one is based on the volume of region.

Based on supremum.

Suppose that the nonnegative extended real-valued function \(f:\bar{\mathbb{R}}^{n}\rightarrow [0,+\infty ]\) is Lebesgue measurable. We shall present the Lebesgue integral based on the concept of supremum by mimicing the lower Riemann integral.

\begin{equation}{\label{ma67}}\tag{21}\mbox{}\end{equation}

Definition \ref{ma67}. Given any Lebesgue measurable set \(E\), the {\bf Lebesgue integral} of nonnegative extended real-valued function \(f\) on \(E\) is defined by

\[\int_{E}fd\nu =\sup_{h\leq f}\int_{E}hd\nu ,\]

where the supremum is taken over all bounded measurable functions \(h\) defined on \(E\) satisfying

\[\nu (\{x:h(x)\neq 0\})<\infty ,\]

where the integral \(\int_{E}hd\nu\) is given in Definition \ref{ma66}. \(\sharp\)

\begin{equation}{\label{rap101}}\tag{22}\mbox{}\end{equation}

Proposition \ref{rap101}. Suppose that the nonnegative extended real-valued functions \(f\) and \(g\) defined on a Lebesgue measurable set \(E\) are Lebesgue-measurable. Then, we have the following properties.

(i)}] If \(\nu (E)=0\), then

\[\int_{E}fd\nu =0.\]

(ii) For any \(c>0\), we have

\[\int_{E}cfd\nu =c\int_{E}fd\nu .\]

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

\[\int_{E}fd\nu\leq\int_{E}gd\nu .\]

If \(f=g\) a.e. on \(E\), then

\[\int_{E}fd\nu =\int_{E}gd\nu .\]

(iv) If \(F\) is a Lebesgue measurable set with \(F\subseteq E\), then

\[\int_{F}fd\nu\leq\int_{E}fd\nu .\]

Proof. The results follow from Propositions \ref{rap143}, \ref{rap145} and \ref{rap110} by considering the definition of Lebesgue integral given in Definition \ref{ma67}. \(\blacksquare\)

\begin{equation}{\label{rap108}}\tag{23}\mbox{}\end{equation}

Proposition \ref{rap108}. Suppose that \(f\) is nonnegative and Lebesgue-measurable on \(E\) with

\[\int_{E}fd\nu <+\infty,\]

Then \(f<+\infty\) a.e. on \(E\).

Proof. We may assume that \(\nu (E)>0\). Suppose that \(f=+\infty\) on a subset \(F\) of \(E\) with \(\nu (F)>0\). Then, for any sufficiently large \(a\), we have

\[\int_{E}fd\nu\geq\int_{F}fd\nu\geq\int_{F}a=a\cdot\nu (F)\rightarrow +\infty,\]

which contradicts \(\int_{E}fd\nu <+\infty\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap147}}\tag{24}\mbox{}\end{equation}

Proposition \ref{rap147}. Suppose that the nonnegative extended real-valued functions \(f\) and \(g\) defined on a Lebesgue measurable set \(E\) are Lebesgue-measurable. Then, we have

\[\int_{E}(f+g)d\nu =\int_{E}fd\nu +\int_{E}gd\nu .\]

Proof. Suppose that \(h\) and \(k\) are bounded and Lebesgue-measurable functions satisfying

\[h\leq f\mbox{ and }\nu (\{x:h(x)\neq 0\})<\infty\]

and

\[k\leq g\mbox{ and }\nu (\{x:k(x)\neq 0\})<\infty.\]

Then, it is clear that \(h+k\) is also bounded and Lebesgue-measurable satisfying

\[h+k\leq f+g\mbox{ and }\nu (\{x:h(x)+k(x)\neq 0\})<\infty.\]

Using Proposition \ref{rap144}, we have

\begin{align*} \int_{E}(f+g)d\nu & =\sup_{l\leq f+g}\int_{E}ld\nu\geq\int_{E}(h+k)d\nu \\ & =\int_{E}hd\nu +\int_{E}kd\nu .\{*}

By taking supremum, we have

\{*} \int_{E}(f+g)d\nu & \geq\sup_{h\leq f}\left (\int_{E}hd\nu +\int_{E}kd\nu\right )
\\ & =\int_{E}kd\nu +\sup_{h\leq f}\int_{E}hd\nu \\ & =\int_{E}kd\nu +\int_{E}fd\nu .\end{align*}

Taking supremum again, we obtain

\begin{align*} \int_{E}(f+g)d\nu & \geq\int_{E}fd\nu +\sup_{k\leq g}\int_{E}kd\nu\\ & =\int_{E}fd\nu +\int_{E}gd\nu .\end{align*}

On the other hand, let \(l\) be bounded and Lebesgue measurable satisfying

\[l\leq f+g\mbox{ and }\nu (\{x:l(x)\neq 0\})<\infty.\]

We define two functions \(h\) and \(k\) by \(h=\min\{f,l\}\) and \(k=l-h\). Then \(h\) and \(k\) are bounded and Lebesgue-measurable satisfying

\[h\leq f\mbox{ and }\nu (\{x:h(x)\neq 0\})<\infty\]

and

\[k\leq g\mbox{ and }\nu (\{x:k(x)\neq 0\})<\infty.\]

By Proposition \ref{rap144} again, we have

\begin{align*} \int_{E}ld\nu & =\int_{E}(h+k)d\nu =\int_{E}hd\nu +\int_{E}kd\nu
\\ & \leq\sup_{h\leq f}\int_{E}hd\nu +\sup_{k\leq g}\int_{E}kd\nu\\ & =\int_{E}fd\nu +\int_{E}gd\nu .\end{align*}

By taking supremum, we obtain

\begin{align*} \int_{E}(f+g)d\nu & =\sup_{l\leq f+g}\int_{E}ld\nu\\ & \leq\int_{E}fd\nu +\int_{E}gd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap122}}\tag{25}\mbox{}\end{equation}

Proposition \ref{rap122}. (Tchebyshev’s inequality). Suppose that the nonnegative extended real-valued function \(f\) defined on a Lebesgue-measurable set \(E\) is Lebesgue-measurable. If \(\alpha >0\), then

\[\nu\left (\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}\right )\leq\frac{1}{\alpha}\int_{E}fd\nu .\]

Proof. By Proposition \ref{rap101}, we have

\begin{align*} \alpha\cdot\nu\left (\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}\right )
& =\int_{\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}}\alpha d\nu
\\ & \leq\int_{\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}}fd\nu\leq\int_{E}fd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap123}}\tag{26}\mbox{}\end{equation}

Proposition \ref{rap123}. Suppose that the nonnegative extended real-valued function \(f\) defined on a Lebesgue-measurable set \(E\) is Lebesgue-measurable. Then

\[\int_{E}fd\nu =0\mbox{ if and only if \(f=0\) a.e. on \(E\)}.\]

 

Proof. Suppose that \(f=0\) a.e. on \(E\). Using Propositions \ref{rap101}, we have

\[\int_{E}fd\nu =\int_{E}0d\nu =0.\]

On the other hand, suppose that \(\int_{E}fd\nu=0\). For any \(\alpha >0\), using the Tchebyshev’s inequality in Proposition \ref{rap122}, we have

\[\nu\left (\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}\right )=0.\]

Since

\[\left\{{\bf x}\in E:f({\bf x})>0\right\}=\bigcup_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})>\frac{1}{k}\right\},\]

we obtain

\[\nu\left (\left\{{\bf x}\in E:f({\bf x})>0\right\}\right )=0.\]

This says that \(f=0\) a.e. on \(E\), and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{rat150}}\tag{27}\mbox{}\end{equation}

Theorem \ref{rat150}. (Fatou’s Lemma). Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) of nonnegative extended real-valued functions defined on the same Lebesgue-measurable set \(E\) are all Lebesgue-measurable satisfying

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E.\]

Then \(f\) is Lebesgue-measurable on \(E\) satisfying

\[\int_{E}fd\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Proof. Let

\[N=\left\{{\bf x}\in E:\lim_{k\rightarrow\infty}f_{k}({\bf x})\neq f({\bf x})\right\}.\]

Then \(\nu (N)=0\). Therefore, we have

\[\int_{E\setminus N}fd\nu =\int_{E\setminus N}fd\nu+\int_{N}fd\nu=\int_{E}fd\nu\]

and

\begin{align*} \int_{E\setminus N}f_{k}d\nu & =\int_{E\setminus N}f_{k}d\nu+\int_{N}f_{k}d\nu\\ & =\int_{E}f_{k}d\nu .\mbox{ for all }k\end{align*}

Based on this observation, we may just assume that

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ for each }x\in E.\]

Let \(h\) be any bounded and Lebesgue-measurable function defined on \(E\) satisfying

\[h\leq f\mbox{ and }\nu (\{x:h({\bf x})\neq 0\})<\infty.\]

Let \(A=\{x:h({\bf x})\neq 0\}\). Then \(\nu (A)<+\infty\). It follows

\begin{align}
\int_{E}hd\nu & =\int_{A}hd\nu+\int_{E\setminus A}hd\nu\nonumber\\ =\int_{A}hd\nu+\int_{E\setminus A}0d\nu=\int_{A}hd\nu.\label{raeq96}\tag{28}
\end{align}

We define

\[h_{k}({\bf x})=\min\left\{h({\bf x}),f_{k}({\bf x})\right\}.\]

Then \(h_{k}\) is bounded by the same bound of \(h\), and is Lebesgue-measurable satisfying

\begin{align*} \lim_{k\rightarrow\infty}h_{k}({\bf x}) & =\min\left\{h({\bf x}),\lim_{k\rightarrow\infty}
f_{k}({\bf x})\right\}\\ & =\min\left\{h({\bf x}),f({\bf x})\right\}\\ & =h({\bf x})\mbox{ for each }{\bf x}\in A.\end{align*}

Using the bounded convergence Theorem\ref{rat96}, we have

\begin{equation}{\label{raeq97}}\tag{29}
\int_{A}hd\nu =\lim_{k\rightarrow\infty}\int_{A}h_{k}d\nu .
\end{equation}

Since \(f_{k}\) is nonnegative, we also have \(h_{k}({\bf x})=0\) for \(x\not\in A\), i.e.,

\begin{equation}{\label{ra1}}\tag{30}
\int_{A}h_{k}d\nu=\int_{E}h_{k}d\nu.
\end{equation}

Since \(h_{k}\leq f_{k}\) for all \(k\), from (\ref{raeq96}), (\ref{ra1}) and (\ref{raeq97}), we obtain

\begin{align*}
\int_{E}hd\nu & =\int_{A}hd\nu =\lim_{k\rightarrow\infty}\int_{A}h_{k}d\nu=\liminf_{k\rightarrow\infty}\int_{A}h_{k}d\nu\\
& =\liminf_{k\rightarrow\infty}\int_{E}h_{k}d\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .
\end{align*}

By taking supremum over all \(h\) satisfying \(h\leq f\), we obtain

\[\int_{E}fd\nu=\sup_{h\leq f}\int_{E}hd\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu.\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rat98}}\tag{31}\mbox{}\end{equation}

Theorem \ref{rat98}. (Monotone Convergence Theorem). Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) of nonnegative extended real-valued functions defined on the same Lebesgue-measurable set \(E\) are all Lebesgue-measurable satisfying \(f_{k}\leq f\) for all \(k\) and

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E.\]

Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

In particular, if the sequence \(\{f_{k}\}_{k=1}^{\infty}\) is increasing, then the equality still holds true.

Proof. We first note that the limit function \(f\) is Lebesgue-measurable on \(E\). Since \(f_{k}\leq f\) for all \(k\), it follows

\[\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu\leq\int_{E}fd\nu .\]

By Fatou’s Lemma \ref{rat150}, we have

\begin{align*} \int_{E}fd\nu & \leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu\\ & \leq
\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu\leq\int_{E}fd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{raeq100}}\tag{32}\mbox{}\end{equation}

Proposition \ref{raeq100}. Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) of nonnegative extended real-valued functions defined on the same Lebesgue-measurable set \(E\) are all Lebesgue-measurable satisfying

\[\sum_{k=1}^{\infty}f_{k}({\bf x})=g({\bf x})\mbox{ a.e. on }E.\]

Then

\[\int_{E}gd\nu=\int_{E}\sum_{k=1}^{\infty}f_{k}d\nu =\sum_{k=1}^{\infty}\int_{E}f_{k}d\nu .\]

Proof. Let

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

Then, the sequence \(\{g_{m}\}_{m=1}^{\infty}\) is increasing satisfying

\[\lim_{m\rightarrow\infty}g_{m}({\bf x})=g({\bf x})\mbox{ a.e. on }E.\]

Using the monotone convergence Theorem \ref{rat98}, we have

\begin{align*} \int_{E}\sum_{k=1}^{\infty}f_{k}d\nu & =\int_{E}gd\nu=\lim_{m\rightarrow\infty}\int_{E}g_{m}d\nu
\\ & =\lim_{m\rightarrow\infty}\int_{E}\sum_{k=1}^{m}f_{k}
\\ & =\lim_{m\rightarrow\infty}\sum_{k=1}^{m}\int_{E}f_{k}=\sum_{k=1}^{\infty}\int_{E}f_{k}.\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap132}}\tag{33}\mbox{}\end{equation}

Proposition \ref{rap132}. Suppose that the extended real-valued function \(f\) defined on \(E\) is nonnegative and Lebesgue-measurable. Let \(\{E_{k}\}_{k=1}^{\infty}\) be a disjoint sequence of Lebesgue-measurable sets satisfying \(E=\bigcup_{k=1}^{\infty}E_{k}\). Then, we have

\[\int_{E}fd\nu =\sum_{k=1}^{\infty}\int_{E_{k}}fd\nu .\]

Proof. We first have

\[f({\bf x})=\sum_{k=1}^{\infty}\chi_{E_{k}}({\bf x})\cdot f({\bf x})\mbox{ for }{\bf x}\in E,\]

Using Proposition~\ref{raeq100}, we obtain

\begin{align*}
\int_{E}fd\nu & =\int_{E}\sum_{k=1}^{\infty}\chi_{E_{k}}({\bf x})\cdot f({\bf x})d\nu\\
& =\sum_{k=1}^{\infty}\int_{E}\chi_{E_{k}}({\bf x})\cdot f({\bf x})d\nu\\
& =\sum_{k=1}^{\infty}\int_{E_{k}}f({\bf x})d\nu.
\end{align*}

This completes the proof. \(\blacksquare\)

We can prove a more general version of Fatou’s lemma.

\begin{equation}{\label{ra2}}\tag{34}\mbox{}\end{equation}

Theorem \ref{ra2}. (Fatou’s Lemma). Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) of nonnegative extended real-valued functions defined on the same Lebesgue-measurable set \(E\) are all Lebesgue-measurable. Then, we have

\[\int_{E}\left (\liminf_{k\rightarrow\infty}f_{k}\right )d\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Proof. We define

\[g_{k}({\bf x})=\inf_{i\geq k}f_{i}({\bf x}).\]

Then, each \(g_{k}\) is measurable. Since \(g_{k}\leq f_{k}\) for all \(k\), it is clear that

\[\int_{E}g_{k}d\nu\leq\int_{E}f_{k}d\nu .\]

Since \(0\leq g_{1}\leq g_{2}\leq\cdots\), by the definition of limit inferior, we have

\begin{align*} \lim_{k\rightarrow\infty}g_{k} & =\sup_{k}g_{k}=\sup_{k}\inf_{i\geq k}f_{i}\\ =\liminf_{k\rightarrow\infty}f_{k}.\end{align*}

By the monotone convergence Theorem \ref{rat98}, we obtain

\begin{align*}
\int_{E}\left (\liminf_{k\rightarrow\infty} f_{k}\right )d\nu & =\int_{E}\left (\lim_{k\rightarrow\infty}g_{k}\right )d\nu
=\lim_{k\rightarrow\infty}\int_{E}g_{k}d\nu\\
& =\liminf_{k\rightarrow\infty}\int_{E}g_{k}d\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .
\end{align*}

This completes the proof. \(\blacksquare\)

Definition. A nonnegative Lebesgue measurable function \(f\) is called Lebesgue-integrable over the Lebesgue measurable set \(E\) when

\[\int_{E}fd\nu <+\infty .\]

\begin{equation}{\label{rap178}}\tag{35}\mbox{}\end{equation}

Proposition \ref{rap178}. Suppose that \(f\) is a nonnegative and Lebesgue-integrable extended real-valued function defined on the Lebesgue-measurable set \(E\). Given \(\epsilon >0\), there exists \(\delta >0\) such that, for each set \(A\subseteq E\) with \(\nu (A)<\delta\), we have

\[\int_{A}fd\nu <\epsilon .\]

Proof. Suppose that \(f\) is bounded by \(M\). If we take \(\delta =\epsilon /M\), then

\[\int_{A}fd\nu\leq M\cdot\nu (A)<M\cdot\delta =\epsilon .\]

Suppose that \(f\) is not bounded. Then, we define

\[f_{k}({\bf x})=\left\{\begin{array}{ll}
f({\bf x}) & \mbox{if \(f({\bf x})\leq k\)}\\
k & \mbox{otherwise}.
\end{array}\right .\]

It is clear that each \(f_{k}\) is bounded and \(f_{k}\) converges to \(f\) at each point. By the monotone convergence Theorem \ref{rat98}, we have

\[\lim_{k\rightarrow\infty}\int_{A}f_{k}d\nu =\int_{A}fd\nu .\]

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

\[\left |\int_{A}f_{N}d\nu-\int_{A}fd\nu\right |<\frac{\epsilon}{2},\]

which implies

\[\int_{A}f_{N}d\nu >\int_{A}fd\nu -\frac{\epsilon}{2},\mbox{ i.e., }\int_{A}(f-f_{N})d\nu <\frac{\epsilon}{2}.\]

We take \(\delta <\epsilon /2N\). If \(\nu (A)<\delta\), then

\begin{align*}
\int_{A}fd\nu & =\int_{A}(f-f_{N})d\nu +\int_{A}f_{N}d\nu\\
& \leq\int_{A}(f-f_{N})d\nu +N\cdot\nu (A) <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{align*}

This completes the proof. \(\blacksquare\)

We can also study the Lebesgue integral based on the concept of infimum by mimicing the upper Riemann integral, which is presented below

Definition. Given any Lebesgue measurable set \(E\), the Lebesgue integral of nonnegative extended real-valued function \(f\) on \(E\) is defined by

\[\int_{E}fd\nu =\inf_{h\geq f}\int_{E}hd\nu ,\]

where the infimum is taken over all bounded measurable functions \(h\) defined on \(E\) satisfying

\[\nu (\{x:h(x)\neq 0\})<\infty ,\]

where the integral \(\int_{E}hd\nu\) is given in Definition \ref{ma66}. \(\sharp\)

Based on the Volume of the Region.

This approach is based on the notion that the integral of a nonnegative function should represent the volume of the region under the graph of function. Let

\[G(f,E)=\left\{({\bf x},f({\bf x}))\in\bar{\mathbb{R}}^{n+1}:{\bf x}\in E\mbox{ and }f({\bf x})<+\infty\right\}\]

and

\[R(f,E)=\left\{({\bf x},y)\in\bar{\mathbb{R}}^{n+1}:{\bf x}\in E, 0\leq y\leq f({\bf x})\mbox{ if }f({\bf x})<+\infty\mbox{ and }0\leq y<+\infty\mbox{ if }f({\bf x})=+\infty\right\},\]

where \(G(f,E)\) is called the graph of \(f\) on \(E\), and \(R(f,E)\) is called the region under \(f\) on \(E\). We denote by \(\hat{\nu}\) and \(\nu\) the Lebesgue measures based on \(\bar{\mathbb{R}}^{n+1}\) and \(\bar{\mathbb{R}}^{n}\), respectively. If \(R(f,E)\) is a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n+1}\), its Lebesgue measure \(\nu (R(f,E))\) is called the Lebesgue integral of \(f\) on \(E\), and we write
\[\int_{E}f({\bf x})d\nu =\hat{\nu}(R(f,E)).\]

Theorem. Let \(f\) be a nonnegative extended real-valued function defined on a Lebesgue-measurable set \(E\). Then, the Lebesgue integral \(\int_{E}fd\nu\) exists if and only if \(f\) is Lebesgue-measurable. \(\sharp\)

\begin{equation}{\label{rap108}}\tag{36}\mbox{}\end{equation}

Proposition \ref{rap108}. We have the following properties.

(i) If \(f\) and \(g\) are Lebesgue measurable and \(0\leq g\leq f\) on \(E\), then

\[\int_{E}gd\nu\leq\int_{E}fd\nu .\]

(ii) Let \(F\) and \(E\) be Lebesgue-measurable with \(F\subseteq E\). If \(f\) is nonnegative and Lebesgue-measurable on \(E\), then

\[\int_{F}fd\nu\leq\int_{E}fd\nu .\]

(iii) If \(f\) is nonnegative and Lebesgue measurable on \(E\) with \(\int_{E}fd\nu <+\infty\), then \(f<+\infty\) a.e. on \(E\).

Proof. If \(0\leq g\leq f\), then \(R(g,E)\subseteq R(f,E)\), which proves part (i). Also, if \(F\subseteq E\), then \(R(f,F)\subseteq R(f,E)\), which proves part (ii). To prove part (iii), we may assume that \(\nu (E)>0\). If \(f=+\infty\) on a subset \(F\) of \(E\) with \(\nu (F)>0\), then, by parts (i) and (ii), we have

\[\int_{E}fd\nu\geq\int_{F}fd\nu\geq\int_{F}a=a\cdot\nu (F)\]

for any sufficiently large \(a\), since \(f=+\infty\) on \(F\), which contradicts \(\int_{E}fd\nu <+\infty\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{ral103}}\tag{37}\mbox{}\end{equation}

Proposition \ref{ral103}. Given a subset \(E\) of \(\bar{\mathbb{R}}^{n}\) and \(0\leq\alpha\leq +\infty\), we define

\[\left\{\begin{array}{ll}
E_{\alpha}=\left\{({\bf x},y)\in\bar{\mathbb{R}}^{n+1}:{\bf x}\in E\mbox{ and }0\leq y\leq a\right\} & \mbox{if \(\alpha\) is finite}\\
E_{\infty}=\left\{({\bf x},y)\in\bar{\mathbb{R}}^{n+1}:{\bf x}\in E\mbox{ and }0\leq y<+\infty\right\} & \mbox{if \(\alpha =+\infty\)}\\
\end{array}\right .\]

If \(E\) is a Lebesugue-measurable as a subset of \(\bar{\mathbb{R}}^{n}\), then \(E_{\alpha}\) is also a Lebesgue-measurable as a subset of \(\bar{\mathbb{R}}^{n+1}\). Moreover, we have \(\hat{\nu}(E_{\alpha})=\alpha\cdot\nu (E)\), where \(+\infty\cdot 0\) is interpreted as \(0\). \(\sharp\)

\begin{equation}{\label{rap105}}\tag{38}\mbox{}\end{equation}

Proposition \ref{rap105}. If \(f\) is a nonnegative Lebesgue measurable function defined on \(E\) with
$0\leq\nu (E)\leq +\infty$, then \(\hat{\nu}(G(f,E))=0\). \(\sharp\)

\begin{equation}{\label{rat109}}\tag{39}\mbox{}\end{equation}

Theorem \ref{rat109}. If \(f\) is a nonnegative measurable function taking values \(\alpha_{1},\cdots ,\alpha_{n}\) on disjoint Lebesgue-measurable sets \(E_{1},\cdots ,E_{n}\), respectively, with \(E=\bigcup_{i=1}^{n}E_{i}\), then

\[\int_{E}fd\nu =\sum_{i=1}^{n}\alpha_{i}\nu (E_{i}).\]

Proof. According to Proposition \ref{ral103}, for each \(E_{i}\), we can consider the set \(E_{i,\alpha_{i}}\). Since \(E_{i}\) are disjoint Lebesgue-measurable sets, Proposition \ref{ral103} says that \(E_{i,\alpha_{i}}\) are also disjoint Lebesgue-measurable sets and \(\hat{\nu}(E_{i,\alpha_{i}})=\alpha_{i}\cdot\nu (E_{i})\). By definition, it is clear to see \(R(f,E)=\bigcup_{i=1}^{n}E_{i,\alpha_{i}}\). Therefore, we have

\begin{align*} \int_{E}fd\nu & =\nu (R(f,E))=\nu\left (\bigcup_{i=1}^{n}E_{i,\alpha_{i}}\right )\\ & =
\sum_{i=1}^{n}\hat{\nu}(E_{i,\alpha_{i}})=\sum_{i=1}^{n}\alpha_{i}\nu (E_{i}).\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap114}}\tag{40}\mbox{}\end{equation}

Proposition \ref{rap114}. Suppose that \(f\) is nonnegative and Lebesgue-measurable function defined on \(E\). If \(E\) is the countable union of disjoint Lebesgue-measurable sets \(\{E_{j}\}_{j=1}^{\infty}\) with \(E=\bigcup_{j=1}^{\infty}E_{j}\), then

\[\int_{E}fd\nu =\sum_{j=1}^{\infty}\int_{E_{j}}fd\nu .\]

Proof. Since the sets \(E_{j}\) are disjoint and Lebesgue-measurable, we see that the sets \(R(f,E_{j})\) are also disjoint and Lebesgue-measurable. Since \(R(f,E)=\bigcup_{j=1}^{\infty}R(f,E_{j})\), Proposition \ref{rap67} says

\[\int_{E}fd\nu =\hat{\nu}(R(f,E))=\sum_{j=1}^{\infty}\hat{\nu}(R(f,E_{j}))=\sum_{j=1}^{\infty}\int_{E_{j}}fd\nu .\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rat104}}\tag{41}\mbox{}\end{equation}

Theorem \ref{rat104}. (Monotone Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative Lebesgue-measurable
functions defined on a Lebesgue-measurable set \(E\) satisgying \(f_{k}(x)\uparrow f(x)\) on \(E\) and Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}d\nu .\]

Proof. We first have that \(f\) is Lebesgue-measurable. We also have

\[R(f_{k},E)\cup G(f,E)\uparrow R(f,E).\]

Therefore, using Proposition \ref{rap107} and \ref{rap105}, we obtain

\begin{align*}
\int_{E}fd\nu & =\hat{\nu}(R(f,E))\\
& =\lim_{k\rightarrow\infty}\hat{\nu}\left (R(f_{k},E)\cup G(f,E)\right )\\
& =\lim_{k\rightarrow\infty}\hat{\nu}(R(f_{k},E))\\
& =\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu .
\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rat112}}\tag{42}\mbox{}\end{equation}

Theorem \ref{rat112}. Let \(f\) be a nonnegative extended real-valued function defined on a Lebesgue-measurable set \(E\). Then, we have

\begin{equation}{\label{raeq30}}\tag{43}
\int_{E}fd\nu =\sup\sum_{k}\left [\nu (E_{k})\cdot\inf_{{\bf x}\in E_{k}}f({\bf x})\right ],
\end{equation}

where the supremum is taken over all decompositions \(E=\bigcup_{k=1}^{n}E_{k}\) of \(E\) into the union of a finite number of disjoint Lebesgue-measurable sets \(E_{k}\).

Proof. Give any decomposition \(E=\bigcup_{k=1}^{n}E_{k}\) of \(E\), we consider the Lebesgue-measurable function \(g\) taking values

\[\alpha_{k}=\int_{{\bf x}\in E_{k}}f({\bf x})\]

on \(E_{k}\) for \(k=1,\cdots ,n\). Since \(0\leq g\leq f\), Proposition \ref{rap108} and Theorem \ref{rat109} say

\begin{align*} \sum_{k=1}^{n}\left [\nu (E_{k})\cdot\inf_{{\bf x}\in E_{k}}f({\bf x})\right ] & =
\sum_{k=1}^{n}\alpha_{k}\nu (E_{k})\\ & =\int_{E}gd\nu\leq\int_{E}fd\nu .\end{align*}

Therefore, we obtain

\[\sup\sum_{k}\left [\nu (E_{k})\cdot\inf_{{\bf x}\in E_{k}}f({\bf x})\right ]\leq\int_{E}fd\nu .\]

For the opposite inequality, for \(k=1,2,\cdots\), we define the sets \(E_{jk}\) for \(j=0,1,\cdots ,k2^{k}\) by

\[E_{0k}=\left\{{\bf x}:f({\bf x})\geq k\right\}\]

and

\[E_{jk}=\left\{{\bf x}:\frac{j-1}{2^{k}}\leq f({\bf x})<\frac{j}{2^{k}}\right\}\]

for \(j=1,2,\cdots ,k2^{k}\). Then, we have \(E=\bigcup_{j}E_{jk}\) for each \(k=1,2,\cdots\). We also define the Lebesgue-measurable functions as follows:

\[f_{k}=\sum_{j}\left [\inf_{{\bf x}\in E_{jk}}f({\bf x})\right ]\chi_{E_{jk}} .\]

Then, we have \(0\leq f_{k}\) and \(f_{k}\uparrow f\) on \(E\). By the monotone convergence theorem, we also have \(\int_{E}f_{k}d\nu\rightarrow\int_{E}fd\nu\). Since

\[\int_{E}f_{k}d\nu=\sum_{j}\left [\nu (E_{jk})\cdot\inf_{{\bf x}\in E_{jk}}f({\bf x})\right ],\]

we obtain

\begin{align*} \sup\sum_{j}\left [\nu (E_{j})\cdot\inf_{{\bf x}\in E_{j}}f({\bf x})\right ] & \geq
\sum_{j}\left [\nu (E_{jk})\cdot\inf_{{\bf x}\in E_{jk}}f({\bf x})\right ]\\ & =\int_{E}f_{k}d\nu\end{align*}

for each \(k=1,2,\cdots\). By taking \(k\rightarrow\infty\), we obtain

\[\sup\sum_{j=1}^{n}\left [\nu (E_{j})\cdot\inf_{{\bf x}\in E_{j}}f({\bf x})\right ]\geq\int_{E}fd\nu .\]

This completes the proof. \(\blacksquare\)

We observe that formula (\ref{raeq30}) resembles the definition of the Riemann integral when the sets \(E_{k}\) are taken to be sub-intervals.

\begin{equation}{\label{rac113}}\tag{44}\mbox{}\end{equation}

Corollary \ref{rac113}. Let \(f\) be a nonnegative extended real-valued function defined on a Lebesgue-measurable set \(E\). If \(\nu (E)=0\), then \(\int_{E}fd\nu =0\).

Proof. The result follows immediately from Theorem \ref{rat112}. \(\blacksquare\)

Now, we can strengthen result (i) of Proposition \ref{rap108} as follows.

\begin{equation}{\label{rap124}}\tag{45}\mbox{}\end{equation}

Proposition \ref{rap124}. If \(f\) and \(g\) are nonnegative and Lebesgue-measurable functions defined on \(E\) with \(g\leq f\) a.e. on \(E\), then

\[\int_{E}gd\nu\leq\int_{E}fd\nu .\]

Therefore, if \(g=f\) a.e. on \(E\), then

\[\int_{E}gd\nu =\int_{E}fd\nu .\]

Proof. Let \(N=\{{\bf x}:g({\bf x})>f({\bf x})\}\). Then, we have \(\nu (N)=0\) and \(E=F\cup N\) satisfying \(g({\bf x})\leq f({\bf x})\) for each \({\bf x}\in F\).
Using Proposition \ref{rap114} and Corollary \ref{rac113}, we obtain

\[\int_{E}fd\nu =\int_{F}fd\nu +\int_{N}fd\nu =\int_{F}fd\nu\geq
\int_{F}gd\nu =\int_{E}gd\nu .\]

This completes the proof. \(\blacksquare\)

Proposition. (Tchebyshev’s Inequality). Let \(f\) be a nonnegative and Lebesgue-measurable extended real-valued function defined on a Lebesgue-measurable set \(E\). If \(\alpha >0\), then

\[\nu\left (\left\{{\bf x}\in E:f({\bf x})>\alpha\right\}\right )\leq\frac{1}{\alpha}\int_{E}fd\nu .\]

Proof. By Proposition \ref{rap108}, the argument of the proof of Proposition \ref{rap122} is still valid. \(\blacksquare\)

\begin{equation}{\label{rap140}}\tag{46}\mbox{}\end{equation}

Proposition \ref{rap140}. Let \(f\) be a nonnegative and Lebesgue-measurable extended real-valued function defined on a Lebesgue-measurable set \(E\). Then \(\int_{E}fd\nu =0\) if and only if \(f=0\) a.e. on \(E\).

Proof. By Proposition \ref{rap124}, the argument of the proof of Proposition \ref{rap123} is still valid. \(\blacksquare\)

\begin{equation}{\label{rap142}}\tag{47}\mbox{}\end{equation}

Proposition \ref{rap142}. Let \(f\) be a nonnegative and Lebesgue-measurable extended real-valued function defined on a Lebesgue-measurable set \(E\). If \(c\geq 0\), then

\begin{equation}{\label{raeq125}}\tag{48}
\int_{E}cfd\nu =c\int_{E}fd\nu .
\end{equation}

Proof. If \(f\) is a simple function, then \(cf\) is also a simple function. The equality (\ref{raeq125}) follows from Theorem \ref{rat109} immediately. For any nonnegative and Lebesgue-measurable function \(f\), by Theorem \ref{rat127}, there exists a sequence \(\{\phi_{k}\}_{k=1}^{\infty}\) of simple Lebesgue-measurable functions satisfying \(0\leq\phi_{k}\uparrow f\), which also implies \(0\leq c\phi_{k}\uparrow cf\). By the monotone convergence Theorem \ref{rat104}, we have

\begin{align*} \int_{E}cfd\nu & =\lim_{k\rightarrow\infty}\int_{E}c\phi_{k}d\nu \\ & =
\lim_{k\rightarrow\infty}c\int_{E}\phi_{k}d\nu =c\int_{E}fd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

Proposition. Let \(f\) and \(g\) be nonnegative and Lebesgue-measurable functions defined on a Lebesgue-measurable set. Then, we have

\[\int_{E}(f+g)d\nu =\int_{E}fd\nu +\int_{E}gd\nu .\]

Proof. Suppose that \(f\) and \(g\) are simple functions described by

\[f=\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}}\mbox{ and }g=\sum_{j=1}^{m}\beta_{j}\chi_{B_{j}}.\]

Then \(f+g\) is also a simple function described by

\[f+g=\sum_{i,j}(\alpha_{i}+\beta_{j})\chi_{A_{i}\cal B_{j}}.\]

Therefore, we have

\begin{align*}
\int_{E}(f+g)d\nu & =\sum_{i,j}(\alpha_{i}+\beta_{j})\nu (A_{i}\cap B_{j})=\sum_{i}\alpha_{i}\left (\sum_{j}\nu (A_{i}\cap B_{j}\right )+
\sum_{b}\beta_{j}\left (\sum_{i}\nu (A_{i}\cap B_{j}\right )\\
& =\sum_{i}\alpha_{i}\nu (A_{i})+\sum_{j}\beta_{j}\nu (B_{j})=\int_{E}fd\nu +\int_{E}gd\nu .
\end{align*}

For the general nonnegative Lebesgue-measurable functions \(f\) and \(g\), using Theorem \ref{rat127}, there exist two sequences \(\{\phi_{k}\}_{k=1}^{\infty}\) and \(\{\psi_{k}\}_{k=1}^{\infty}\) of simple Lebesgue-measurable functions satisfying \(0\leq\phi_{k}\uparrow f\) and \(0\leq\psi_{k}\uparrow f\). Then, we see that each \(\phi_{k}+\psi_{k}\) is a simple function and \(0\leq\phi_{k}+\psi_{k}\uparrow f+g\). By the monotone convergence Theorem \ref{rat104}, we have

\begin{align*} \int_{E}(f+g)d\nu & =\lim_{k\rightarrow\infty}\int_{E}(\phi_{k}+\psi_{k})d\nu
\\ & =\lim_{k\rightarrow\infty}\left (\int_{E}\phi_{k}d\nu +\int_{E}\psi_{k}d\nu\right )
\\ & =\int_{E}fd\nu +\int_{E}gd\nu .\end{align*}

This completes the proof. \(\blacksquare\)

We also remark that the Fatou’s lemma can also be obtained by the above same arguments.

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

The General Lebesgue Integral.

Given an extended real-valued function \(f:E\rightarrow\bar{\mathbb{R}}\) defined on \(E\subseteq\bar{\mathbb{R}}^{n}\), recall that \(f=f^{+}-f^{-}\). It is obvious that if \(f\) is Lebesgue-measurable, then \(f^{+}\) and \(f^{-}\) are also Lebesgue-measurable.

Definition. Let \(f\) be an extended real-valued function and Lebesgue measurable defined on a Lebesgue-measurable subset \(E\) of \(\bar{\mathbb{R}}^{n}\). Suppose that at least one of the nonnegative functions \(f^{+}\) and \(f^{-}\) is Lebesgue-integrable on \(E\). Then, we define

\[\int_{E}fd\nu =\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu.\]

When the nonnegative functions \(f^{+}\) and \(f^{-}\) are both Lebesgue-integrable on \(E\), we say that \(f\) is Lebesgue-integrable on \(E\). \(\sharp\)

Note that the above definition can only assume that \(f\) is defined a.e. on \(E\). For the sake of simplicity, we still assume that \(f\) is defined everywhere on \(E\).

Proposition. We have the following properties.

(i) Suppose that \(f\) is Lebesgue-integrable on a Lebesgue measurable set \(E\). Then

\[\left |\int_{E}fd\nu\right |\leq\int_{E}|f|d\nu .\]

(ii) Let \(f\) be Lebesgue-measurable on a Lebesgue measurable set \(E\). Then \(f\) is Lebesgue-integrable on \(E\) if and only if \(|f|\) is Lebesgue-integrable on \(E\).

(iii) Suppose that \(f\) is Lebesgue-integrable on \(E\). Then \(f<+\infty\) a.e. on \(E\).

Proof. Since \(\int_{E}fd\nu <+\infty\), we have

\begin{align*} \left |\int_{E}fd\nu\right | & =\left |\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu\right |
\\ & \leq\int_{E}f^{+}d\nu +\int_{E}f^{-}d\nu \\ & =\int_{E}(f^{+}+f^{-})d\nu =\int_{E}|f|d\nu ,\end{align*}

which proves part (i).

To prove part (ii), suppose that \(|f|\) is Lebesgue-integrable on \(E\). Then, it is clear that \(f\) is also Lebesgue-integrable on \(E\). Conversely, suppose that \(f\) is Lebesgue-integrable on \(E\). Then, we have

\[\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu<+\infty.\]

Since

\[\int_{E}f^{+}d\nu<+\infty\mbox{ or }\int_{E}f^{-}d\nu<+\infty,\]

it follows

\[\int_{E}f^{+}d\nu<+\infty\mbox{ and }\int_{E}f^{-}d\nu<+\infty,\]

Therefore, we have

\[\int_{E}|f|d\nu =\int_{E}\left (f^{+}+f^{-}\right )d\nu =\int_{E}f^{+}d\nu +\int_{E}f^{-}d\nu <+\infty ,\]

which says that \(|f|\) is Lebesgue-integrable on \(E\).

To prove part (iii), since \(f\) is Lebesgue-integrable on \(E\), it says that \(|f|\) is also Lebesgue-integrable on \(E\) by part (ii). Therefore, from Proposition \ref{rap108}, we have \(|f|<+\infty\) a.e. on \(E\), which implies \(f<+\infty\) a.e. on \(E\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap148}}\tag{49}\mbox{}\end{equation}

Proposition \ref{rap148}. Suppose that the extended real-valued functions \(f\) and \(g\) are Lebesgue-measurable defined on a Lebesgue-measurable set \(E\). If \(f\) and \(g\) are Lebesgue-integrable on \(E\), then we have the following properties.

(i) Suppose that \(\nu (E)=0\) or \(f=0\) a.e. on \(E\). Then

\[\int_{E}fd\nu =0.\]

(ii) Suppose that \(f\leq g\) a.e. on \(E\). Then

\[\int_{E}fd\nu\leq\int_{E}gd\nu .\]

In particular, if \(f=g\) a.e. on \(E\), then

\[\int_{E}fd\nu =\int_{E}gd\nu .\]

Proof. To prove part (i), since \(f=f^{+}-f^{-}\) and

\[\int_{E}fd\nu =\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu,\]

the results follow immediately from Propositions \ref{rap101} and \ref{rap123}. To prove part (ii), since \(f\leq g\) a.e on \(E\) implies \(0\leq f^{+}\leq g^{+}\) a.e. on \(E\) and \(0\leq f^{-}\geq g^{-}\) a.e. on \(E\). Using Propositions \ref{rap101} and \ref{rap108}, we have

\[\int_{E}f^{+}d\nu\leq\int_{E}g^{+}d\nu\mbox{ and }\int_{E}f^{-}d\nu\geq\int_{E}g^{-}d\nu .\]

By subtracting these inequalities, we obtain the desired inequality, and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{rap149}}\tag{50}\mbox{}\end{equation}

Proposition \ref{rap149}. Let the extended real-valued function \(f\) be Lebesgue measurable defined on a Lebesgue-measurable set \(E\). Suppose that \(f\) is Lebesgue-integrable on \(E\). Then, we have the following properties.

(i) We have

\[\int_{E}(-f)d\nu =-\int_{E}fd\nu .\]

(ii) For any constant \(c\), the function \(cf\) is Lebesgue-integrable on \(E\) and

\[\int_{E}(cf)d\nu =c\int_{E}fd\nu .\]

Proof. To prove part (i), since \((-f)^{+}=f^{-}\) and \((-f)^{-}=f^{+}\) with

\[\int_{E}f^{-}d\nu<+\infty\mbox{ or }\int_{E}f^{+}d\nu<+\infty,\]

we have

\begin{align*} \int_{E}(-f)d\nu & =\int_{E}(-f)^{+}d\nu -\int_{E}(-f)^{-}d\nu \\ & =\int_{E}f^{-}d\nu -\int_{E}f^{+}d\nu =-\int_{E}fd\nu .\end{align*}

To prove part (ii), we first assume that \(c\geq 0\). Then, we have \((cf)^{+}=cf^{+}\) and \((cf)^{-}=cf^{-}\). Therefore, using Proposition \ref{rap101}, we obtain

\begin{align*} \int_{E}(cf)d\nu & =\int_{E}(cf)^{+}d\nu -\int_{E}(cf)^{-}d\nu
\\ & =\int_{E}cf^{+}d\nu -\int_{E}cf^{-}d\nu =c\int_{E}fd\nu .\end{align*}

For any \(c<0\), we have \(c=-|c|\). Therefore, the desired result follows from part (i) and the case of \(c\geq 0\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap150}}\tag{51}\mbox{}\end{equation}

Proposition \ref{rap150}. Suppose that the extended real-valued functions \(f\) and \(g\) are Lebesgue-integrable on the Lebesgue-measurable set \(E\). Then, the function \(f+g\) is Lebesgue-integrable on \(E\) satisfying

\[\int_{E}(f+g)d\nu =\int_{E}fd\nu +\int_{E}gd\nu .\]

Proof. We first want to show that if \(f\) can be represented as \(f=f_{1}-f_{2}\), where \(f_{1}\) and \(f_{2}\) are nonnegative and Lebesgue-integrable functions on \(E\), then

\begin{equation}{\label{raeq130}}\tag{52}
\int_{E}fd\nu =\int_{E}f_{1}d\nu -\int_{E}f_{2}d\nu .
\end{equation}

Since \(f=f^{+}-f^{-}\), we have \(f^{+}+f_{2}=f^{-}+f_{1}\). Therefore, using Proposition \ref{rap147}, we have

\[\int_{E}f^{+}d\nu +\int_{E}f_{2}d\nu =\int_{E}f^{-}d\nu +\int_{E}f_{1}d\nu ,\]

which implies

\[\int_{E}fd\nu =\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu =\int_{E}f_{1}d\nu -\int_{E}f_{2}d\nu .\]

Now, we have

\[f+g=f^{+}-f^{-}+g^{+}-g^{-}=(f^{+}+g^{+})-(f^{-}+g^{-})\equiv f_{1}-f_{2}.\]

Therefore, according to (\ref{raeq130}), we obtain

\begin{align*}
\int_{E}(f+g)d\nu & =\int_{E}(f^{+}+g^{+})d\nu -\int_{E}(f^{-}+g^{-})d\nu\\
& =\int_{E}f^{+}d\nu +\int_{E}g^{+}d\nu -\int_{E}f^{-}d\nu -\int_{E}g^{-}d\nu
=\int_{E}fd\nu +\int_{E}gd\nu .
\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap136}}\tag{53}\mbox{}\end{equation}

Proposition \ref{rap136}. Let \(E_{1}\) and \(E_{2}\) be Lebesgue-measurable sets satisfying \(E_{1}\subseteq E_{2}\). Suppose that the extended real-valued function \(f\) defined on \(E_{2}\) is Lebesgue-measurable. Then, we have that

\[\int_{E_{2}}fd\nu <+\infty\mbox{ implies }\int_{E_{1}}fd\nu <+\infty .\]

Proof. Using Proposition \ref{rap101}, we have

\begin{align*} \int_{E_{1}}fd\nu & =\int_{E_{1}}f^{+}d\nu-\int_{E_{1}}f^{-}d\nu\\ & \leq
\int_{E_{2}}f^{+}d\nu-\int_{E_{2}}f^{-}d\nu=\int_{E_{2}}fd\nu<+\infty.\end{align*}

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap135}}\tag{54}\mbox{}\end{equation}

Proposition \ref{rap135}. Let \(f\) be a Lebesgue-integrable function defined on a Lebesgue-measurable \(E\). Suppose that \(E\) is the countable union of disjoint Lebesgue-measurable sets \(\{E_{k}\}_{k=1}^{\infty}\) with \(E=\bigcup_{k=1}^{\infty}E_{k}\). Then, we have

\[\int_{E}fd\nu =\sum_{k=1}^{\infty}\int_{E_{k}}fd\nu .\]

Proof. Using Proposition \ref{rap136}, we have

\[(\int_{E_{k}}f^{+}d\nu -\int_{E_{k}}f^{-}d\nu=\int_{E_{k}}fd\nu <+\infty\mbox{ for all }k.\]

Using Proposition \ref{rap132}, we also have

\begin{align*}
\int_{E}fd\nu & =\int_{E}f^{+}d\nu -\int_{E}f^{-}d\nu =\sum_{k=1}^{\infty}\int_{E_{k}}f^{+}d\nu -\sum_{k=1}^{\infty}\int_{E_{k}}f^{-}d\nu\\
& =\sum_{k=1}^{\infty}\left (\int_{E_{k}}f^{+}d\nu -\int_{E_{k}}f^{-}d\nu\right )=\sum_{k=1}^{\infty}\int_{E_{k}}fd\nu .
\end{align*}

This completes the proof. \(\blacksquare\)

Theorem.  (Monotone Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\).

(i) Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) is increasing and

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E.\]

We also assume that there exists a Lebesgue-integrable function \(\phi\) on \(E\) satisfying \(f_{k}\geq\phi\) a.e. on \(E\) for all \(k\). Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

(ii) Suppose that the sequence \(\{f_{k}\}_{k=1}^{\infty}\) is decreasing and

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E.\]

We also assume that there exists a Lebesgue-integrable function \(\phi\) on \(E\) satisfying \(f_{k}\leq\phi\) a.e. on \(E\) for all \(k\). Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. To prove part (i), given any subset \(N\) of \(E\) with \(\nu (E)=0\), it is clear to see

\[\int_{E}fd\nu =\int_{E\setminus N}fd\nu\]

and

\[\int_{E}f_{k}d\nu =\int_{E\setminus N}f_{k}d\nu\mbox{ for all }k.\]

We can simply assume

\begin{align*} \lim_{k\rightarrow\infty}f_{k}({\bf x}) & =f({\bf x})\mbox{ and }f_{k}({\bf x})\\ & \geq\phi({\bf x})\mbox{ for all }{\bf x}\in E\mbox{ for all }k.\end{align*}

Then, the sequence \(\{f_{k}-\phi\}_{k=1}^{\infty}\) is nonnegative and increasing satisfying

\[\lim_{k\rightarrow\infty}\left [f_{k}({\bf x})-\phi({\bf x})\right ]=f({\bf x})-\phi({\bf x})\mbox{ for all }{\bf x}\in E.\]

Using Theorem \ref{rat98}, Propositions \ref{rap149} and \ref{rap150}, we obtain

\begin{align*}
\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu -\int_{E}\phi d\nu
& =\lim_{k\rightarrow\infty}\left (\int_{E}f_{k}d\nu -\int_{E}\phi d\nu\right )
\\& =\lim_{k\rightarrow\infty}\int_{E}\left (f_{k}-\phi\right )d\nu\\
& =\int_{E}(f-\phi )d\nu =\int_{E}fd\nu -\int_{E}\phi d\nu .
\end{align*}

Since \(\int_{E}\phi d\nu <+\infty\), we obtain the desired result. Part (ii) can be similarly obtained by considering \(-f_{k}\). This completes the proof. \(\blacksquare\)

Theorem.  (Uniform Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-integrable functions defined on the same Lebesgue-measurable set \(E\) with \(\nu (E)<+\infty\). Suppose that \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(E\). Then \(f\) is Lebesgue-integrable on \(E\) and

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. Since \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(E\) and \(|f|\leq |f_{k}|+|f-f_{k}|\), there exists an integer \(N\) satisfying \(|f|\leq |f_{N}|+1\) on \(E\). Since \(\nu (E)<+\infty\), it says that \(f\) is Lebesgue-integrable on \(E\). We also have

\begin{align*} \left |\int_{E}fd\nu -\int_{E}f_{k}d\nu\right | & =\left |\int_{E}\left (f-f_{k}\right )d\nu \right |\\ & \leq
\int_{E}\left |f-f_{k}\right |d\nu\\ & \leq\nu (E)\cdot\sup_{{\bf x}\in E}\left |f({\bf x})-f_{k}({\bf x})\right |\rightarrow 0\end{align*}

as \(k\rightarrow\infty\) by the uniform convergence of the sequence \(\{f_{k}\}_{k=1}^{\infty}\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rat151}}\tag{55}\mbox{}\end{equation}

Theorem \ref{rat151}. (Fatou’s Lemma). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\). Suppose that there exists a Lebesgue-integrable function \(\phi\) on \(E\) satisfying \(f_{k}({\bf x})\geq\phi({\bf x})\) a.e. on \(E\) for all \(k\). Then, we have

\[\int_{E}\left (\liminf_{k\rightarrow\infty}f_{k}\right )d\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Proof. We can assume that \(f_{k}\geq\phi\) every where on \(E\). Then \(f_{k}-\phi\) is nonnegative on \(E\) for all \(k\). Using the Fatou’s lemma for the nonnegativity as shown in Theorem \ref{ra2}, and Propositions \ref{rap149} and \ref{rap150}, we obtain

\begin{align*}
\int_{E}\left (\liminf_{k\rightarrow\infty}f_{k}\right )d\nu -\int_{E}\phi d\nu & =\int_{E}\left [\liminf_{k\rightarrow\infty}\left (f_{k}-\phi\right )\right ]d\nu\\
& \leq\liminf_{k\rightarrow\infty}\int_{E}(f_{k}-\phi )d\nu \\ & =\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu -\int_{E}\phi d\nu .
\end{align*}

Since \(\int_{E}\phi d\nu <+\infty\), we obtain the desired result. \(\blacksquare\)

Theorem. (Fatou’s Lemma). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\). Suppose that there exists a Lebesgue-integrable function \(\phi\) on \(E\) satisfying \(f_{k}({\bf x})\leq\phi({\bf x})\) a.e. on \(E\) for all \(k\). Then, we have

\[\int_{E}\left (\limsup_{k\rightarrow\infty}f_{k}\right )d\nu\geq\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Proof. Since \(-f_{k}({\bf x})\geq -\phi({\bf x})\) a.e. on \(E\) for all \(k\) and

\[\liminf_{k\rightarrow\infty}(-f_{k})=-\limsup_{k\rightarrow\infty}f_{k},\]

the result follows immediately from Theorem \ref{rat151}. \(\blacksquare\)

\begin{equation}{\label{rat152}}\tag{56}\mbox{}\end{equation}

Theorem \ref{rat152}. (Dominated Convergence Theorem). Let \(\phi\) and \(\{f_{k}\}_{k=1}^{\infty}\) be Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\) satisfying \(|f_{k}({\bf x})|\leq\phi({\bf x})\) a.e on \(E\) and

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})\rightarrow f({\bf x})\mbox{ a.e. on }E.\]

Suppose that \(\phi\) is Lebesgue-integrable on \(E\). Then,we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. We can assume that \(|f_{k}({\bf x})|\leq\phi({\bf x})\) everywhere on \(E\) and

\[f_{k}({\bf x})\rightarrow f({\bf x})\mbox{ for all }{\bf x}\in E.\]

We also have

\[|f({\bf x})|=\lim_{k\rightarrow\infty}|f_{k}({\bf x})|\leq\phi({\bf x})\mbox{ for all }{\bf x}\in E,\]

which implies that \(f\) is Lebesgue-integrable on \(E\). Using \(-\phi\leq f_{k}\leq\phi\), we have that \(\phi -f_{k}\) and \(\phi +f_{k}\) are nonnegative on \(E\). Considering the nonnegative sequence \(\{\phi -f_{k}\}_{k=1}^{\infty}\) and using the Fatou’s Lemma shown in Theorem \ref{rat150}, we have

\begin{align*} \int_{E}\phi d\nu -\int_{E}fd\nu & =\int_{E}\left (\phi -f\right )d\nu
\\ & \leq\liminf_{k\rightarrow\infty}\int_{E}\left (\phi -f_{k}\right )d\nu
\\ & =\int_{E}\phi d\nu -\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu ,\end{align*}

which implies

\[\int_{E}fd\nu\geq\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Considering the nonnegative sequence \(\{\phi +f_{k}\}_{k=1}^{\infty}\) on \(E\), we can similarly obtain

\[\int_{E}fd\nu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu .\]

Since

\[\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu\leq\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu,\]

it follows

\[\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\nu=\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\nu=\int_{E}fd\nu.\]

This completes the proof. \(\blacksquare\)

Corollary. (Bounded Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\) with \(\nu (E)<+\infty\) satisfying

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})\rightarrow f({\bf x})\mbox{ a.e. on }E.\]

Suppose that there is a finite constant \(M\) satisfying \(|f_{k}({\bf x})|\leq M\) a.e. on \(E\). Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. By considering the Lebesgue-integrable function \(\phi({\bf x})=M\) for all \(x\in E\) and using the dominated convergence Theorem \ref{rat152}, we complete the proof. \(\blacksquare\)

Theorem.  (Generalization of Lebesgue Dominated Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-measurable functions defined on the same Lebesgue-measurable set \(E\), and let \(\{\phi_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue-integrable functions on \(E\) which converges a.e. to a Lebesgue-integrable function \(\phi\) on \(E\). We also assume that \(|f_{k}({\bf x})|\leq\phi_{k}({\bf x})\) a.e. on \(E\) and

\[\lim_{k\rightarrow\infty}f_{k}({\bf x})=f({\bf x})\mbox{ a.e. on }E.\]

Suppose that

\[\lim_{k\rightarrow\infty}\int_{E}\phi_{k}d\nu =\int_{E}\phi d\nu .\]

Then, we have

\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\nu =\int_{E}fd\nu .\]

Proof. Using the proof of Theorem \ref{rat152} by replacing the function \(\phi\) as the appropriate function \(\phi_{k}\), we can obtain the generalization, which is left as an exercise. \(\blacksquare\)

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

Iterated Integration.

Let \(f\) be an extended real-valued function defined on a rectangle

\[{\bf I}=\left\{(x,y)\in\mathbb{R}^{2}:a\leq x\leq b\mbox{ and }c\leq y\leq d\right\}.\]

Suppose that \(f\) is continuous on \({\bf I}\). Then, the formula is given by

\begin{align*} \int\!\!\!\!\!\int_{\bf I}f(x,y)dxdy & =\int_{a}^{b}\left [\int_{c}^{d}f(x,y)dy\right ]dx\\ & =\int_{c}^{d}\left [\int_{a}^{b}f(x,y)dx\right ]dy.\end{align*}

Now, we are going to consider the analogous formula for functions of \(n\) variables. Let \({\bf x}\) be a point of an \(n\)-dimensional interval \({\bf I}_{1}\)

\[{\bf I}_{1}=\left\{{\bf x}\in\mathbb{R}^{n}:a_{i}\leq x_{i}\leq b_{i}\mbox{ for }i=1,\cdots ,n\right\},\]

and let \({\bf y}\) be a point of an \(m\)-dimensional interval \({\bf I}_{2}\)

\[{\bf I}_{2}=\left\{{\bf y}\in\mathbb{R}^{m}:c_{j}\leq x_{i}\leq d_{j}\mbox{ for }j=1,\cdots ,m\right\}.\]

Here \({\bf I}_{1}\) and \({\bf I}_{2}\) may be all of \(\mathbb{R}^{n}\) and \(\mathbb{R}^{m}\), respectively. The Cartesian product $latex {\bf I}={\bf I}_{1}
\times {\bf I}_{2}$ is an \((n+m)\)-dimensional interval consisting of points \(({\bf x},{\bf y})\in\mathbb{R}^{n+m}\).

\begin{equation}{\label{rat31}}\tag{57}\mbox{}\end{equation}

Theorem \ref{rat31}. (Fubini’s Theorem on Intervals) Let \(f\) be Lebesgue-integrable on \({\bf I}={\bf I}_{1}\times {\bf I}_{2}\). Then, we have the following properties.

(i) For almost every \({\bf x}_{0}\in {\bf I}_{1}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue measurable and integrable on \({\bf I}_{2}\) as a function of \({\bf y}\).

(ii) We have that

\[\int_{{\bf I}_{2}}f({\bf x},{\bf y})d\nu ({\bf y})\]

is Lebesgue-measurable and integrable on \({\bf I}_{1}\) as a function of \({\bf x}\). We also have

\[\int\!\!\!\!\!\int_{\bf I}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y})
=\int_{{\bf I}_{1}}\left [\int_{{\bf I}_{2}}f({\bf x},{\bf y})
d\nu ({\bf y})\right ]d\nu ({\bf x}).\]

Since the roles of \({\bf x}\) and \({\bf y}\) in Theorem \ref{rat31} can be interchanged, if \(f\) is Lebesgue-integrable on \({\bf I}={\bf I}_{1}\times {\bf I}_{2}\), then

\begin{align*} \int\!\!\!\!\!\int_{\bf I}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y})
& =\int_{{\bf I}_{1}}\left [\int_{{\bf I}_{2}}f({\bf x},{\bf y})d\nu ({\bf y})\right ]d\nu ({\bf x})
\\ & =\int_{{\bf I}_{2}}\left [\int_{{\bf I}_{1}}f({\bf x},{\bf y})d\nu ({\bf x})\right ]d\nu ({\bf y}).\end{align*}

We shall extend the Fubini’s theorem to functions defined on subsets of \(\bar{\mathbb{R}}^{n+m}\).

Proposition. Suppose that the extended real-valued function \(f\) defined on \(\bar{\mathbb{R}}^{n+m}\) is Lebesgue-measurable. Then, for almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\), the function \(f({\bf x}_{0},{\bf y})\) is a Lebesgue-measurable function of \({\bf y}\in\bar{\mathbb{R}}^{m}\). In particular, if \(E\) is a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n+m}\), then the set

\[E_{\bf x_{0}}=\left\{{\bf y}\in\bar{\mathbb{R}}^{m}:({\bf x}_{0},{\bf y})\in E\right\}\]

is Lebesgue-measurable in \(\bar{\mathbb{R}}^{m}\) for almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\). \(\sharp\)

Theorem.  (Fubini’s Theorem). Let \(f\) be Lebesgue-integrable on a Lebesgue measurable subset \(E\) of \(\bar{\mathbb{R}}^{n+m}\), and let

\[E_{{\bf x}_{0}}=\left\{{\bf y}\in\bar{\mathbb{R}}^{m}:({\bf x}_{0},{\bf y})\in E\right\}.\]

Then, we have the following properties.

(i) For almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue-measurable on \(E_{{\bf x}_{0}}\) as a function of \({\bf y}\).

(ii) Suppose that \(f\) is Lebesgue-integrable on \(E\). Then, for almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue-integrable on \(E_{{\bf x}_{0}}\) with respect to \({\bf y}\). Moreover, the function

\[\int_{E_{{\bf x}_{0}}}f({\bf x},{\bf y})d\nu ({\bf y})\]

is a Lebesgue-integrable function of \({\bf x}\) and

\[\int\!\!\!\!\!\int_{E}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y})=\int_{\mathbb{R}^{n}}\left [
\int_{E_{\bf x}}f({\bf x},{\bf y})d\nu ({\bf y})\right ]d\nu ({\bf x}).\]

According to the Fubini’s theorem, If \(f\) is Lebesgue-integrable on \(E\), then it is equal to the corresponding iterated integral. However, the converse is not true. Let \({\bf I}=[0,1]\times [0,1]\). We can construct a function \(f\) satisfying

\[\int_{0}^{1}\left [\int_{0}^{1}f(x,y)dy\right ]dx=0=\int_{0}^{1}\left [\int_{0}^{1}f(x,y)dx\right ]dy\]

and

\[\int\!\!\!\!\!\int_{\bf I}|f(x,y)|dxdy=+\infty .\]

Hence, the finiteness of the iterated integrals does not in general imply that of the multiple integral.

Theorem. (Tonelli’s Theorem on Intervals). Let the extended real-valued function \(f\) be nonnegative and Lebesgue-measurable on \({\bf I}={\bf I}_{1}\times{\bf I}_{2}\subseteq\bar{\mathbb{R}}^{n+m}\). Then, we have the following properties.

(i) For almost every \({\bf x}_{0}\in {\bf I}_{1}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue measurable and integrable on \({\bf I}_{2}\) as a function of \({\bf y}\).

(ii) We have that

\[\int_{{\bf I}_{2}}f({\bf x},{\bf y})d\nu ({\bf y})\]

is Lebesgue measurable and integrable on \({\bf I}_{1}\) as a function of \({\bf x}\). We also have

\[\int\!\!\!\!\!\int_{\bf I}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y})=\int_{{\bf I}_{1}}\left [
\int_{{\bf I}_{2}}f({\bf x},{\bf y})d\nu ({\bf y})\right ]d\nu ({\bf x}).\]

An extension of Tonelli’s theorem to functions defined on arbitrary Lebesgue measurable sets \(E\) is presented below.

\begin{equation}{\label{rat32}}\tag{58}\mbox{}\end{equation}

Theorem \ref{rat32}. (Tonelli’s Theorem on Measurable Sets). Let \(f\) be nonnegative and Lebesgue measurable on a Lebesgue measurable subset \(E\) of \(\bar{\mathbb{R}}^{n+m}\), and let

\[E_{{\bf x}_{0}}=\left\{{\bf y}\in\bar{\mathbb{R}}^{m}:({\bf x}_{0},{\bf y}\in E\right\}.\]

Then, we have the following properties.

(i) For almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue-measurable on \(E_{{\bf x}_{0}}\) as a function of \({\bf y}\).

(ii) If \(f({\bf x},{\bf y})\) is Lebesgue-integrable on \(E\), then for almost every \({\bf x}_{0}\in\bar{\mathbb{R}}^{n}\), the function \(f({\bf x}_{0},{\bf y})\) is Lebesgue-integrable on \(E_{{\bf x}_{0}}\) with respect to \({\bf y}\). Moreover, the function \(\int_{E_{{\bf x}_{0}}}f({\bf x},{\bf y})d\nu ({\bf y})\) is a Lebesgue-integrable function of \({\bf x}\) and

\[\int\!\!\!\!\!\int_{E}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y})=\int_{\mathbb{R}^{n}}\left [
\int_{E_{\bf x}}f({\bf x},{\bf y})d\nu ({\bf y})\right ]d\nu ({\bf x}).\]

Since the roles of \({\bf x}\) and \({\bf y}\) in Theorem \ref{rat32} can be interchanged, if \(f({\bf x},{\bf y})\) is nonnegative and Lebesgue-measurable on
${\bf I}={\bf I}_{1}\times {\bf I}_{2}$, then

\begin{align*} \int\!\!\!\!\!\int_{\bf I}f({\bf x},{\bf y})d\nu ({\bf x},{\bf y}) & =\int_{{\bf I}_{1}}\left [
\int_{{\bf I}_{2}}f({\bf x},{\bf y})d\nu ({\bf y})\right ]d\nu ({\bf x})\\ & =\int_{{\bf I}_{2}}\left [
\int_{{\bf I}_{1}}f({\bf x},{\bf y})d\nu ({\bf x})\right ]d\nu ({\bf y}).\end{align*}

Therefore, if \(f({\bf x},{\bf y})\) is nonnegative and Lebesgue-measurable, then the finiteness of any one of Fubini’s three integrals implies that of the other two integrals. It also says that, for any Lebesgue-measurable function \(f\), the finiteness of one of these integrals for \(|f|\) implies that \(f\) is Lebesgue-integrable and that all three Fubini’s integrals of \(f\) are equal.

If \(f\) and \(g\) are Lebesgue-measurable on \(\bar{\mathbb{R}}^{n}\), their convolution is denoted and defined by

\[(f*g)({\bf x})=\int_{\bar{\mathbb{R}}^{n}}f({\bf x}-{\bf t})g({\bf t})d\nu ({\bf t})\]
provided the integral exists. Then, we have \(f*g=g*f\).

Proposition. If \(f\) and \(g\) are Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\), then \((f*g)({\bf x})\) exists a.e. on \(\bar{\mathbb{R}}^{n}\), \(f*g\) is
Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\) and

\[\int_{\bar{\mathbb{R}}^{n}}|f*g|d\nu\leq\left (\int_{\bar{\mathbb{R}}^{n}}|f|d\nu\right )\left (\int_{\bar{\mathbb{R}}^{n}}|g|d\nu\right ).\]

Corollary. If \(f\) and \(g\) are nonnegative and Lebesgue-measurable on \(\bar{\mathbb{R}}^{n}\), then \((f*g)({\bf x})\) exists a.e. on \(\bar{\mathbb{R}}^{n}\), \(f*g\) is Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\) and

\[\int_{\bar{\mathbb{R}}^{n}}(f*g)d\nu\leq\left (\int_{\bar{\mathbb{R}}^{n}}fd\nu\right )
\left (\int_{\bar{\mathbb{R}}^{n}}gd\nu\right ).\]

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

Inequalities.

Let \(E\) be a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n}\). For \(0<p<+\infty\), the space \(L^{p}(E)\) denotes the collection of Lebesgue-measurable function \(f\) satisfying

\[\int_{E}|f|^{p}d\nu <+\infty.\]

More precisely, we have

\[L^{p}=\left\{f:\int_{E}|f|^{p}d\nu <+\infty\right\}.\]

We also define

\[\parallel f\parallel_{p}=\left (\int_{E}|f|^{p}d\nu\right )^{1/p}.\]

Then, we have

\[L^{p}=\left\{f:\parallel f\parallel_{p}<+\infty\right\}.\]

In order to define \(L^{\infty}(E)\), we introduce the concept of essential supremum as follows.

Definition. Let \(f\) be Lebesgue-measurable on a Lebesgue-measurable set \(E\) with \(\nu (E)>0\).

• Suppose that \(\nu (\{{\bf x}\in E:f({\bf x})>\alpha\})>0\) for all real number \(\alpha\). The essential supremum of \(f\) on \(E\) is denoted and defined as \(\mbox{ess sup}_{E}f=+\infty\).
• Suppose that \(\nu (\{{\bf x}\in E:f({\bf x})>\alpha\})=0\) for some real number \(\alpha\). Then, we consider the set \[Z=\left\{\alpha\in\mathbb{R}:\nu (\{{\bf x}\in E:f({\bf x})>\alpha\})=0\right\}\neq\emptyset\] and the essential supremum of \(f\) on \(E\) is denoted and defined by \begin{align*} \mbox{ess sup}_{E}f & =\inf Z\\ & =\inf\left\{\alpha\in\mathbb{R}:\nu (\{{\bf x}\in E:f({\bf x})>\alpha\})=0\right\}.\end{align*}

We have

\[\mbox{ess sup}_{E}f=\inf\left\{\alpha\in\mathbb{R}:f({\bf x})\leq\alpha\mbox{ a.e. on }E\right\},\]

which also says that \(\mbox{ess sup}_{E}f\) is the smallest number \(M\) satisfying \(f({\bf x})\leq M\) a.e. on \(E\). A Lebesgue-measurable function \(f\) is said to be essentially bounded on \(E\) when \(\mbox{ess sup}_{E}|f|<+\infty\). The space of all Lebesgue-measurable functions that are essentially bounded on \(E\) is denoted by \(L^{\infty}\). More precisely, we have

\[L^{\infty}=\left\{f:\mbox{ess sup}_{E}|f|<+\infty\right\}.\]

We also define

\[\parallel f\parallel_{\infty}=\mbox{ess sup}_{E}|f|.\]

Therefore, we have

\[|f({\bf x})|\leq\parallel f\parallel_{\infty}\mbox{ a.e. on }E\]

and

\[L^{\infty}(E)=\left\{f:\parallel f\parallel_{\infty}<+\infty\right\}.\]

The motivation for this notation is given below.

Proposition. Suppose that \(\nu (E)<+\infty\). Then, we have

\[\parallel f\parallel_{\infty}=\lim_{p\rightarrow\infty}\parallel f\parallel_{p}.\]

Proof. Let \(M=\parallel f\parallel_{\infty}\). For any \(M'<M\), the definition of essential supremum says that the following set

\[A=\left\{{\bf x}\in E:|f({\bf x})|>M’\right\}\]

has a positive Lebesgue measure. Then, we have

\begin{align}
\parallel f\parallel_{p} & =\left (\int_{E}|f|^{p}d\nu\right )^{1/p}\geq\left (\int_{A}|f|^{p}d\nu\right )^{1/p}\nonumber\\ & \geq\left (\int_{A}(M’)^{p}d\nu\right )^{1/p}=\left ((M’)^{p}\cdot\nu (A)\right )^{1/p}\nonumbner\\ & =M’\cdot\nu (A)^{1/p}.\label{raeq160}\tag{59}
\end{align}

Using (\ref{raeq160}), we have

\begin{align*} \liminf_{p\rightarrow\infty}\parallel f\parallel_{p} & \geq\liminf_{p\rightarrow\infty}
M’\cdot\nu (A)^{1/p}\\ & =M’\cdot\lim_{p\rightarrow\infty}\nu (A)^{1/p}=M’.\end{align*}

Since \(M’\) can be any real number satisfying \(M'<M\), we also have

\[\liminf_{p\rightarrow\infty}\parallel f\parallel_{p}\geq M.\]

On the other hand, since

\[|f({\bf x})|\leq\parallel f\parallel_{\infty}=M\mbox{ a.e. on }E,\]

we have

\begin{align*} \parallel f\parallel_{p} & =\left (\int_{E}|f|^{p}d\nu\right )^{1/p}\leq\left (\int_{E}M^{p}d\nu\right )^{1/p}\\ & =
\left (M^{p}\cdot\nu (E)\right )^{1/p}=M\cdot\nu (E)^{1/p},\end{align*}

which implies

\begin{align*} \limsup_{p\rightarrow\infty}\parallel f\parallel_{p} & \leq\limsup_{p\rightarrow\infty}
M\cdot\nu (E)^{1/p}\\ & =M\cdot\lim_{p\rightarrow\infty}\nu (E)^{1/p}=M.\end{align*}

Therefore, we obtain
\begin{align*} \liminf_{p\rightarrow\infty}\parallel f\parallel_{p}
& =\limsup_{p\rightarrow\infty}\parallel f\parallel_{p}\\ & =M=\parallel f\parallel_{\infty}.\end{align*}

This completes the proof. \(\blacksquare\)

Proposition. Suppose that \(0<p_{1}<p_{2}\leq +\infty\) and \(\nu (E)<+\infty\). Then, we have \(L^{p_{2}}\subseteq L^{p_{1}}\).

Proof. We write \(E=E_{1}\cup E_{2}\), where

\[E_{1}=\{{\bf x}\in E:|f({\bf x})|\leq 1\}\mbox{ and }E_{2}=\{{\bf x}\in E:|f({\bf x})|>1\}.\]

Therefore, for \(0<p<+\infty\), we have

\begin{align*} \int_{E}|f|^{p}d\nu & =\int_{E_{1}}|f|^{p}d\nu +\int_{E_{2}}|f|^{p}d\nu\\ & \leq\nu (E_{1})+\int_{E_{2}}|f|^{p}d\nu .\end{align*}

We see that the second integral \(\int_{E_{2}}|f|^{p}d\nu\) is increasing with respect to \(p\), since \(|f|>1\) on \(E_{2}\).

• Suppose that \(f\in L^{p_{2}}\) for \(0<p_{2}<+\infty\). Then \(f\in L^{p_{1}}\) for \(0<p_{1}<p_{2}<+\infty\).
• Suppose that \(p_{2}=+\infty\). Then \(f\) is a bounded function on a set of finite measure. This also says that \(f\in L^{p_{1}}\).

Therefore, we have the inclusion \(L^{p_{2}}\subseteq L^{p_{1}}\). This completes the proof. \(\blacksquare\)

Proposition. Suppose that \(f,g\in L^{p}\) for \(p>0\). Then \(f+g\in L^{p}\) and \(cf\in L^{p}\) for any constant \(c\). \(\sharp\)

Proposition. (Young’s Inequality). Let \(y=f(x)\) be a real-valued continuous and strictly increasing function for \(x\geq 0\) satisfying \(f(0)=0\). Then, for \(a,b>0\), we have

\[ab\leq\int_{0}^{a}f(x)dx +\int_{0}^{b}f^{-1}(y)dy,\]

where the integrals are Riemann integrals. The equality holds if and only if \(b=f(a)\).

Proof. A geometric proof is obvious. We recall that the graph of \(f\) also serves as that of \(f^{-1}\) if we interchange the \(x\) and \(y\) axes. If we interpret each integral as an area, then the inequality follows immediate. \(\blacksquare\)

Given \(f(x)=x^{\alpha}\) for \(\alpha >0\), we have \(f^{-1}(y)=y^{1/\alpha}\). In this case, the Young’s inequality becomes

\[ab\leq\frac{a^{1+\alpha}}{1+\alpha}+\frac{b^{1+1/\alpha}}{1+1/\alpha}.\]

By setting \(p=1+\alpha\) and \(q=1+1/\alpha\), we obtain

\begin{equation}{\label{raeq162}}\tag{60}
ab\leq\frac{a^{p}}{p}+\frac{b^{q}}{q}\mbox{ and }\frac{1}{p}+\frac{1}{q}=1
\end{equation}

for \(a,b\geq 0\) and \(1<p<+\infty\). The number \(p\) and \(q\) that satisfy \(1/p+1/q=1\) for \(p,q>1\) are called conjugate exponents. We shall adopt the convention for \(p=1\) and \(q=+\infty\).

\begin{equation}{\label{rat165}}\tag{61}\mbox{}\end{equation}

Theorem \ref{rat165}. (Holder’s Inequality). Suppose that \(1\leq p\leq +\infty\) and \(1/p+1/q=1\). Then, we have

\[\parallel fg\parallel_{1}\leq\parallel f\parallel_{p}\cdot\parallel g\parallel_{q}.\]

More precisely, we have the following inequalities.

• For \(1<p<+\infty\), we have
  \[\int_{E}|fg|d\nu\leq\left (\int_{E}|f|^{p}d\nu\right )^{1/p}\cdot\left (\int_{E}|g|^{q}d\nu\right )^{1/q},\]
• For \(p=1\), we have
  \begin{align*} \int_{E}|fg|d\nu & \leq\parallel g\parallel_{\infty}\cdot\int_{E}|f|d\nu\\ & =\left (\mbox{ess sup}_{E}|g|\right )\cdot\int_{E}|f|d\nu\end{align*}
• For \(p=+\infty\), we have
  \begin{align*} \int_{E}|fg|d\nu & \leq\parallel f\parallel_{\infty}\cdot\int_{E}|g|d\nu\\ & =\left (\mbox{ess sup}_{E}|f|\right )\cdot\int_{E}|g|d\nu.\end{align*}

Proof. For \(p=1\), since \(|g({\bf x})|\leq\parallel g\parallel_{\infty}\) a.e. on \(E\), we have

\begin{align*} \int_{E}|fg|d\nu & \leq\int_{E}\parallel g\parallel_{\infty}\cdot |f|d\nu
\\ & =\parallel g\parallel_{\infty}\cdot\int_{E}|f|d\nu.\end{align*}

The inequality corresponding to the cases of \(p=+\infty\) can be similarly obtained. Now, we assume \(1<p<+\infty\). Suppose that any one of \(\parallel f\parallel_{p}\) and \(\parallel g\parallel_{q}\) is zero. Then \(fg\) is zero a.e. on \(E\). Therefore, the inequality is obvious. Now, we consider the case of

\[\parallel f\parallel_{p}=1=\parallel g\parallel_{q}.\]

Using inequality (\ref{raeq162}), we have

\[|fg|\leq\frac{|f|^{p}}{p}+\frac{|g|^{q}}{q}.\]

Therefore, we obtain

\begin{align}
\parallel fg\parallel_{1} & =\int_{E}|fg|d\nu\leq\int_{E}\left (\frac{|f|^{p}}{p}+\frac{|g|^{q}}{q}\right )d\nu
\nonumber\\ & =\frac{\parallel f\parallel_{p}^{p}}{p}+\frac{\parallel g\parallel_{q}^{q}}{q}
=\frac{1}{p}+\frac{1}{q}=1.\label{raeq163}\tag{62}
\end{align}

For the general case, we set

\[\hat{f}=\frac{f}{\parallel f\parallel_{p}}\mbox{ and }\hat{g}=\frac{g}{\parallel g\parallel_{q}}.\]

Then

\[\parallel \hat{f}\parallel_{p}=1=\parallel \hat{g}\parallel_{q}.\]

Using the above case by referring to (\ref{raeq163}), we have

\[\int_{E}|\hat{f}\hat{g}|d\nu\leq 1.\]

Therefore, we obtain

\begin{align*} \parallel fg\parallel_{1} & =\int_{E}|fg|d\nu \\ & =
\parallel f\parallel_{p}\cdot\parallel g\parallel_{q}\cdot\int_{E}|\hat{f}\hat{g}|d\nu
\\ & \leq \parallel f\parallel_{p}\cdot\parallel g\parallel_{q}.\end{align*}

This completes the proof. \(\blacksquare\)

By taking \(p=q=2\) in Theorem \ref{rat165}, we obtain the Schwartz’s inequality given below

\[\int_{E}|fg|d\nu\leq\left (\int_{E}f^{2}d\nu\right )^{1/2}\cdot\left (\int_{E}g^{2}d\nu\right )^{1/2}.\]

Theorem.  (Minkowski’s Inequality). Suppose that \(1\leq p\leq +\infty\). Then, we have

\[\parallel f+g\parallel_{p}\leq\parallel f\parallel_{p}+\parallel g\parallel_{p}.\]

More precisely, we have

\[\left (\int_{E}|f+g|^{p}\right )^{1/p}\leq\left (\int_{E}|f|^{p}d\nu\right )^{1/p}
+\left (\int_{E}|g|^{p}d\nu\right )^{1/p}\mbox{ for }1\leq p<+\infty\]

and

\[\mbox{ess sup}_{E}|f+g|\leq\mbox{ess sup}_{E}|f|+\mbox{ess sup}_{E}|g|\mbox{ for }p=+\infty .\]

Proof. For \(p=1\), we have

\begin{align*} \parallel f+g\parallel_{p} & =\int_{E}|f+g|d\nu\\ & \leq\int_{E}|f|d\nu+\int_{E}|g|d\nu
\\ & =\parallel f\parallel_{1}+\parallel g\parallel_{1}.\end{align*}

For \(p=+\infty\), we have

\[|f({\bf x})|\leq\parallel f\parallel_{\infty}\mbox{ a.e. on }E\mbox{ and }|g({\bf x})|\leq\parallel g\parallel_{\infty}\mbox{ a.e. on }E.\]

Therefore, we obtain

\begin{align*} |f({\bf x})+g({\bf x})| & \leq |f({\bf x})|+|g({\bf x})|\\ & \leq\parallel f\parallel_{\infty}+\parallel g\parallel_{\infty}\mbox{ a.e. on \(E\)},\end{align*}

which implies

\[\parallel f+g\parallel_{\infty}\leq\parallel f\parallel_{\infty}+\parallel g\parallel_{\infty}.\]

Now, we consider the case of \(1<p<+\infty\). Suppose that \(\parallel f+g\parallel_{p}=0\). Then, there is nothing to prove. Suppose that \(\parallel f+g\parallel_{p}=+\infty\). Then, either \(\parallel f\parallel_{p}=+\infty\) or \(\parallel g\parallel_{p}=+\infty\), since \(\parallel f\parallel_{p}<+\infty\) and \(\parallel g\parallel_{p}<+\infty\) implies \(\parallel f+g\parallel_{p}<+\infty\). Therefore, we remain to consider the case of \(0<\parallel f+g\parallel_{p} <+\infty\). Now, we have

\begin{align}
\parallel f+g\parallel_{p}^{p} & =\int_{E}|f+g|^{p}d\nu =\int_{E}|f+g|^{p-1}\cdot |f+g|d\nu\nonumber\\
& \leq\int_{E}|f+g|^{p-1}\cdot |f|d\nu+\int_{E}|f+g|^{p-1}\cdot |g|d\nu .\label{raeq177}\tag{63}
\end{align}

Let \(q=p/(p-1)\). Since

\begin{align*} \parallel (f+g)^{p-1}\parallel_{q} & =\left [\int_{E}\left (f+g\right )^{(p-1)q}d\nu \right ]^{1/q}
\\ & =\left [\int_{E}\left (f+g\right )^{p}d\nu \right ]^{(p-1)/p}\\ & =\parallel f+g\parallel_{p}^{p-1},\end{align*}

applying the H\”{o}lder’s inequality in Theorem \ref{rat165}, we obtain

\begin{align}
\int_{E}|f+g|^{p-1}\cdot |g|d\nu & \leq\parallel g\parallel_{p}\cdot\parallel (f+g)^{p-1}
\parallel_{q}\nonumber\\ & =\parallel f+g\parallel_{p}^{p-1}\cdot\parallel g\parallel_{p},\label{raeq178}\tag{64}
\end{align}

Similarly, we can obtain

\begin{equation}{\label{raeq179}}\tag{65}
\int_{E}|f+g|^{p-1}\cdot |f|d\nu\leq\parallel f+g\parallel_{p}^{p-1}\cdot\parallel f\parallel_{p}.
\end{equation}

Therefore, using (\ref{raeq177}), (\ref{raeq178}) and (\ref{raeq179}), we have

\begin{align}
\parallel f+g\parallel_{p}^{p} & =\int_{E}|f+g|^{p}d\nu\nonumber\\ & \leq\parallel f+g\parallel_{p}^{p-1}\cdot
\left (\parallel f\parallel_{p}+\parallel g\parallel_{p}\right ).\label{raeq180}\tag{66}
\end{align}

By dividing \(\parallel f+g\parallel_{p}^{p-1}\) on both sides, we complete the proof. \(\blacksquare\)

The Minkowski’s inequality fails for \(0<p<1\). To see this, we take \(E=(0,1)\), \(f=\chi_{(0,1/2)}\) and \(g=\chi_{(1/2,1)}\). Then, we have

\[\parallel f+g\parallel_{p}=1\mbox{ and }\parallel f\parallel_{p}+\parallel g\parallel_{p}=2^{-1/p}+2^{-1/p}=2^{1-1/p}<1.\]

Theorem. Let the extended real-valued function \(f\) be Lebesgue-measurable on \(E\). Suppose that \(1\leq p\leq +\infty\) and \(1/p+1/q=1\). We take a collection of extended real-valued functions by

\[{\cal G}=\left\{g:\parallel g\parallel_{q}\leq 1\right\}.\]

Then, we have

\[\parallel f\parallel_{p}=\sup_{g\in {\cal G}}\int_{E}fgd\nu .\]

Proof. Using the H\”{o}lder’s inequality, since \(\parallel g\parallel_{q}\leq 1\) for any \(g\), we have

\begin{align*} \sup_{g\in {\cal G}}\int_{E}fgd\nu & \leq\sup_{g\in {\cal G}}\int_{E}|fg|d\nu\\ & \leq\sup_{g\in {\cal G}}\left (\parallel f\parallel_{p}
\cdot\parallel g\parallel_{q}\right )\leq\parallel f\parallel_{p}.\end{align*}

For the opposite inequality, we assume \(1<p<+\infty\). We consider three cases. Suppose that \(\parallel f\parallel_{p}=0\). Then \(f=0\) a.e. on \(E\). The result is obvious. Suppose that \(0<\parallel f\parallel_{p}<+\infty\). We first consider the case of \(f\geq 0\). Now, we assume \(\parallel f\parallel_{p}=1\).
Let \(g’=f^{p/q}\). Then, we obtain \(\parallel g’\parallel_{q}=1\) and

\begin{align*} \int_{E}fg’d\nu & =\int_{E}f\cdot f^{q/p}d\nu =\int_{E}f^{(p+q)/q}d\nu
\\ & =\int_{E}f^{p}d\nu =\int_{E}|f|^{p}d\nu =\parallel f\parallel_{p}^{p}=1.\end{align*}

Therefore, we have

\begin{equation}{\label{raeq166}}\tag{67}
\parallel f\parallel_{p}=1=\int_{E}fg’d\nu .
\end{equation}

For the general case, we set \(\hat{f}=f/\parallel f\parallel_{p}\geq 0\). Then, we have \(\parallel \hat{f}\parallel_{p}=1\). By (\ref{raeq166}), we obtain

\begin{align}
\parallel f\parallel_{p} & =\parallel f\parallel_{p}\cdot\parallel \hat{f}\parallel_{p}\nonumber\\ & =\parallel f\parallel_{p}\cdot\int_{E}\hat{f}g’d\nu =\int_{E}fg’d\nu .\label{raeq170}\tag{68}
\end{align}

To ignore the restriction \(f\geq 0\), we apply the above result to the case of \(|f|\). Therefore, from (\ref{raeq170}), we have

\begin{equation}{\label{raeq175}}\tag{69}
\parallel f\parallel_{p}=\parallel |f|\parallel_{p}=\int_{E}|f|g’d\nu =\int_{E}f\hat{g}’d\nu ,
\end{equation}

where \(\hat{g}’=g’\cdot\mbox{sign}(f)\) and

\[\mbox{sign}(x)=\left\{\begin{array}{ll}
1 & \mbox{if \(x>0\)}\\
-1 & \mbox{if \(x<0\)}.
\end{array}\right .\]

Since \(\parallel\hat{g}’\parallel_{q}=\parallel g’\parallel_{q}=1\), we have \(\hat{g}’\in {\cal G}\). Using (\ref{raeq175}), we also have

\[\parallel f\parallel_{p}=\int_{E}f\hat{g}’d\nu\leq\sup_{g\in {\cal G}}\int_{E}fgd\nu .\]

Therefore, we obtain the desired result. Suppose that \(\parallel f\parallel_{p}=+\infty\). We first consider the case of \(f\geq 0\). We define the functions \(f_{k}\) on \(E\) by

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

Then, we have \(0\leq f_{k}\leq k\) for all \(k\). We also have

\[\lim_{k\rightarrow\infty}f_{k}=f,\mbox{ i.e.},\lim_{k\rightarrow\infty}\parallel f_{k}\parallel_{p}=\parallel f\parallel_{p}=+\infty.\]

Since \(0<\parallel f_{k}\parallel_{p}<+\infty\), using (\ref{raeq170}), there exists \(g’_{k}=f_{k}^{p/q}\geq 0\) satisying \(\parallel g’_{k}\parallel_{q}=1\) and

\[\parallel f_{k}\parallel_{p}=\int_{E}f_{k}g’_{k}d\nu .\]

Since \(f\geq f_{k}\geq 0\) and \(g’_{k}\geq 0\), for each \(k\), we obtain

\begin{equation}{\label{raeq171}}\tag{70}
\int_{E}fg’_{k}d\nu\geq\int_{E}f_{k}g’_{k}d\nu =\parallel f_{k}\parallel_{p}
\end{equation}

Since \(\parallel g’_{k}\parallel_{q}=1\), it follows \(g’_{k}\in {\cal G}\) for each \(k\). Therefore, we have

\begin{equation}{\label{raeq173}}\tag{71}
\sup_{g\in {\cal G}}\int_{E}fgd\nu\geq\int_{E}fg’_{k}d\nu
\end{equation}

Using (\ref{raeq171}) and (\ref{raeq173}), we obtain

\begin{align}
\sup_{g\in {\cal G}}\int_{E}fgd\nu & \geq\lim_{k\rightarrow\infty}\int_{E}fg’_{k}d\nu\nonumber\\
& \geq\lim_{k\rightarrow\infty}\int_{E}f_{k}g’_{k}d\nu\nonumber\\
& =\lim_{k\rightarrow\infty}\parallel f_{k}\parallel_{p}=\parallel f\parallel_{p}=+\infty.\label{ma68}\tag{72}
\end{align}

To ignore the restriction \(f\geq 0\), we apply the above result to the case of \(|f|\). Therefore, using (\ref{ma68}), there exists \(g’_{k}\geq 0\) with \(\parallel g’_{k}\parallel_{q}=1\) satisfying

\begin{align*} +\infty & =\parallel f\parallel_{p}\leq\lim_{k\rightarrow\infty}\int_{E}|f|g’_{k}d\nu\\ & =\lim_{k\rightarrow\infty}\int_{E}f\hat{g}’_{k}d\nu ,\end{align*}

where \(\hat{g}’_{k}=g’_{k}\cdot\mbox{sign}(f)\), which implies

\begin{equation}{\label{raeq172}}\tag{73}
\lim_{k\rightarrow\infty}\int_{E}f\hat{g}’_{k}d\nu =+\infty .
\end{equation}

Since \(\parallel\hat{g}’_{k}\parallel_{q}=\parallel g’_{k}\parallel_{q}=1\) for each \(k\), we have \(\hat{g}’_{k}\in {\cal G}\) and

\[\sup_{g\in {\cal G}}\int_{E}fgd\nu\geq\int_{E}f\hat{g}’_{k}d\nu\mbox{ for each }k.\]

Using (\ref{raeq172}), we obtain

\[\sup_{g\in {\cal G}}\int_{E}fgd\nu\geq\lim_{k\rightarrow\infty}\int_{E}f\hat{g}’_{k}d\nu=+\infty =\parallel f\parallel_{p}.\]

Therefore, we obtain the desired result. The cases of \(p=1\) and \(p=+\infty\) are left as exercises. \(\blacksquare\)

Let \(a=\{a_{k}\}_{k=1}^{\infty}\) be a sequence of real or complex numbers. We define

\[\parallel a\parallel_{p}=\left (\sum_{k=1}^{\infty}|a_{k}|^{p}\right )^{1/p}\mbox{ for }0<p<+\infty\]

and

\[\parallel a\parallel_{\infty}=\sup_{k}|a_{k}|.\]

We say that \(a\in l^{p}\) when \(\parallel a\parallel_{p}<+\infty\) for \(0<p<+\infty\), and \(a\in l^{\infty}\) when \(\parallel a\parallel_{\infty}<+\infty\).

Proposition. We have the following properties

(i) Suppose that \(0<p_{1}<p_{2}\leq +\infty\). Then, we have the inclusion \(l^{p_{1}}\subseteq l^{p_{2}}\).

(ii) Suppose that \(a\in l^{p}\) for some \(p<+\infty\). Then, we have

\[\lim_{p\rightarrow\infty}\parallel a\parallel_{p}=\parallel a\parallel_{\infty}.\]

Given \(a=\{a_{k}\}_{k=1}^{\infty}\) and \(b=\{b_{k}\}_{k=1}^{\infty}\), we adopt the notations
\[ab=\{a_{k}b_{k}\}_{k=1}^{\infty}\mbox{ and }a+b=\{a_{k}+b_{k}\}_{k=1}^{\infty}.\]

Theorem. (Holder’s Inequality). Suppose that \(1\leq p\leq +\infty\) and \(1/p+1/q=1\). Given \(a=\{a_{k}\}_{k=1}^{\infty}\) and \(b=\{b_{k}\}_{k=1}^{\infty}\), we have

\[\parallel ab\parallel_{1}\leq\parallel a\parallel_{p}\cdot\parallel b\parallel_{q}.\]

More precisely, we also have

\[\sum_{k=1}^{\infty}|a_{k}b_{k}|\leq\left (\sum_{k=1}^{\infty}|a_{k}|^{p}\right )^{1/p}\cdot
\left (\sum_{k=1}^{\infty}|b_{k}|^{q}\right )^{1/q}\mbox{ for }1\leq p<+\infty\]

and

\[\sum_{k=1}^{\infty}|a_{k}b_{k}|\leq\left (\sup_{k}|a_{k}|\right )\cdot\sum_{k=1}^{\infty}|b_{k}|.\]

Theorem. (Minkowski’s Inequality). Suppose that \(1\leq p\leq +\infty\) and \(1/p+1/q=1\). Given \(a=\{a_{k}\}_{k=1}^{\infty}\) and \(b=\{b_{k}\}_{k=1}^{\infty}\), we have

\[\parallel a+b\parallel_{p}\leq\parallel a\parallel_{p}+\parallel b\parallel_{q}.\]

More precisely, we also have

\[\sum_{k=1}^{\infty}\left (|a_{k}+b_{k}|^{p}\right )^{1/p}\leq\left (\sum_{k=1}^{\infty}|a_{k}|^{p}\right )^{1/p}+
\left (\sum_{k=1}^{\infty}|b_{k}|^{q}\right )^{1/q}\mbox{ for }1\leq p<+\infty\]

and

\[\sup_{k}|a_{k}+b_{k}|\leq\sup_{k}|a_{k}|+\sup_{k}|b_{k}|.\]

Definition. Let the real-valued function \(f\) be finite on an open interval \((a,b)\). We say that \(f\) is convex on \((a,b)\) when, for any \(x,y\in (a,b)\) and \(\lambda\in [0,1]\), we have

\begin{equation}{\label{raeq215}}\tag{111}
f(\lambda x+(1-\lambda )y)\leq\lambda f(x)+(1-\lambda )f(y).
\end{equation}

Equivalently, the function \(f\) is convex on \((a,b)\) if and only if, for any \(x_{1},x_{2}\in (a,b)\) and \(p_{1},p_{2}\in\mathbb{R}_{+}\) with \(p_{1}+p_{2}>0\), we have

\begin{equation}{\label{raeq213}}\tag{112}
f\left (\frac{p_{1}x_{1}+p_{2}x_{2}}{p_{1}+p_{2}}\right )\leq\frac{p_{1}f(x_{1})+p_{2}f(x_{2})}{p_{1}+p_{2}}.
\end{equation}

When we look at the graph of \(f\) in \(\mathbb{R}^{2}\), the condition (\ref{raeq215}) can be formulated geometrically by saying that each point on the chord between \((x,f(x))\) and \((y,f(y))\) is above the graph of \(f\). An important property of the chords of a convex function is given below.

\begin{equation}{\label{ral216}}\tag{113}\mbox{}\end{equation}

Lemma \ref{ral216}. Let \(f\) be a convex function on \((a,b)\). Given any points \(x_{1},y_{1},x_{2},y_{2}\) in \((a,b)\) satisfying \(x_{1}\leq x_{2}<y_{2}\) and \(x_{1}<y_{1}\leq y_{2}\), we have

\[\frac{f(y_{1})-f(x_{1})}{y_{1}-x_{1}}\leq\frac{f(y_{2})-f(x_{2})}{y_{2}-x_{2}}.\]

Let \(f\) be a convex function on \((a,b)\). Given \(x_{0}\in (a,b)\), the line \(y=m(x-x_{0})+f(x_{0})\) passing through \((x_{0},f(x_{0}))\) is called a  supporting line at \(x_{0}\) if it always lies below the graph of \(f\), i.e., if

\begin{equation}{\label{raeq222}}\tag{114}
f(x)\geq m(x-x_{0})+f(x_{0}).
\end{equation}

From Lemma \ref{ral216}, we see that such a line is a supporting line if and only if \(D^{-}f(x_{0})\leq m\leq D^{+}f(x_{0})\).

Proposition. (Jensen’s Inequality). Suppose that the real-valued function \(f\) is convex on \((a,b)\). Given points \(x_{k}\in (a,b)\) for \(k=1,\cdots ,n\) and \(p_{k}\in\mathbb{R}_{+}\) for \(k=1,\cdots ,n\) satisfying \(\sum_{k=1}^{n}p_{k}>0\), we have

\[f\left (\frac{\displaystyle
\sum_{k=1}^{n}p_{k}x_{k}}{\displaystyle \sum_{k=1}^{n}p_{k}}\right )
\leq\frac{\displaystyle \sum_{k=1}^{n}p_{k}f(x_{k})}{\displaystyle \sum_{k=1}^{n}p_{k}}.\]

Proof. The result follows immediately by repeated application of inequality (\ref{raeq213}). \(\blacksquare\)

Proposition. Let \(f\) be a real-valued function defined on \((a,b)\). Suppose that \(f’\) exists and is increasing on \((a,b)\). Then \(f\) is convex on \((a,b)\). In particular, if \(f”\) exists and is nonnegative on \((a,b)\), then \(f\) is convex on \((a,b)\).

\begin{equation}{\label{raeq220}}\tag{115}\mbox{}\end{equation}

Proposition \ref{raeq220}. Suppose that the real-valued function \(f\) is convex on \((a,b)\). Then \(f\) is continuous on \((a,b)\). Moreover, \(f’\) exists except at most in a countable set and is increasing.

Proposition. Suppose that the real-valued function \(f\) is convex on \([a,b]\). Then, it satisfies the Lipschitz condition on ever closed sub-interval of \((a,b)\). In particular, if \(a<x_{1}<x_{2}<b\), then

\[f(x_{2})-f(x_{1})=\int_{x_{1}}^{x_{2}}f’d\nu .\]

\begin{equation}{\label{rat224}}\tag{116}\mbox{}\end{equation}

Theorem \ref{rat224}. (Jensen’s Inequality). Let \(f\) and \(g\) be Lebesgue measurable functions that are finite a.e. on a subset \(A\) of \(\mathbb{R}^{n}\). Suppose that \(fg\) and \(g\) are Lebesgue-integrable on \(A\) satisfying \(g\geq 0\) and \(\int_{A}gd\nu >0\). If the real-valued function \(\phi\) is convex on an interval containing the range of \(f\), then

\[\phi\left (\frac{\displaystyle \int_{A}fgd\nu}{\displaystyle \int_{A}gd\nu}\right )
\leq\frac{\displaystyle \int_{A}\phi (f)gd\nu}{\displaystyle \int_{A}gd\nu}.\]

Proof. We define the number \(\gamma\) by

\begin{equation}{\label{raeq223}}\tag{117}
\gamma =\frac{\displaystyle \int_{A}fgd\nu}{\displaystyle \int_{A}gd\nu}.
\end{equation}

Then, we see that \(\gamma\) is finite. Now, we take an open interval \((a,b)\) satisfying \(\gamma\in (a,b)\), \(a<f({\bf x})<b\) for every \({\bf x}\) with \(f({\bf x})<+\infty\), and \(\phi\) is convex on \((a,b)\). Proposition \ref{raeq220} says that \(\phi\) is continuous on \((a,b)\). It follows that \(\phi (f)\) is Lebesgue-measurable. By the H\”{o}lder’s inequality in Theorem \ref{rat165}, we have

\begin{equation}{\label{raeq221}}\tag{118}
\int_{A}|\phi (f)g|d\nu\leq\left (\mbox{ess sup}_{A}|\phi (f)|\right )\cdot\int_{A}|g|d\nu .
\end{equation}

Since \(\phi\) is continuous on \((a,b)\), it is bounded on \((a,b)\). We also see that \(\phi (f)\) is bounded on \((a,b)\). Since \(g\) is Lebesgue-integrable on \(A\), the inequality (\ref{raeq221}) shows that \(\phi (f)g\) is Lebesgue-integrable on \(A\). Let \(m\) be the slope of a supporting line at \(\gamma\). Given any \(t\in (a,b)\), using (\ref{raeq222}), we have

\[\phi (\gamma )+m\cdot (t-\gamma )\leq\phi (t).\]

Therefore, for almost every \({\bf x}\in A\), we obtain

\[\phi (\gamma )+m\cdot\left (f({\bf x})-\gamma\right )\leq\phi (f({\bf x})).\]

Multiplying \(g({\bf x})\geq 0\) on both side of the above inequality and integrating the result with respect to \({\bf x}\), we obtain

\[\phi (\gamma )\int_{A}gd\nu +m\cdot\left (\int_{A}fgd\nu -\gamma\int_{A}gd\nu\right )\leq\int_{A}\phi (f)gd\nu .\]

Since

\[\int_{A}fgd\nu -\gamma\int_{A}gd\nu =0\]

by (\ref{raeq223}), we also obtain

\[\phi (\gamma )\int_{A}gd\nu\leq\int_{A}\phi (f)gd\nu ,\]

which implies the desired Jensen’s inequality, since \(\int_{A}gd\nu >0\). This completes the proof. \(\blacksquare\)

Corollary. We have the following properties.

(i) Let \(\phi\) be a convex function defined on \((-\infty ,+\infty )\), and let \(f\) be Lebesgue-integrable on \([0,1]\). Then, we have

\[\int_{[0,1]}\phi (f)d\nu\geq\phi\left (\int_{[0,1]}fd\nu\right ).\]

(ii) Suppose that \(f\) is Lebesgue-integrable on \([0,1]\). Then, we have

\[\int_{[0,1]}\exp (f)d\nu\geq\exp\left (\int_{[0,1]}fd\nu\right ).\]

(iii) Suppose that \(f\) is nonnegative and Lebesgue-integrable on \([0,1]\). Then, we have

\[\int_{[0,1]}\log (f)d\nu\leq\log\left (\int_{[0,1]}fd\nu\right ).\]

Proof. Part (i) follows from Theorem \ref{rat224} by taking \(A=[0,1]\) and \(g(x)=1\) on \([0,1]\). Part (ii) follows from part (i), since the exponential function is convex. Since \(\log (x)\) is a concave function, part (iii) follows from part (i) by taking \(\phi (x)=-\log (x)\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{g}}\tag{G}\mbox{}\end{equation}

Differentiation and Integration.

Definition. Let \(E\) be a subset of \(\mathbb{R}\), and let \({\cal I}\) be a collection of intervals. We say that \({\cal I}\) covers \(E\) in the sense of Vitali when, for each \(\epsilon >0\) and any \(x\in E\), there exists an interval \(I\in {\cal I}\) satisfying \(x\in I\) and \(v(I)<\epsilon\), where \(v(I)\) denotes the length of interval \(I\). \(\sharp\)

The following lemma is very useful for investigating the derivatives.

\begin{equation}{\label{ral201}}\tag{74}\mbox{}\end{equation}

Lemma \ref{ral201}. (Vitali Covering Lemma). Let \(E\) be a subset of \(\mathbb{R}\) with finite outer measure, and let \({\cal I}\) be a collection of intervals which covers \(E\) in the sense of Vitali. Then, given any \(\epsilon >0\), there exists a finite disjoint collection \(\{I_{1},\cdots ,I_{n}\}\) of intervals in \({\cal I}\) satisfying

\[\nu^{*}\left (E\setminus\left (\bigcup_{i=1}^{n}I_{i}\right )\right )<\epsilon .\]

We denote by \(Q\) an \(n\)-dimensional cube with edges parallel to the coordinate axes. The diameter of \(Q\) is the number

\[\mbox{dia}(Q)=\sup\left\{\parallel {\bf x}-{\bf y}\parallel :{\bf x},{\bf y}\in Q\right\}.\]

Then, we are going to consider the Vitali covering lemma in \(\mathbb{R}^{n}\).

Definition. A family \({\cal Q}\) of cubes is said to cover a set \(E\subseteq\mathbb{R}^{n}\) in the Vitali sense when, for every \({\bf x}\in E\) and \(\eta <0\), there is a cube in \({\cal Q}\) containing \({\bf x}\) whose diameter is less that \(\eta\). \(\sharp\)

Lemma. (Vitali Covering Lemma). Suppose that \(E\subseteq\mathbb{R}^{n}\) is covered in the Vitali sense by a family \({\cal Q}\) of cubes with \(0<\nu (E)<+\infty\). Then, given any \(\epsilon >0\), there is a countable sequence \(\{Q_{k}\}_{k=1}^{\infty}\) of disjoint cubes in \({\cal Q}\) satisfying

\[\nu\left (E\setminus\bigcup_{k=1}^{\infty}Q_{k}\right )=0\mbox{ and }\sum_{k=1}^{\infty}\nu (Q_{k})<(1+\epsilon )\nu (E).\]

\begin{equation}{\label{rac199}}\tag{75}\mbox{}\end{equation}

Theorem \ref{rac199}. Suppose that \(E\subseteq\mathbb{R}^{n}\) is covered in the Vitali sense by a family \({\cal Q}\) of cubes with \(0<\nu (E <+\infty\). Then, given any \(\epsilon >0\), there is a finite collection \(\{Q_{k}\}_{k=1}^{n}\) of disjoint cubes in \({\cal Q}\) satisfying

\[\nu\left (E\setminus\bigcup_{k=1}^{n}Q_{k}\right )<\epsilon\mbox{ and }\sum_{k=1}^{n}\nu (Q_{k})<(1+\epsilon )\nu (E).\]

By Carath\'{e}odory’s Theorem \ref{rat45}, we have

\[\nu (E)=\nu\left (E\setminus\bigcup_{k=1}^{n}Q_{k}\right )+\nu\left (E\cap\left (\bigcup_{k=1}^{n}Q_{k}\right )\right ).\]

Suppose that \(E\) and \(\{Q_{k}\}_{k=1}^{n}\) are in Theorem \ref{rac199}. Then, we have

\[\nu (E)<\epsilon +\nu\left (E\cap\left (\bigcup_{k=1}^{n}Q_{k}\right )\right )<\epsilon +\sum_{k=1}^{n}\nu (Q_{k}).\]

We shall consider the sense in which the differentiation is the inverse operation of integration. In other words, we are going to develop the fundamental theorem of calculus based on the Lebesgue integral.

Let \(f:[a,b]\rightarrow\mathbb{R}\) be a real-valued function defined on a bounded closed interval \([a,b]\). We can also consider the limit inferior and limit superior of function \(f\). Now, for \(y\in [a,b]\), we define

\[\limsup_{x\rightarrow y+}f(x)=\inf_{\delta >0}\sup_{0<x-y<\delta}f(x)
\mbox{ and }\limsup_{x\rightarrow y-}f(x)=\inf_{\delta >0}\sup_{0<y-x<\delta}f(x)\]

\[\liminf_{x\rightarrow y+}f(x)=\sup_{\delta >0}\inf_{0<x-y<\delta}f(x)
\mbox{ and }\liminf_{x\rightarrow y-}f(x)=\sup_{\delta >0}\inf_{0<y-x<\delta}f(x)\]

\[\limsup_{x\rightarrow y}f(x)=\inf_{\delta >0}\sup_{0<|x-y|<\delta}f(x)
\mbox{ and }\liminf_{x\rightarrow y}f(x)=\sup_{\delta >0}\inf_{0<|x-y|<\delta}f(x).\]

Let
\[p(\delta)=\sup_{0<|x-y|<\delta}f(x)\mbox{ and }q(\delta)=\inf_{0<|x-y|<\delta}f(x).\]

Then, we see that \(p\) is increasing and \(q\) is decreasing. It follows

\[\limsup_{x\rightarrow y}f(x)=\inf_{\delta >0}\sup_{0<|x-y|<\delta}f(x)=
\inf_{\delta >0}p(\delta)=\lim_{\delta\rightarrow 0+}p(\delta)
=\lim_{\delta\rightarrow 0+}\sup_{0<|x-y|<\delta}f(x)\]

and
\[\liminf_{x\rightarrow y}f(x)=\sup_{\delta >0}\inf_{0<|x-y|<\delta}f(x)=\sup_{\delta >0}q(\delta)
=\lim_{\delta\rightarrow 0+}q(\delta)=\lim_{\delta\rightarrow 0+}\inf_{0<|x-y|<\delta}f(x).\]

It is easy to see
\[\sup_{0<x-y<\delta}f(x)\leq\sup_{0<|x-y|<\delta}f(x)\mbox{ and }
\sup_{0<y-x<\delta}f(x)\leq\sup_{0<|x-y|<\delta}f(x),\]

which also says
\[\limsup_{x\rightarrow y+}f(x)\leq\limsup_{x\rightarrow y}f(x)\mbox{ and }
\limsup_{x\rightarrow y-}f(x)\leq\limsup_{x\rightarrow y}f(x).\]

It is also easy to see
\[\inf_{0<x-y<\delta}f(x)\geq\inf_{0<|x-y|<\delta}f(x)\mbox{ and }
\inf_{0<y-x<\delta}f(x)\geq\inf_{0<|x-y|<\delta}f(x),\]

which says
\[\liminf_{x\rightarrow y+}f(x)\geq\liminf_{x\rightarrow y}f(x)\mbox{ and }
\liminf_{x\rightarrow y-}f(x)\geq\liminf_{x\rightarrow y}f(x).\]

Proposition. Let \(f:[a,b]\rightarrow\mathbb{R}\) be a real-valued function defined on a closed interval \([a,b]\). Then, we have the following properties..

(i) \(\limsup_{x\rightarrow y}f(x)\leq A\) if and only if, given \(\epsilon >0\), there is a \(\delta >0\) such that, for all \(x\) with \(0<|x-y|<\delta\), we have
$f(x)\leq A+\epsilon$.

(ii) \(\limsup_{x\rightarrow y}f(x)\geq A\) if and only if, given \(\epsilon >0\) and \(\delta >0\) there is an \(x\) satisfying \(0<|x-y|<\delta\) and \(f(x)\geq A-\epsilon\).

(iii) \(\liminf_{x\rightarrow y}\leq\limsup_{x\rightarrow y}f(x)\) with equality if and only if the limit \(\lim_{x\rightarrow y}f(x)\) exists.

(iv) If \(\limsup_{x\rightarrow y}f(x)=A\) and \(\{x_{n}\}\) is a sequence satisfying \(x_{n}\neq y\) and \(y=\lim_{n} x_{n}\), then \(\limsup_{n}f(x_{n})\leq A\).

(v) If \(\limsup_{x\rightarrow y}f(x)=A\), then there is a sequence \(\{x_{n}\}\) satisfying \(x_{n}\neq y\) with \(y=\lim_{n}x_{n}\) and \(A=\lim_{n}f(x_{n})\).

(vi) For a real number \(l\), we have \(l=\lim_{x\rightarrow y}f(x)\) if and only if \(l=\lim_{n}f(x_{n})\) for every sequence \(\{x_{n}\}\) with \(x_{n}\neq y\)
and \(y=\lim_{n}x_{n}\). \(\sjarp\)

We say that \(f\) is lower semi-continuous at \(y\) when, given \(\epsilon >0\), there exists \(\delta >0\) satisfying \(f(y)<f(x)+\epsilon\) for all \(x\) with \(|x-y|<\delta\). The function \(f\) is upper semi-continuous at \(y\)  when the function \(-f\) is lower semi-continuous at \(y\). In other words, given \(\epsilon >0\), there exists \(\delta >0\) satisfying \(f(y)+\epsilon<f(x)\) for all \(x\) with \(|x-y|<\delta\).

Proposition. Let \(f:[a,b]\rightarrow\mathbb{R}\) be a real-valued function defined on a closed interval \([a,b]\). Then, we have the following properties.

(i) The function \(f\) is lower semi-continuous at \(y\) if and only if \(f(y)\leq\liminf_{x\rightarrow y}f(x)\) for \(f(y)\neq -\infty\).

(ii) The function \(f\) is upper semi-continuous at \(y\) if and only if \(f(y)\geq\limsup_{x\rightarrow y}f(x)\) for \(f(y)\neq\infty\).

(iii) The function \(f\) is continuous at \(y\) if and only if it is both lower and upper semi-continuous at \(y\).

(iv) The function \(f\) is lower semi-continuous on the open interval \((a,b)\) if and only if the set \(\{x:f(x)>\lambda\}\) is open for each \(\lambda\in\mathbb{R}\).

(v) The function \(f\) defined on the closed interval \([a,b]\) is lower semi-continuous if and only if there is a monotone increasing sequence \(\{\phi_{n}\}\) of lower semi-continuous step functions on \([a,b]\) satisfying \(f(x)=\lim_{n}\phi_{n}(x)\) for each \(x\in [a,b]\). \(\sharp\)

Definition. Let \(f\) be a real-valued function defined on a compact interval \([a,b]\). For \(x\in (a,b)\), we define four Dini derivatives as follows.

\begin{align*}
D^{+}f(x) & =\limsup_{h\rightarrow 0+}\frac{f(x+h)-f(x)}{h}\\
D^{-}f(x) & =\limsup_{h\rightarrow 0+}\frac{f(x)-f(x-h)}{h}=\limsup_{h\rightarrow 0-}\frac{f(x+h)-f(x)}{h}\\
D_{+}f(x) & =\liminf_{h\rightarrow 0+}\frac{f(x+h)-f(x)}{h}\\
D_{-}f(x) & =\liminf_{h\rightarrow 0+}\frac{f(x)-f(x-h)}{h}=\liminf_{h\rightarrow 0-}\frac{f(x+h)-f(x)}{h}.
\end{align*}

We say that \(f\) is  differentiable at \(x\) when

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

The common value is denoted by \(f'(x)\) and is called the derivative of \(f\) at \(x\). \(\sharp\)

It is clear to see

\[D^{+}f(x)\geq D_{+}f(x)\mbox{ and }D^{-}f(x)\geq D_{-}f(x).\]

\begin{equation}{\label{rat180}}\tag{76}\mbox{}\end{equation}

Theorem \ref{rat180}. Let \(f\) be an increasing real-valued function defined on the compact interval \([a,b]\). Then \(f'(x)\) exists and is nonnegative a.e. on \((a,b)\), and the derivative \(f’\) is Lebesgue measurable satisfying

\[0\leq\int_{[a,b]}f'(x)d\nu\leq f(b)-f(a).\]

Theorem. Let the real-valued function \(f\) defined on \(\mathbb{R}\) be increasing and finite on an open interval \((a,b)\). Then \(f'(x)\) exists and is nonnegative a.e. on \((a,b)\), and the derivative \(f’\) is Lebesgue-measurable satisfying

\[0\leq\int_{[a,b]}f'(x)d\nu\leq f(b-)-f(a+).\]

\begin{equation}{\label{h}}\tag{H}\mbox{}\end{equation}

Functions of Bounded Variation.

Let \(f\) be a real-valued function defined on the compact interval \([a,b]\). We denote by \(\mathfrak{P}([a,b])\) the family of all partitions of \([a,b]\). Given any partition \({\cal P}=\{a=x_{0},x_{1},x_{2},\cdots ,x_{n}=b\}\) of \([a,b]\), we define

\begin{align*}
& \parallel{\cal P}\parallel=\max_{i}(x_{i}-x_{i-1})\\
& p(f,{\cal P})=\sum_{k=1}^{n}\left (f(x_{k})-f(x_{k-1})\right )^{+}\mbox{ and }P_{a}^{b}(f)=\sup_{{\cal P}\in\mathfrak{P}([a,b])} p(f,{\cal P})\\
& n(f,{\cal P})=\sum_{k=1}^{n}\left (f(x_{k})-f(x_{k-1})\right )^{-}\mbox{ and }N_{a}^{b}(f)=\sup_{{\cal P}\in\mathfrak{P}([a,b])} n(f,{\cal P})\\
& t(f,{\cal P})=p(f,{\cal P})+n(f,{\cal P})=\sum_{k=1}^{n}\left |f(x_{k})-f(x_{k-1})\right |\mbox{ and }T_{a}^{b}(f)=\sup_{{\cal P}\in\mathfrak{P}([a,b])} t(f,{\cal P}).
\end{align*}

It is clear to see

\[f(b)-f(a)=p(f,{\cal P})-n(f,{\cal P}).\]

and

\[P_{a}^{b}(f)\leq T_{a}^{b}(f)\leq P_{a}^{b}(f)+N_{a}^{b}(f).\]

We say that the function \(f\) is of bounded variation when \(T_{a}^{b}(f)<+\infty\).

Example. We say that the real-valued function defined on \([a,b]\) satisfies the Lipschitz condition when there exists a constant \(k\) satisfying

\[\left |f(x)-f(y)\right |\leq k\cdot |x-y|\mbox{ for all }x,y\in [a,b].\]

We can show that if a function defined on \([a,b]\) satisfies the Lipschitz condition, then it is of bounded variation satisfying \(T_{a}^{b}(f)\leq k\cdot (b-a)\). \(\sharp\)

\begin{equation}{\label{rap211}}\tag{77}\mbox{}\end{equation}

Proposition \ref{rap211}. Suppose that \(f\) is of bounded variation on \([a,b]\). Then, we have

\[T_{a}^{b}(f)=P_{a}^{b}(f)+N_{a}^{b}(f)\mbox{ and }f(b)-f(a)=P_{a}^{b}(f)-N_{a}^{b}(f).\]

Proposition. Suppose that the real-valued function \(f\) is of bounded variation on \([a,b]\). Then, it is bounded on \([a,b]\).

Proposition. Suppose that the real-valued functions \(f\) and \(g\) be of bounded variation on \([a,b]\). Then, we have the following properties.

(i) Given any constant \(c\), the real-valued function \(cf\) is of bounded variation on \([a,b]\).

(ii) The real-valued functions \(f+g\) and \(fg\) are of bounded variation on \([a,b]\).

(iii) If there exists an \(\epsilon >0\) satisfying \(|g(x)|>\epsilon\) for all \(x\in [a,b]\), then \(f/g\) is of bounded variation on \([a,b]\).

Proposition. Let \(f\) be a real-valued function defined on \([a,b]\). Then, we have following properties.

(i) Suppose that \([c,d]\) is a subinterval of \([a,b]\). Then, we have \(T_{c}^{d}(f)\leq T_{a}^{b}(f)\).

(ii) Suppose that \(c\in (a,b)\). Then, we have

\[T_{a}^{b}(f)=T_{a}^{c}(f)+T_{c}^{b}(f).\]

\begin{equation}{\label{rat181}}\tag{78}\mbox{}\end{equation}

Theorem \ref{rat181}. A function \(f\) is of bounded variation on \([a,b]\) if and only if \(f\) is the difference of two increasing real-valued functions on \([a,b]\).

Proposition. Every real-valued function of bounded variation on \([a,b]\) has at most a countable number of discontinuities. In other words, it is continuous a.e. on \([a,b]\).

Proposition. Suppose that the real-valued function \(f\) is continuous on \([a,b]\). Then, we have

\[T_{a}^{b}(f)=\lim_{\parallel{\cal P}\parallel\rightarrow 0}t(f,{\cal P}).\]

In other words, given any \(k\) satisfying \(k<T_{a}^{b}(f)\), there exists \(\delta >0\) such that \(\parallel{\cal P}\parallel<\delta\) implies \(t(f,{\cal P})>k\).

\begin{equation}{\label{rap186}}\tag{79}\mbox{}\end{equation}

Proposition \ref{rap186}. Suppose that \(f\) is of bounded variation on \([a,b]\). Then \(f'(x)\) exists a.e. on \((a,b)\).

Proof. The result follows immediately from Theorems \ref{rat180} and \ref{rat181}. \(\blacksquare\)

\begin{equation}{\label{rap210}}\tag{80}\mbox{}\end{equation}

Proposition \ref{rap210}. Suppose that \(f\) is of bounded variation on \([a,b]\). Define \(T(x)=T_{a}^{x}(f)\) for \(a\leq x\leq b\). Then, we have \(T'(x)=|f'(x)|\) a.e. on \((a,b)\).

Definition. Suppose that the real-valued function \(f\) is finite on the compact interval \([a,b]\). Then \(f\) is said to be absolutely continuous on \([a,b]\) when, given any \(\epsilon >0\), there exists \(\delta >0\) such that, for every finite collection \(\{[a_{i},b_{i}]\}_{i=1}^{n}\) of non-overlapping sub-intervals of \([a,b]\),

\[\sum_{i=1}^{n}(b_{i}-a_{i})<\delta\mbox{ implies }\sum_{i=1}^{n}\left |f(b_{i})-f(a_{i})\right |<\epsilon .\]

It is obvious that an absolutely continuous function is also continuous.

\begin{equation}{\label{rap200}}\tag{81}\mbox{}\end{equation}

Proposition \ref{rap200}. Suppose that the real-valued function \(f\) is absolutely continuous on \([a,b]\). Then, it is of bounded variation on \([a,b]\).

Proof. By definition, we can take \(\delta >0\) satisfying

\[\sum_{i}|f(b_{i})-f(a_{i})|\leq 1\]

for any collection of non-overlapping sub-intervals satisfying \(\sum_{i}(b_{i}-a_{i})<\delta\). Then, the variation of \(f\) over any sub-interval of \([a,b]\) with length less than \(\delta\) is at most \(1\). We consider the partition \({\cal P}_{0}=\{a=x_{0},x_{1},\cdots ,x_{K}=b\}\) that divides \([a,b]\) into \(K\) sub -intervals such that each length is less than \(\delta\). Then, we have \(t(f,{\cal P}_{0})\leq K\). Therefore, for any partition \({\cal P}\) of \([a,b]\), which is finer than \({\cal P}_{0}\), we also have \(t(f,{\cal P})\leq K\), since the partition \({\cal P}\) just further divides the sub-interval \([x_{k-1},x_{k}]\) satisfying

\[\sum_{i}(b_{i}^{(k)}-a_{i}^{(k)})<\delta\mbox{ and }\bigcup_{i}[a_{i}^{(k)},b_{i}^{(k)}]=[x_{k-1},x_{k}].\]

This shows \(T_{a}^{b}(f)\leq K\), and the proof is complete. \(\blacksquare\)

\begin{equation}{\label{rap204}}\tag{82}\mbox{}\end{equation}

Proposition \ref{rap204}. Suppose that the real-valued function \(f\) is absolutely continuous on \([a,b]\). Then \(f'(x)\) exists a.e. on \((a,b)\).

Proof. The result follows immediately from Propositions \ref{rap186} and \ref{rap200}. \(\blacksquare\)

\begin{equation}{\label{rap208}}\tag{83}\mbox{}\end{equation}

Proposition \ref{rap208}. Suppose that the real-valued function \(f\) is absolutely continuous on \([a,b]\) and \(f'(x)=0\) a.e. on \((a,b)\). Then, it is a constant function on \([a,b]\).

Proof. We need to show that \(f(a)=f(c)\) for any \(c\in [a,b]\). Let \(E\) be a subset of \([a,c]\) with \(\nu (E)=c-a\) and \(f'(x)=0\) for each \(x\in E\). Let \(\eta\) and \(\epsilon\) be any positive numbers. For any \(x\in E\), since

\[\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x)=0,\]

there is an sufficiently small interval \([x,x+h]\) contained in$[a,c]$ satisfying

\begin{equation}{\label{raeq202}}\tag{84}
|f(x+h)-f(x)|\leq\eta h.
\end{equation}

Using the Vitali covering Lemma \ref{ral201}, we can find a finite collection \(\{[x_{i},y_{i}]\}_{i=1}^{n}\) of non-overlapping intervals with \(y_{i}=x_{i}+h_{i}\) for some \(h_{i}>0\), which cover \(E\) satisfying

\begin{equation}{\label{ma69}}\tag{85}
\nu^{*}\left (E\setminus\left (\bigcup_{i=1}^{n}[x_{i},y_{i}]\right )\right )<\delta ,
\end{equation}

where \(\delta\) is taken to be the corresponding positive number to \(\epsilon\) in the definition of absolute continuity of \(f\). We rename \(x_{k}\) satisfying \(x_{k}\leq x_{k+1}\). Then

\[y_{0}\equiv a\leq x_{1}<y_{1}\leq x_{2}<\cdots\leq y_{n}\leq c\equiv x_{n+1}\]

Using (\ref{ma69}), we have

\[\sum_{i=0}^{n}|x_{i+1}-y_{i}|<\delta .\]

Using the absolute continuity of \(f\), we also have

\begin{equation}{\label{raeq203}}\tag{86}
\sum_{i=0}^{n}\left |f(x_{i+1})-f(y_{i})\right |<\epsilon .
\end{equation}

Since \(y_{i}=x_{i}+h_{i}\), using (\ref{raeq202}), we obtain

\begin{equation}{\label{raeq204}}\tag{87}
\sum_{i=1}^{n}\left |f(y_{i})-f(x_{i})\right |\leq\eta\sum_{i=1}^{n}(y_{i}-x_{i})<\eta (c-a).
\end{equation}

Therefore, from (\ref{raeq203}) and (\ref{raeq204}), we also obtain

\begin{align*}
\left |f(c)-f(a)\right | & =\left |\sum_{i=0}^{n}\left [f(x_{i+1})-f(y_{i})\right ]+\sum_{i=1}^{n}\left [f(y_{i})-f(x_{i})\right ]\right |\\
& \leq\sum_{i=0}^{n}\left |f(x_{i+1})-f(y_{i})\right |+\sum_{i=1}^{n}\left |f(y_{i})-f(x_{i})\right |\\ & \leq\epsilon +\eta (c-a).
\end{align*}

Since \(\eta\) and \(\epsilon\) can be any positive numbers, we conclude \(f(c)-f(a)=0\). This completes the proof. \(\blacksquare\)

\begin{equation}{\label{i}}\tag{I}\mbox{}\end{equation}

The Real-Valued Indefinite Integrals.

Suppose that \(f\) is Riemann-integrable on a compact interval \([a,b]\). Its indefinite integral is defined by

\[F(x)=\int_{a}^{x}f(y)dy\mbox{ for }x\in [a,b].\]

When \(f\) is continuous, the fundamental theorem of calculus asserts \(F’=f\). We shall study the analogue of this result for Lebesgue-integrable function \(f\).

\begin{equation}{\label{rap185}}\tag{88}\mbox{}\end{equation}

Proposition \ref{rap185}. Suppose that \(f\) is Lebesgue-integrable on the compact interval \([a,b]\). Then, the indefinite integral

\[F(x)=\int_{[a,x]}f(t)d\nu (t)\mbox{ for }x\in [a,b]\]

is a uniformly continuous function of bounded variation on \([a,b]\).

Proof. Since

\begin{align*} F(x) & =\int_{[a,x]}f(t)d\nu (t)\\ & =\int_{[a,x]}f^{+}(t)d\nu (t)-\int_{[a,x]}f^{-}(t)d\nu (t)\\ & \equiv F_{1}(x)-F_{2}(x),\end{align*}

where

\[F_{1}(x)=\int_{[a,x]}f^{+}(t)d\nu (t)\mbox{ and }F_{2}(x)=\int_{[a,x]}f^{-}(t)d\nu (t),\]

Proposition \ref{rap178} says that \(F_{1}\) and \(F_{2}\) are uniformly continuous on \([a,b]\), which also says that \(F\) is uniformly continuous on \([a,b]\). To show that \(F\) is of bounded variation, let \(a=x_{0}<x_{1}<\cdots <x_{n}=b\) be a partition of \([a,b]\). Then, we have

\begin{align*}
\sum_{k=1}^{n}\left |F(x_{k})-F(x_{k-1})\right |& =\sum_{k=1}^{n}\left |\int_{[x_{k-1},x_{k}]}f(t)d\nu (t)\right |\\
& \leq\sum_{k=1}^{n}\int_{[x_{k-1},x_{k}]}\left |f(t)\right |d\nu (t)\\ & =\int_{a}^{b}\left |f(t)\right |d\nu (t),
\end{align*}

which implies

\[T_{a}^{b}(F)\leq\int_{a}^{b}\left |f(t)\right |d\nu (t)<+\infty .\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap183}}\tag{89}\mbox{}\end{equation}

Lemma \ref{rap183}. We have the following properties.

(i) Given any subset \(E\) of \(\bar{\mathbb{R}}^{n}\) and any \(\epsilon >0\), there exists a closed subset \(F\) of \(\bar{\mathbb{R}}^{n}\) satisfying

\[F\subseteq E\mbox{ and }\nu^{*} (E)\leq\nu^{*}(F)+\epsilon.\]

Let \(\mathfrak{F}\) be a collection of all closed sets in \(\bar{\mathbb{R}}^{n}\) contained \(E\), i.e,

\[\mathfrak{F}=\{F:F\mbox{ is closed in \(\bar{\mathbb{R}}^{n}\) and }F\subseteq E\}.\]

Then, we have

\[\nu^{*} (E)=\sup_{F\in\mathfrak{F}}\nu^{*} (F).\]

(ii) Given any subset \(E\) of \(\bar{\mathbb{R}}^{n}\), there exists a set \(F=\bigcup_{k=1}^{\infty}F_{k}\), where each \(F_{k}\) is closed in \(\bar{\mathbb{R}}^{n}\), satisfying \(F\subseteq E\) and \(\nu^{*} (E)=\nu^{*} (F)\). \(\sharp\)

The above Lemma \ref{rap183} can refer to the page Lebesgue Measurable Sets and Functions.

\begin{equation}{\label{rap189}}\tag{90}\mbox{}\end{equation}

Proposition \ref{rap189}. Suppose that \(f\) is Lebesgue-integrable on the compact interval \([a,b]\) satisfying

\[F(x)=\int_{[a,x]}f(t)d\nu (t)=0\mbox{ for }x\in [a,b].\]

Then \(f(t)=0\) a.e. on \([a,b]\).

Proof. Suppose that \(f(x)>0\) on a set \(E\subseteq [a,b]\) with \(\nu (E)>0\). We are going to lead to a contradiction. Lemma \ref{rap183} says that there exists a closed set \(F\subseteq E\) satisfying \(\nu (F)>0\), since we can take \(\epsilon >0\) satisfying \(\nu^{*} (E)>\epsilon\). Since \(f(x)>0\) on \(F\), it follows

\[\int_{F}f(t)d\nu (t)\neq 0.\]

Let

\[O=(a,b)\setminus F=(a,b)\cap F^{c}.\]

Then \(O\) is an open set, which says that the open set \(O\) is a disjoint union of a countable collection of open intervals \(\{(a_{k},b_{k})\}_{k=1}^{\infty}\). Using Proposition \ref{rap135}, we have

\begin{equation}{\label{raeq184}}\tag{91}
\int_{O}f(t)d\nu (t)=\sum_{k=1}^{\infty}\int_{(a_{k},b_{k})}f(t)d\nu (t).
\end{equation}

Suppose that \(F(b)\neq 0\). Then, we immediately have a contradiction for \(F(x)=0\) for all \(x\in [a,b]\). Now, we assume

\begin{align*} 0 & =F(b)=\int_{[a,b]}f(t)d\nu (t)=\int_{(a,b)}f(t)d\nu (t)\\ & =\int_{F}f(t)d\nu (t)+\int_{O}f(t)d\nu (t),\end{align*}

which implies

\[\int_{O}f(t)d\nu (t)=-\int_{F}f(t)d\nu (t)\neq 0.\]

From (\ref{raeq184}), we have

\[\int_{(a_{k},b_{k})}f(t)d\nu (t)\neq 0\mbox{ for some }k,\]

which also says

\[\int_{[a,b_{k}]}f(t)d\nu (t)=\int_{(a,b_{k})}f(t)d\nu (t)\neq 0.\]

Therefore, we get a contradiction. Similarly, if \(f\) is negative on a set of positive measure, then we can also lead to a contradiction. This completes the proof. \(\blacksquare\)

Let \(f\) be a Lebesgue-integrable function on the compact interval \([a,b]\). Then, we can define

\[F(x)=p+\int_{[a,x]}f(t)d\nu (t)\]

for some constant \(p\). By taking \(x=a\), we obtain \(p=F(a)\). Therefore, we can always define the function

\[F(x)=F(a)+\int_{[a,x]}f(t)d\nu (t).\]

\begin{equation}{\label{rap190}}\tag{92}\mbox{}\end{equation}

Proposition \ref{rap190}. Suppose that \(f\) is bounded and Lebesgue-integrable on the compact interval \([a,b]\). Let

\[F(x)=F(a)+\int_{[a,x]}f(t)d\nu (t)\mbox{ for }x\in [a,b].\]

Then \(F'(x)=f(x)\) a.e. on \((a,b)\).

Proof. Proposition \ref{rap185} says that \(F\) is a continuous function of bounded variation on \([a,b]\). Proposition \ref{rap186} also says that \(F'(x)\) exists a.e. on the open interval \((a,b)\). Assume that \(f\) is bounded by \(K\), i.e., \(|f(x)|\leq K\) for all \(x\in [a,b]\). We define

\[f_{n}(x)=\frac{F(x+h)-F(x)}{h}\mbox{ with }h=\frac{1}{n}.\]

Then, we have

\begin{align*} f_{n}(x) & =\frac{1}{h}\left [\int_{[a,x+h]}f(t)d\nu (t)-\int_{[a,x]}f(t)d\nu (t)\right ]\\ & =\frac{1}{h}\int_{[x,x+h]}f(t)d\nu (t),\end{align*}

which implies

\begin{align*} |f_{n}(x)| & \leq\frac{1}{h}\int_{[x,x+h]}|f(t)|d\nu (t)\\ & \leq\frac{1}{h}\int_{[x,x+h]}Kd\nu (t)=K.\end{align*}

Since

\begin{align*} \lim_{n\rightarrow\infty}f_{n}(x) & =\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}\\ & =F'(x)\mbox{ a.e. on }(a,b),\end{align*}

using the bounded convergence Theorem \ref{rat96}, we have

\begin{align}
\int_{[a,c]}F'(x)d\nu (x) & =\lim_{n\rightarrow\infty}\int_{[a,c]}f_{n}(x)d\nu (x)
\nonumber\\ & =\lim_{h\rightarrow 0}\frac{1}{h}\int_{[a,c]}\left [F(x+h)-F(x)\right ]d\nu (x).\label{raeq187}\tag{93}\end{align}

Since \(F\) is continuous on \([a,b]\), it says that \(F\) is Riemann integrable on \([a,b]\). Proposition \ref{rap7} says that the Lebesgue integral of \(F\) on any compact interval \(I\subseteq [a,b]\) is equal to the Riemann integral of \(F\) on \(I\). Now, for any \(c\in [a,b]\), we have

\begin{align*}
& \int_{[a,c]}\left [F(x+h)-F(x)\right ]d\nu (x)=\int_{a}^{c}\left [F(x+h)-F(x)\right ]dx\\
& \quad\quad\quad\quad\mbox{ (The Riemann and Lebesgue integrals are identical)}\\
& \quad =\int_{a}^{c}F(x+h)dx-\int_{a}^{c}F(x)dx\mbox{ (for Riemann integral)}\\
& \quad =\int_{a+h}^{c+h}F(x)dx-\int_{a}^{c}F(x)dx\mbox{ (change of variables for Riemann integral)}\\
& \quad =\left (\int_{a+h}^{c}F(x)dx+\int_{c}^{c+h}F(x)dx\right )-\left (\int_{a}^{a+h}F(x)dx+\int_{a+h}^{c}F(x)dx\right )\\
& \quad =\int_{c}^{c+h}F(x)dx-\int_{a}^{a+h}F(x)dx,
\end{align*}

which also implies

\begin{align}
& \lim_{h\rightarrow 0}\frac{1}{h}\int_{[a,c]}\left [F(x+h)-F(x)\right ]d\nu (x)=\lim_{h\rightarrow 0}\frac{1}{h}\left [\int_{c}^{c+h}F(x)dx\right ]
-\lim_{h\rightarrow 0}\frac{1}{h}\left [\int_{a}^{a+h}F(x)dx\right ]\nonumber\\
& \quad=F(c)-F(a)=\int_{[a,c]}f(x)d\nu (x).\label{raeq188}\tag{94}
\end{align}

Using (\ref{raeq187}) and (\ref{raeq188}), we obtain

\[\int_{[a,c]}\left [F'(t)-f(t)\right ]d\nu (t)=0\]

for any \(c\in [a,b]\). Proposition~\ref{rap189} says that \(F'(x)=f(x)\) a.e. on \((a,b)\). This completes the proof. \(\blacksquare\)

The more general result without considering boundedness is given below.

\begin{equation}{\label{rat206}}\tag{95}\mbox{}\end{equation}

Theorem \ref{rat206}. Suppose that \(f\) is Lebesgue-integrable on the compact interval \([a,b]\). Let

\[F(x)=F(a)+\int_{[a,x]}f(t)d\nu (t)\mbox{ for }x\in [a,b].\]

Then \(F'(x)=f(x)\) a.e. on \((a,b)\).

Proof. We first assume \(f\geq 0\). Define

\[f_{n}(x)=\left\{\begin{array}{ll}
f(x), & \mbox{if \(f(x)\leq n\)}\\
n, & \mbox{if \(f(x)>n\)}.
\end{array}\right .\mbox{ and }F_{n}(x)=\int_{[a,x]}f_{n}(t)d\nu (t).\]

Then, each \(f_{n}\) is bounded and Lebesgue-measurable on \([a,b]\) and

\[\lim_{n\rightarrow\infty}f_{n}(x)=f(x)\mbox{ for all }x\in [a,b].\]

Using Proposition \ref{rap190}, we have \(F’_{n}(x)=f_{n}(x)\) a.e. on \((a,b)\). Since \(f-f_{n}\geq 0\) for each \(n\), the function defined by

\[G_{n}(x)=\int_{[a,x]}\left [f(t)-f_{n}(t)\right ]d\nu (t)\]

is an increasing function of \(x\). Theorem \ref{rat180} says \(G’_{n}(x)\geq 0\) a.e. on \((a,b)\). Since

\[F(x)=F(a)+G_{n}(x)+F_{n}(x),\]

it follows

\[F'(x)=G’_{n}(x)+F’_{n}(x)=G’_{n}(x)+f_{n}(x)\geq f_{n}(x)\mbox{ a.e. on }(a,b),\]

which implies \(F'(x)\geq f(x)\) a.e. on \((a,b)\) by taking limit. Therefore, we obtain

\[\int_{[a,b]}F'(t)d\nu (t)\geq\int_{[a,b]}f(t)d\nu (t)=F(b)-F(a).\]

Since \(F\) is increasing, using Theorem \ref{rat180}, we also have

\[\int_{[a,b]}F'(t)d\nu (t)\leq F(b)-F(a).\]

Therefore, we obtain

\[\int_{[a,b]}F'(t)d\nu (t)=F(b)-F(a)=\int_{[a,b]}f(t)d\nu (t),\]

which implies

\[\int_{[a,b]}\left [F'(t)-f(t)\right ]d\nu (t)=0.\]

Since \(F'(x)-f(x)\geq 0\) a.e. on \((a,b)\), we conclude \(F'(x)-f(x)=0\) a.e. on \((a,b)\), i.e., \(F'(x)=f(x)\) a.e. on \((a,b)\). For the general case, we have \(f(x)=f^{+}(x)-f^{-}(x)\). Now, we write

\[G(x)=F(a)+\int_{[a,x]}f^{+}(t)d\nu (t)\mbox{ and }H(x)=\int_{[a,x]}f^{-}(t)d\nu (t).\]

Then, we have \(G'(x)=f^{+}(x)\) and \(H'(x)=f^{-}(x)\) a.e. on \((a,b)\). Since \(F(x)=G(x)-H(x)\), we obtain

\[F'(x)=G'(x)-H'(x)=f^{+}(x)-f^{-}(x)=f(x)\mbox{ a.e on }(a,b).\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{rap155}}\tag{96}\mbox{}\end{equation}

Proposition \ref{rap155}. Suppose that \(f\) is Lebesgue-integrable on the compact interval \([a,b]\). Then, the indefinite integral

\[F(x)=\int_{[a,x]}fd\nu\]

is absolutely continuous on \([a,b]\).

Proof. Given any non-overlapping subintervals \(\{[a_{i},b_{i}]\}_{i=1}^{n}\), we have

\begin{align*} \sum_{i=1}^{n}\left |F(b_{i})-F(a_{i})\right | & =\sum_{i=1}^{n}\left |\int_{[a,b_{i}]}fd\nu
-\int_{[a,a_{i}]}fd\nu\right |\\ & =\sum_{i=1}^{n}\left |\int_{[a_{i},b_{i}]}fd\nu\right |
\\ & \leq\int_{\bigcup_{i}[a_{i},b_{i}]}|f|d\nu .\end{align*}

Given any \(\epsilon >0\), for

\[\nu\left (\bigcup_{i=1}^{n}[a_{i},b_{i}]\right )=\sum_{i=1}^{n}(b_{i}-a_{i})<\delta ,\]

Proposition \ref{rap178} says

\[\sum_{i=1}^{n}\left |F(b_{i})-F(a_{i})\right |\leq\int_{\bigcup_{i}[a_{i},b_{i}]}|f|d\nu<\epsilon.\]

This completes the proof. \(\blacksquare\)

The converse is shown below.

\begin{equation}{\label{rat209}}\tag{97}\mbox{}\end{equation}

Theorem \ref{rat209}. Suppose that the real-valued function \(f\) is absolutely continuous on the compact interval \([a,b]\). Then \(f’\) exists a.e. on the open interval \((a,b)\), and \(f’\) is Lebesgue-integrable on \([a,b]\) satisfying

\[f(x)-f(a)=\int_{[a,x]}f’d\nu\]

for \(x\in [a,b]\).

Proof. Propositions \ref{rap200} and \ref{rap204} says that \(f'(x)\) exists on \((a,b)\) and \(f\) is of bounded variation on \([a,b]\). In this case, using Proposition \ref{rat181}, we can write \(f(x)=f_{1}(x)-f_{2}(x)\) such that \(f_{1}\) and \(f_{2}\) are increasing on \([a,b]\). We also have

\[\left |f'(x)\right |=\left |f’_{1}(x)-f’_{2}(x)\right |\leq |f’_{1}(x)|+|f’_{2}(x)|=f’_{1}(x)+f’_{2}(x).\]

Using Theorem \ref{rat180}, we obtain

\begin{align*} \int_{[a,b]}\left |f'(x)\right |d\nu & \leq\int_{[a,b]}f’_{1}(x)d\nu +\int_{[a,b]}f’_{2}(x)d\nu
\\ & \leq f_{1}(b)-f_{1}(a)+f_{2}(b)-f_{2}(a),\end{align*}

which says that \(f'(x)\) is Lebesgue-integrable on \([a,b]\). We define

\[F(x)=\int_{[a,x]}f'(t)d\nu .\]

Proposition \ref{rap155} says that \(F\) is absolutely continuous on \([a,b]\), which also says that the function \(g=f-F\) is absolutely continuous on \([a,b]\). Therefore, using Theorem \ref{rat206}, we obtain

\[g'(x)=f'(x)-F'(x)=0\mbox{ a.e. on }(a,b).\]

From Proposition \ref{rap208}, we also see that \(g\) is a constant function on \([a,b]\). In this case, we say \(g(x)=K\) for all \(x\in [a,b]\). Therefore, we have

\[K=g(a)=f(a)-F(a)=f(a)\]

and

\[f(x)=F(x)+g(x)=F(x)+f(a)=\int_{[a,x]}f'(t)d\nu +f(a).\]

This completes the proof. \(\blacksquare\)

Proposition. Suppose that the real-valued function is of bounded variation on the compact interval \([a,b]\). Then \(f\) can be written as \(f=g+h\), where \(g\) is absolutely continuous on \([a,b]\) and \(h\) is a function on \([a,b]\) satisfying \(h'(x)=0\) a.e. on the open interval \((a.b)\). Moreover, suppose that \(f=g_{0}+h_{0}\) is another decomposition with the same desired properties. Then \(g-g_{0}=h_{0}-h\) is a constant function on \([a,b]\).

Proof. Proposition \ref{rap186} says that \(f'(x)\) exists a.e. on \((a,b)\). We define

\[g(x)=\int_{[a,x]}f’d\nu\mbox{ and }h=f-g.\]

Proposition \ref{rap155} says that \(g\) is absolutely continuous on \([a,b]\). Also, using Theorem \ref{rat206}, we have

\[h’=f’-g’=f’-f’=0\mbox{ a.e. on }(a,b).\]

This shows that \(f=g+h\) is the desired decomposition. Suppose that \(f=g_{0}+h_{0}\) is another decomposition. Then, we have \(g-g_{0}=h_{0}-h\). We write \(G=g-g_{0}=h_{0}-h\). Then \(G\) is absolutely continuous on \([a,b]\) and \(G'(x)=0\) a.e. on \((a,b)\). Proposition \ref{rap208} says that \(G\) is a constant function on \([a,b]\). This completes the proof. \(\blacksquare\)

Proposition. Suppose that the real-valued function \(f\) is absolutely continuous on the compact interval \([a,b]\). We consider the functions

\[T(x)=T_{a}^{x}(f),\quad P(x)=P_{a}^{x}(f)\mbox{ and }N(x)=N_{a}^{x}(f)\]

on \([a,b]\). Then, the functions \(T\), \(P\) and \(N\) are absolutely continuous on \([a,b]\). Moreover, we have

\[T(x)=\int_{[a,x]}|f’|d\nu,\quad P(x)=\int_{[a,x]}(f’)^{+}d\nu\mbox{ and }N(x)=\int_{[a,x]}(f’)^{-}d\nu .\]

Proof. We first show that \(T\) is absolutely continuous on \([a,b]\). Let \([c,d]\) be a sub-interval of \([a,b]\), and let \({\cal P}=\{x_{k}\}_{k=0}^{n}\) be a partition of \([c,d]\). Then, we have

\begin{align}
T(d)-T(c) & =T_{c}^{d}(f)=\sup_{\cal P}\sum_{k=1}^{n}\left |f(x_{k})-f(x_{k-1})\right |
\nonumber\\ & =\sup_{\cal P}\sum_{k=1}^{n}\left |\int_{[x_{k-1},x_{k}]}f’d\nu\right |\leq\int_{[c,d]}|f’|d\nu .\label{ma70}\tag{98}
\end{align}

Let \(\{[a_{i},b_{i}]\}_{i=1}^{n}\) be a collection of non-overlapping subintervals of \([a,b]\). Then, using (\ref{ma70}), we have

\begin{equation}{\label{raeq208}}\tag{99}
\sum_{i=1}^{n}\left [T(b_{i})-T(a_{i})\right ]\leq\int_{\bigcup_{i=1}^{n}[a_{i},b_{i}]}|f’|d\nu .
\end{equation}

Using Proposition \ref{rap178}, the inequality (\ref{raeq208}) says that \(T\) is absolutely continuous on \([a,b]\). Since \(T(a)=0\), Theorem \ref{rat209} and
Proposition \ref{rap210} say

\[T(x)=\int_{[a,x]}T’d\nu =\int_{[a,x]}|f’|d\nu .\]

Using Proposition \ref{rap211}, since

\[T(x)=P(x)+N(x)\mbox{ and }f(x)-f(a)=P(x)-N(x),\]

we obtain

\begin{equation}{\label{rap212}}\tag{100}
P(x)=\frac{1}{2}\left (T(x)+f(x)-f(a)\right )\mbox{ and }N(x)=\frac{1}{2}\left (T(x)-f(x)+f(a)\right ),
\end{equation}

which shows that \(P\) and \(N\) are absolutely continuous on \([a,b]\). Moreover, from (\ref{rap212}), we obtain

\begin{align*}
P(x) & =\frac{1}{2}\left (T(x)+f(x)-f(a)\right )\\ & =\frac{1}{2}\left (\int_{[a,x]}|f’|d\nu +\int_{[a,x]}f’d\nu\right )\\
& =\frac{1}{2}\int_{[a,x]}\left (|f’|+f’\right )d\nu \\ & =\frac{1}{2}\int_{[a,x]}\left ((f’)^{+}+(f’)^{-}+(f’)^{+}-(f’)^{-}\right )d\nu\\ & =\int_{a}^{x}(f’)^{+}d\nu .
\end{align*}

Similarly, we can show that

\[N(x)=\frac{1}{2}\left (T(x)-f(x)+f(a)\right )=\int_{a}^{x}(f’)^{-}d\nu .\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{j}}\tag{J}\mbox{}\end{equation}

The Set-Valued Indefinite Integrals.

Let \(f\) be a Lebesgue-integrable function defined on a Lebesgue measurable subset \(A\) of \(\bar{\mathbb{R}}^{n}\). We define the indefinite integral of \(f\) to be the set function

\[F(E)=\int_{E}fd\nu\]

for any Lebesgue measurable subset of \(A\). We are going to discuss the continuity of indefinite integral. We recall that the diameter of a set \(E\) is the number

\[\mbox{dia}(E)=\sup\left\{\parallel {\bf x}-{\bf y}\parallel :{\bf x},{\bf y}\in E\right\}.\]

Definition. Let \(F\) be a set function that assigns each subset \(E\) of \(\bar{\mathbb{R}}^{n}\) to a real number \(F(E)\).

• A set function \(F\) is called continuous when, given any \(\epsilon >0\), there exists a \(\delta >0\) such that \(\mbox{dia}(E)<\delta\) implies \(|F(E)|<\epsilon\).
• A set function \(F\) is called absolutely continuous when, given any \(\epsilon >0\), there exists a \(\delta >0\) such that \(\nu (E)<\delta\) implies \(|F(E)|<\epsilon\).

If the set function \(F\) is absolutely continuous, then it is continuous. However, the converse is not true in general.

\begin{equation}{\label{rat35}}\tag{101}\mbox{}\end{equation}

Theorem \ref{rat35}. Suppose that the extended real-valued function \(f\) is Lebesgue-integrable on a Lebesgue-measurable subset \(A\) of \(\bar{\mathbb{R}}^{n}\). Then, the indefinite integral \(F\) defined by

\[F(E)=\int_{E}fd\nu\]

is a absolutely continuous set function.

Proof. We first assume \(f\geq 0\). Given a fixed constant \(K\), we define

\[g({\bf x})=\left\{\begin{array}{ll}
f ({\bf x}) & \mbox{if \(f({\bf x})\leq K\)}\\
K & \mbox{if \(f({\bf x})>K\)}.
\end{array}\right .\]

Let \(h=f-g\geq 0\). If \(K\) is large, then the function values of \(h\) will be small, Given any \(\epsilon >0\), we may choose a large \(K\) satisfying

\[0\leq\int_{A}hd\nu<\frac{\epsilon}{2},\]

which implies

\[0\leq\int_{E}hd\nu <\frac{\epsilon}{2}\mbox{ for any }E\subseteq A.\]

Now, we take \(\delta =\epsilon /2K\). Since \(0\leq g\leq K\), for any \(E\) with \(\nu (E)<\delta\), we have

\[0\leq\int_{E}gd\nu\leq K\cdot\nu (E)<K\cdot\delta =\frac{\epsilon}{2}.\]

Therefore, we obtain

\begin{equation}{\label{ma71}}\tag{102}
0\leq\int_{E}fd\nu =\int_{E}gd\nu +\int_{E}hd\nu <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .
\end{equation}

Now, given any \(\epsilon >0\), using (\ref{ma71}), there exist \(\delta_{1}>0\) and \(\delta_{2}>0\) satisfying, for any \(E\subseteq A\) with

\[\nu (E)<\delta =\min\{\delta_{1},\delta_{2}\},\]

we have

\[\int_{E}f^{+}d\nu <\frac{\epsilon}{2}\mbox{ and }\int_{E}f^{-}d\nu <\frac{\epsilon}{2}.\]

Therefore, we obtain

\begin{align*} |F(E)| & =\left |\int_{E}fd\nu \right |\leq\int_{E}|f|d\nu \\ & =\int_{E}f^{+}d\nu +\int_{E}f^{-}d\nu \\ & <\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .\end{align*}

This completes the proof. \(\blacksquare\)

Let the extended real-valued function \(f\) be Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\), and let \(F\) be its corresponding indefinite integral defined by

\[F(Q)=\int_{Q}fd\nu.\]

We denote by \(Q\) an \(n\)-dimensional cube with edges parallel to the coordinate axes. Given \({\bf x}\in\mathbb{R}^{n}\), we consider those \(Q\) centered at \({\bf x}\). We are going to ask whether the average

\[\frac{F(Q)}{\nu (Q)}=\frac{1}{\nu (Q)}\int_{Q}f({\bf y})d\nu\]

converges to \(f({\bf x})\) as \(Q\) contracts to \({\bf x}\). If this is the case, we write

\[\lim_{Q\downarrow {\bf x}}\frac{F(Q)}{\nu (Q)}=f({\bf x}),\]

and say that the {\bf indefinite integral \(F\) of \(f\) is differentiable at \({\bf x}\) with derivative \(f({\bf x})\)}. For the case of \(n=1\), the question is reduced to ask whether

\begin{equation}{\label{raeq191}}\tag{103}
\lim_{h\rightarrow 0}\frac{1}{2h}\int_{[x-h,x+h]}f(y)d\nu=f(x).
\end{equation}

Suppose that \(f\) is Riemann integrable on \([a,b]\). Then (\ref{raeq191}) is essentially equivalent to

\[\lim_{h\rightarrow 0}\frac{1}{h}\left [\int_{x}^{x+h}f(y)dy\right ]=f(x);\]

that is,

\[\frac{d}{dx}\left [\int_{a}^{x}f(y)dy\right ]=f(x).\]

\begin{equation}{\label{rat193}}\tag{104}\mbox{}\end{equation}

Theorem \ref{rat193}. (Lebesgue’s Differentiation Theorem). Suppose that the extended real-valued function \(f\) is Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\). Then, its indefinite integral \(F\) defined by

\[F(Q)=\int_{Q}fd\nu\]

is differentiable with derivative \(f({\bf x})\) at almost every \({\bf x}\in\bar{\mathbb{R}}^{n}\); that is,

\[\lim_{Q\downarrow {\bf x}}\frac{F(Q)}{\nu (Q)}=f({\bf x})\mbox{ a.e. on }\bar{\mathbb{R}}^{n}.\]

Definition. A Lebesgue-measurable function defined on \(\bar{\mathbb{R}}^{n}\) is said to be locally Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\) when it is Lebesgue-integrable on every bounded measurable subset of \(\bar{\mathbb{R}}^{n}\). \(\sharp\)

\begin{equation}{\label{rat195}}\tag{105}\mbox{}\end{equation}

Theorem \ref{rat195}. (Lebesgue’s Differentiation Theorem). Suppose that the extended real-valued function \(f\) is locally Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\). Then, its indefinite integral \(F\) defined by

\[F(Q)=\int_{Q}fd\nu\]

is differentiable with derivative \(f({\bf x})\) at almost every \({\bf x}\in\bar{\mathbb{R}}^{n}\); that is,

\[\lim_{Q\downarrow {\bf x}}\frac{F(Q)}{\nu (Q)}=f({\bf x})\mbox{ a.e. on }\bar{\mathbb{R}}^{n}.\]

Proof. It suffices to show that the conclusion holds true a.e. on every open ball. Fix an open ball \(O\), we define a new function

\[g({\bf x})=\left\{\begin{array}{ll}
f({\bf x}) & \mbox{if \({\bf x}\in O\)}\\
0 & \mbox{if \({\bf x}\not\in O\)}.
\end{array}\right .\]

Then, this new function \(g\) is Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\). Using Theorem \ref{rat193}, its indefinite integral \(G\) defined by

\[G(Q)=\int_{Q}gd\nu\]

is differentiable a.e. on \(O\), which also says that that the indefinite integral \(F\) is differentiable a.e. on \(O\). This completes the proof. \(\blacksquare\)

Definition. Let \(E\) be a Lebesgue measurable set. A point \({\bf x}\in E\) is called a point of density of \(E\) when

\[\lim_{Q\downarrow {\bf x}}\frac{\nu (E\cap Q)}{\nu (Q)}=1.\]

A point \({\bf x}\in E\) is called a point of dispersion of \(E\) when

\[\lim_{Q\downarrow {\bf x}}\frac{\nu (E\cap Q)}{\nu (Q)}=0.\]

Since

\[\frac{\nu (E\cap Q)}{\nu (Q)}+\frac{\nu (E^{c}\cap Q)}{\nu (Q)}=\frac{\nu (Q)}{\nu (Q)}=1,\]

it says that every point of density of \(E\) is a point of dispersion of \(E^{c}\), and vice versa.

\begin{equation}{\label{rap194}}\tag{106}\mbox{}\end{equation}

Proposition \ref{rap194}. Let \(E\) be a Lebesgue measurable set. Then almost every point of \(E\) is a point of density of \(E\).

Proof. For \(Q\subseteq E\), let

\[F(Q)=\int_{Q}\chi_{E}d\nu.\]

Using Theorem \ref{rat193}, we have

\[\lim_{Q\downarrow {\bf x}}\frac{F(Q)}{\nu (Q)}=\chi_{E}({\bf x})\mbox{ a.e. on }E.\]

It is obvious that

\[\frac{F(Q)}{\nu (Q)}=\frac{1}{\nu (Q)}\int_{Q}\chi_{E}d\nu =\frac{\nu (E\cap Q)}{\nu (Q)}.\]

Therefore, we have

\begin{align*} \lim_{Q\downarrow {\bf x}}\frac{\nu (E\cap Q)}{\nu (Q)} & =
\lim_{Q\downarrow {\bf x}}\frac{F(Q)}{\nu (Q)}\\ & =\chi_{E}({\bf x})\mbox{ a.e. on }E.\end{align*}

This completes the proof. \(\blacksquare\)

Suppose that \(f\) is locally Lebesgue-integrable. Theorem \ref{rat195} says

\[\lim_{Q\downarrow {\bf x}}\left [\frac{F(Q)}{\nu (Q)}-f({\bf x})\right ]=0\mbox{ a.e. on }\bar{\mathbb{R}}^{n},\]

which implies

\begin{equation}{\label{ma72}}\tag{107}
\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left [f({\bf y})-f({\bf x})\right ]d\nu ({\bf y})=0\mbox{ a.e. on }\bar{\mathbb{R}}^{n}.
\end{equation}

In general, we have the following definition.

Definition. We say that a point \({\bf x}\) is a Lebesgue point of \(f\) when

\[\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})=0\mbox{ for any }Q.\]

The collection of all Lebesgue points of \(f\) is called a Lebesgue set.

\begin{equation}{\label{rat196}}\tag{108}\mbox{}\end{equation}

Theorem \ref{rat196}. Let \(f\) be locally Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\). Then, almost every point of \(\bar{\mathbb{R}}^{n}\) is a Lebesgue point of \(f\).

Proof. Let \(\{r_{k}\}_{k=1}^{\infty}\) be a sequence of rational numbers, and let \(Z_{k}\) be a set such that, for any \({\bf x}\in Z_{k}\), we have

\[\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-r_{k}\right |d\nu ({\bf y})\neq\left |f({\bf x})-r_{k}\right |.\]

Let \(g({\bf y})=f({\bf y})-r_{k}\). Then \(|g|\) is locally Lebesgue-integrable. From (\ref{ma72}), we have

\[\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}|g({\bf y})|d\nu ({\bf y})=|g({\bf x})|\mbox{ a.e. on }\bar{\mathbb{R}}^{n};\]

that is,

\[\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-r_{k}\right |
d\nu ({\bf y})=\left |f({\bf x})-r_{k}\right |\mbox{ a.e. on }\bar{\mathbb{R}}^{n},\]

which says that \(\nu (Z_{k})=0\). Let \(Z=\bigcup_{k=1}^{\infty}Z_{k}\). Then, we also have \(\nu (Z)=0\). Therefore, we have

\begin{equation}{\label{ma73}}\tag{109}
\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-r_{k}\right |d\nu ({\bf y})=\left |f({\bf x})-r_{k}\right |\mbox{ for }{\bf x}\not\in Z.
\end{equation}

Now, given any \(Q\), \({\bf x}\) and \(r_{k}\), we have

\begin{align*}
\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y}) & \leq\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-r_{k}\right |d\nu ({\bf y})
+\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf x})-r_{k}\right |d\nu ({\bf y})\\
& =\left |f({\bf x})-r_{k}\right |+\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-r_{k}\right |d\nu ({\bf y}).
\end{align*}

Therefore, using (\ref{ma73}), we obtain

\begin{equation}{\label{raeq197}}\tag{110}
\limsup_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})\leq 2\left |f({\bf x})-r_{k}\right |
\mbox{ for }{\bf x}\not\in Z
\end{equation}

for each \(r_{k}\). Let \(\hat{Z}=\{{\bf x}:f({\bf x})=+\infty\}\). Then \(\nu (\hat{Z})=0\). Let \(Z_{0}=Z\cup\hat{Z}\). Then \(\nu (Z_{0})=0\). For any \({\bf x}\not\in Z_{0}\), by the denseness of \(\mathbb{R}\), we can take \(r_{k}\) such that \(|f({\bf x})-r_{k}|\) is arbitrary small, which implies, by (\ref{raeq197}),

\begin{align*} \limsup_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})\\ & \leq 0\\ & \leq
\liminf_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y}).\end{align*}

Therefore, we obtain

\[\lim_{Q\downarrow {\bf x}}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})=0\]

for any \({\bf x}\not\in Z_{0}\) with \(\nu (Z_{0})=0\). This completes the proof. \(\blacksquare\)

So far, the set \(Q\) contracting to \({\bf x}\) has been taken as cube centered at \({\bf x}\) with edges parallel to the coordinate axes. Many other sets can be used.

Definition. A family \(\{S_{\alpha}\}_{\alpha\in I}\) of Lebesgue measurable sets is said to be shrink regularly to \({\bf x}\) when the following conditions are satisfied.

• The diameters of the sets \(S_{\alpha}\) tends to zero.
• If \(Q\) is the smallest cube with center \({\bf x}\) containing \(S_{\alpha}\) for all \(\alpha\in I\), then there is a constant \(K\) independent of \(S_{\alpha}\) satisfying \[\nu (Q)\leq K\cdot\nu (S_{\alpha})\mbox{ for all }\alpha\in I.\]

We also remark that the sets \(S_{\alpha}\) need not contain \({\bf x}\).

Theorem.  Let \(f\) be locally Lebesgue-integrable on \(\bar{\mathbb{R}}^{n}\). For each Lebesgue point \({\bf x}\) of \(f\), we have

\[\frac{1}{\nu (S_{\alpha})}\int_{S_{\alpha}}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})\rightarrow 0\]

for any family \(\{S_{\alpha}\}_{\alpha\in I}\) which shrink regularly to \({\bf x}\). We also have

\[\frac{1}{\nu (S_{\alpha})}\int_{S_{\alpha}}f({\bf y})d\nu ({\bf y})\rightarrow f({\bf x})\mbox{ a.e. on }\bar{\mathbb{R}}^{n}.\]

Proof. Suppose that \(S_{\alpha}\subseteq Q\). Then, we have

\[\int_{S_{\alpha}}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})\leq
\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y}).\]

Since \(\{S_{\alpha}\}_{\alpha\in I}\) shrinks regularly to \({\bf x}\) and \(Q\) is the smallest cube with center \({\bf x}\) containing \(S_{\alpha}\) for each \(\alpha\in I\), it follows

\begin{align*}
\frac{1}{\nu (S_{\alpha})}\int_{S_{\alpha}}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})
& \leq\frac{\nu (Q)}{\nu (S_{\alpha})}\frac{1}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y})\\
& \leq\frac{K}{\nu (Q)}\int_{Q}\left |f({\bf y})-f({\bf x})\right |d\nu ({\bf y}).
\end{align*}

If \({\bf x}\) is a Lebesgue point of \(f\), then the above expression tends to zero. By Theorem \ref{rat196}, we also have

\[\frac{1}{\nu (S_{\alpha})}\int_{S_{\alpha}}f({\bf y})d\nu ({\bf y})\rightarrow f({\bf x})\mbox{ a.e. on }\bar{\mathbb{R}}^{n}.\]

This completes the proof. \(\blacksquare\)

\begin{equation}{\label{k}}\tag{K}\mbox{}\end{equation}

The Lebesgue-Stieltjes Integrals.

Let \(\nu\) be a finite Borel measure on \(\bar{\mathbb{R}}\). We can associate a function \(F\) defined by

\[F(x)=\nu ((-\infty ,x]).\]

The function \(F\) is called the cumulative distribution function of \(\nu\) and is increasing on \(\bar{\mathbb{R}}\). We have

\[\nu ((a,b])=F(b)-F(a).\]

Since \((a,b]\) is the intersection of the sets \((a,b+\frac{1}{n}]\) for all \(n\), we have

\[\nu ((a,b])=\lim_{n\rightarrow\infty}\nu\left (\left (a,b+\frac{1}{n}\right ]\right ).\]

Therefore, we obtain

\[F(b)-F(a)=\lim_{n\rightarrow\infty}\left [F\left (b+\frac{1}{n}\right )-F(a)\right ],\]

which implies

\[F(b)=\lim_{n\rightarrow\infty}F\left (b+\frac{1}{n}\right )=F(b+),\]

which says that the cumulative distribution function is right-continuous. Similarly, we have

\begin{align*} \nu (\{b\}) & =\lim_{n\rightarrow\infty}\nu\left (\left (b-\frac{1}{n},b\right ]\right )
\\ & =\lim_{n\rightarrow\infty}\left [F(b)-F\left (b-\frac{1}{n}\right )\right ]\\ & =F(b)-F(b-).\end{align*}

This says that \(F\) is continuous at \(b\) if and only if the singleton set \(\{b\}\) has measure zero. Since

\[\emptyset =\bigcap_{n=1}^{\infty}(-\infty ,-n),\]

we have

\[\lim_{n\rightarrow -\infty}F(n)=0,\]

which implies

\[\lim_{x\rightarrow -\infty}F(x)=0\]

because of the monotonicity of \(F\). The results are summarized below.

Proposition. Let \(\nu\) be a finite Borel measure on \(\bar{\mathbb{R}}\). Then, its cumulative distribution function \(F\) is an increasing, bounded and right-continuous function. Moreover, we have

\[\lim_{x\rightarrow -\infty}F(x)=0.\]

Proposition. Let \(F\) be an increasing and right-continuous function defined on \(\bar{\mathbb{R}}\). Suppose that \((a,b]\subseteq\bigcup_{k=1}^{\infty}(a_{i},b_{i}]\). Then, we have

\[F(b)-F(a)\leq\sum_{k=1}^{\infty}\left [F(b_{i})-F(a_{i})\right ].\]

Proposition. Let \(F\) be an increasing and right-continuous function defined on \(\bar{\mathbb{R}}\). Then, there is a unique Borel measure \(\nu\) satisfying

\[\nu ((a,b])=F(b)-F(a)\]

for all \(a,b\in\bar{\mathbb{R}}\).

Proposition. Each increasing, bounded and right-continuous function \(F\) defined on \(\bar{\mathbb{R}}\) is the cumulative distribution function of a unique Borel measure provided that

\[\lim_{x\rightarrow -\infty}F(x)=0.\]

Definition. Let \(f\) be a nonnegative Borel measurable function defined on \(\bar{\mathbb{R}}\), and let \(F\) be an increasing and right-continuous function defined on \(\bar{\mathbb{R}}\). Then, we define the Lebesgue-Stieltjes integral of \(f\) with respect to \(F\) to be

\[\int_{X}fdF=\int_{X}fd\nu ,\]

where \(\nu\) is the Borel measure having \(F\) as its cumulative distribution function. \(\sharp\)

Suppose that \(F\) is any increasing function. Then, there exists a unique function \(F^{*}\) that is increasing, right-continuous and \(F^{*}(x)=F(x)\) in which \(F\) is right-continuous at \(x\). In this case, we define the Lebesgue-Stieltjes integral of \(f\) with respect to \(F\) by

\[\int_{X}fdF=\int_{X}fdF^{*}.\]

Suppose that \(F\) is increasing and right-continuous. Then the following Lebesgue-Stieltjes integral

\[\int_{[a,b]}fdF\]

is identical with the Riemann-Stieltjes integral. The Lebesgue-Steiltjes integral is only defined when \(F\) is monotone (or more generally of bounded variation), while the Riemann-Stieltjes integral can exist when \(F\) is not of bounded variation, e.g., when \(F\) is continuous and \(f\) is of bounded variation.