Jose Garcia y Ramos (1852-1912) was a Spanish painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Open and Closed Sets.
It is convenient to extend the system of real number \(\mathbb{R}\) by the addition of two elements \(\infty\) (or \(+\infty\)) and \(-\infty\). This enlarged set is called the extended real number system and is denoted by \(\bar{\mathbb{R}}\). We extend the definition of \(<\) to \(\bar{\mathbb{R}}\) by assuming \(-\infty <x<\infty\) for each \(x\in\mathbb{R}\). We define
\[x+\infty =\infty +x=\infty\mbox{ and }x-\infty =-\infty +x=-\infty\]
for any \(x\in\mathbb{R}\). The operation \(\infty -\infty\) is undefined. We also define
\[x\cdot\infty =\infty\cdot x=\left\{\begin{array}{ll}
\infty & \mbox{if \(0<x\leq\infty\)}\\
0 & \mbox{if \(x=0\)}\\
-\infty & \mbox{if \(-\infty\leq x<0\).}
\end{array}\right .\]
It may seem strange to define \(0\cdot\infty =0\). However, we can verify without difficulty that, with this definition, the commutative, associative, and distributive laws hold in \([0,\infty ]\) without any restriction. We shall define and study the Lebesgue measure of sets in the finite-dimensional Euclidean space \(\bar{\mathbb{R}}^{n}\). Given \({\bf x}\in\bar{\mathbb{R}}^{n}\), the set
\[B({\bf x};\delta )=\left\{{\bf y}\in\bar{\mathbb{R}}^{n}:\parallel {\bf x}-{\bf y}\parallel <\delta\right\}\]
is called the open ball with center \({\bf x}\) and radius \(\delta\). Given a subset \(E\) of \(\bar{\mathbb{R}}^{n}\), a point \({\bf x}\in E\) is called an interior point of \(E\) when there exists \(\delta >0\) satisfying \(B({\bf x};\delta )\subseteq E\). The collections of all interiors points of \(E\) is denoted by \(\mbox{int}(E)\). We say that the set \(E\) is open when \(E=\mbox{int}(E)\). In other words, the set \(E\) is open if and only if, for each \({\bf x}\in E\), there exists \(\delta >0\) satisfying \(B({\bf x};\delta )\subseteq E\). By convention, we define the empty set \(\emptyset\) to be open. The whole space \(\bar{\mathbb{R}}^{n}\) is clearly open, and the open ball \(B({\bf x};\delta )\) is also open.
A subset \(E\) of \(\bar{\mathbb{R}}^{n}\) is called closed when its complement set \(E^{c}\) is an open set in \(\bar{\mathbb{R}}^{n}\). Note that the empty set \(\emptyset\) and the whole space \(\bar{\mathbb{R}}^{n}\) are closed sets. A point \({\bf x}\in\bar{\mathbb{R}}^{n}\) is called a limit point of a set \(E\) when there is a sequence \(\{{\bf x}_{n}\}_{n=1}^{\infty}\) in \(E\) satisfying \({\bf x}_{n}\rightarrow {\bf x}\) as \(n\rightarrow\infty\). The union of \(E\) and all the limit points of \(E\) is called the closure of \(E\) and is denoted by \(\mbox{cl}(E)\).
Proposition. We have the following properties.
(i) We have
\[\mbox{cl}(B({\bf x};\delta ))=\left\{{\bf y}\in\bar{\mathbb{R}}^{n}:\parallel {\bf x}-{\bf y}\parallel\leq\delta\right\}.\]
(ii) The set \(E\) is closed if and only if \(E=\mbox{\em cl}(E)\). In other words, the set \(E\) is closed if and only if each point of \(E\) is a limit point of \(E\).
(iii) The closure \(\mbox{\em cl}(E)\) is closed.
(iv) The closure \(\mbox{\em cl}(E)\) is the smallest closed set containing \(E\); that is, if \(F\) is another closed set with \(E\subseteq F\), then \(\mbox{\em cl}(E)\subseteq F\). \(\sharp\)
Proposition. We have the following properties.
(i) The union of any number of open subsets of \(\bar{\mathbb{R}}^{n}\) is also an open subset of \(\bar{\mathbb{R}}^{n}\).
(ii) The intersection of any number of closed subsets of \(\bar{\mathbb{R}}^{n}\) is also a closed subset of \(\bar{\mathbb{R}}^{n}\).
(iii) The intersection of a finite number of open subsets of \(\bar{\mathbb{R}}^{n}\) is also an open subset of \(\bar{\mathbb{R}}^{n}\).
(iv) The union of a finite number of closed subsets of \(\bar{\mathbb{R}}^{n}\) is also a closed subset of \(\bar{\mathbb{R}}^{n}\). \(\sharp\)
\begin{equation}{\label{rad16}}\tag{1}\mbox{}\end{equation}
Definition \ref{rad16}. A subset \(E\) of \(\bar{\mathbb{R}}^{n}\) is of type \(\mathfrak{F}_{\sigma}\) when it can be written as a countable union of closed sets. A subset \(E\) of \(\bar{\mathbb{R}}^{n}\) is of type \(\mathfrak{G}_{\delta}\) when it can be written as a countable intersection of open sets. \(\sharp\)
The complement of a \(\mathfrak{G}_{\delta}\) set is an \(\mathfrak{F}_{\sigma}\) set, and vice versa. Any open (resp. closed) subsets of \(\bar{\mathbb{R}}^{n}\) is of type \(\mathfrak{G}_{\delta}\) (resp. \(\mathfrak{F}_{\sigma}\)). These two special types of sets will be very useful later in the measure approximation of general sets.
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Lebesgue Outer Measure.
In \(\bar{\mathbb{R}}^{n}\), the \(n\)-dimensional closed interval is defined by
\begin{align*} {\bf I} & =\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:a_{i}\leq x_{i}\leq b_{i}\mbox{ for }i=1,\cdots ,n\right\}\\ & =I_{1}\times I_{2}\times\cdots I_{n},\end{align*}
where \(I_{i}=[a_{i},b_{i}]\) is a closed interval in \(\mathbb{R}\). The volume of \({\bf I}\) is given by
\[v({\bf I})=\prod_{i=1}^{n}(b_{i}-a_{i}).\]
Definition. Given an arbitrary subset \(E\) of \(\bar{\mathbb{R}}^{n}\), let \({\cal S}=\{{\bf I}^{(1)},{\bf I}^{(2)},\cdots\}\) be a countable collection of \(n\)-dimensional closed intervals. We say that \({\cal S}\) covers \(E\) when
\[E\subseteq\bigcup_{k=1}^{\infty}{\bf I}^{(k)}.\]
In this case, we also say that \({\cal S}\) is a cover of \(E\). The volume of \({\cal S}\) is defined by
\[\sigma ({\cal S})=\sum_{{\bf I}^{(k)}\in {\cal S}}v({\bf I}^{(k)})=\sum_{k=1}^{\infty}v({\bf I}^{(k)}).\]
Definition. Given an arbitrary subset \(E\) of \(\bar{\mathbb{R}}^{n}\), let \(\mathfrak{C}\) be a collection of all covers of \(E\). The Lebesgue outer measure of \(E\) is denoted and defined by
\begin{equation}{\label{raeq51}}\tag{2}
\nu^{*}(E)=\inf_{{\cal S}\in\mathfrak{C}}\sigma ({\cal S}).
\end{equation}
It is clear to see \(0\leq\nu^{*} (E)<\infty\).
\begin{equation}{\label{rap62}}\tag{3}\mbox{}\end{equation}
Proposition \ref{rap62}. We have the following properties.
(i) For any interval \(I\) in \(\mathbb{R}\), we have \(v(I)=\nu^{*} (I)\).
(ii) If \(E_{1}\subseteq E_{2}\), then \(\nu^{*} (E_{1})\leq\nu^{*} (E_{2})\).
(iii) If \(E=\bigcup_{i=1}^{\infty}E_{i}\) is a countable union of sets, then
\[\nu^{*} (E)\leq\sum_{i=1}^{\infty}\nu^{*} (E_{i}).\]
Proposition. We have the following properties.
(i) Suppose that \(\nu^{*}(E)=0\). Then, we also have \(\nu^{*}(F)=0\) for any \(F\subseteq E\).
(ii) Suppose that \(\nu^{*}(E_{i})=0\) for all \(i=1,2,\cdots\). Let \(E=\bigcup_{i=1}^{\infty}E_{i}\). Then, we have \(\nu^{*}(E)=0\). In particular, any countable subset of \(\bar{\mathbb{R}}^{n}\) has Lebesgue outer measure zero, since the singleton set has Lebesgue outer measure zero.
(iii) The Cantor set is an uncountable set with Lebesgue outer measure zero. \(\sharp\)
\begin{equation}{\label{rap18}}\tag{4}\mbox{}\end{equation}
Proposition \ref{rap18}. We have the following properties.
(i) Given any subset \(E\) of \(\bar{\mathbb{R}}^{n}\) and any \(\epsilon >0\), there exists an open subset \(O\) of \(\bar{\mathbb{R}}^{n}\) satisfying
\[E\subseteq O\mbox{ and }\nu^{*}(O)\leq\nu^{*} (E)+\epsilon.\]
Let \(\mathfrak{O}\) be a collection of all open sets in \(\bar{\mathbb{R}}^{n}\) containing \(E\), i.e,
\[\mathfrak{O}=\left\{O:O\mbox{ is open in \(\bar{\mathbb{R}}^{n}\) and }E\subseteq O\right\}.\]
Then, we have
\begin{equation}{\label{raeq60}}\tag{5}
\nu^{*}(E)=\inf_{O\in\mathfrak{O}}\nu^{*} (O).
\end{equation}
(ii) Given any subset \(E\) of \(\bar{\mathbb{R}}^{n}\), there exists a set \(G=\bigcap_{k=1}^{\infty}O_{k}\), where each \(O_{k}\) is open in \(\bar{\mathbb{R}}^{n}\), satisfying \(E\subseteq G\) and \(\nu^{*} (E)=\nu^{*} (G)\).
Proof. To prove part (i), given any \(\epsilon >0\), according to the concept of infimum regarding the Lebesgue outer measure, there exists a cover \(\mathfrak{C}=\{{\bf I}^{(k)}\}_{k=1}^{\infty}\) of \(E\) satisfying
\begin{equation}{\label{ma54}}\tag{6}
E\subseteq\bigcup_{k=1}^{\infty}{\bf I}^{(k)}\mbox{ and }\sum_{k=1}^{\infty}v({\bf I}^{(k)})\leq\nu^{*}(E)+\frac{\epsilon}{2}.
\end{equation}
For each \({\bf I}^{(k)}\), we can also take an interval \({\bf J}^{(k)}\) satisfying
\begin{equation}{\label{ma53}}\tag{7}
{\bf I}^{(k)}\subseteq\mbox{int}({\bf J}^{(k)})\mbox{ and }v({\bf J}^{(k)})\leq v({\bf I}^{(k)})+\epsilon 2^{-k-1}.
\end{equation}
Now, we define \(O=\bigcup_{k=1}^{\infty}\mbox{int}({\bf J}^{(k)})\). Then \(O\) is and open subset of \(\bar{\mathbb{R}}^{n}\) and
\[E\subseteq O\subseteq\bigcup_{k=1}^{\infty}{\bf J}^{(k)}.\]
Therefore, using (\ref{ma53}) and (\ref{ma54}) (in order), we obtain
\begin{align*} \nu^{*}(O) & \leq\sum_{k=1}^{\infty}v({\bf J}^{(k)})\\ & \quad\mbox{ (by the definition of Lebesgue outer measure)}\\
& \leq\sum_{k=1}^{\infty}v({\bf I}^{(k)})+\epsilon\sum_{k=1}^{\infty}2^{-k-1}\\
& \leq\nu^{*}(E)+\frac{\epsilon}{2}+\frac{\epsilon}{2}\mbox{ (using )}\\
& =\nu^{*}(E)+\epsilon .
\end{align*}
Since \(\epsilon\) can be any positive number, we must have \(\nu^{*}(O)\leq\nu^{*}(E)\), which also implies
\[\inf_{O\in\mathfrak{O}}\nu^{*} (O)\leq\nu^{*}(E).\]
On the other hand, since \(E\subseteq O\), we have \(\nu^{*}(E)\leq\nu^{*}(O)\), which implies
\[\nu^{*}(E)\leq\inf_{O\in\mathfrak{O}}\nu^{*} (O).\]
Therefore, we obtain the equality (\ref{raeq60}).
To prove part (ii), given each positive integer \(k\), part (i) says that there exists an open set \(O_{k}\) satisfying \(E\subseteq O_{k}\) and \(\nu^{*}(O_{k})\leq\nu^{*}(E)+1/k\). We define \(G=\bigcap_{k=1}^{\infty}O_{k}\). Then, we have \(E\subseteq G\). Therefore, we obtain
\begin{align*} \nu^{*}(E) & \leq\nu^{*}(G)\leq\nu^{*}(O_{k})\\ & \leq\nu^{*}(E)+1/k\end{align*}
for each \(k\). This shows that \(\nu^{*}(E)=\nu^{*}(G)\) by taking \(k\rightarrow\infty\), and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{rap183}}\tag{8}\mbox{}\end{equation}
Proposition \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 \(\nu^{*} (E)=\nu^{*} (F)\). \(\sharp\)
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Lebesgue-Measurable Sets.
We introduce the concept of measurable set,
\begin{equation}{\label{ma58}}\tag{9}\mbox{}\end{equation}
Definition \ref{ma58}. A subset \(E\) of \(\bar{\mathbb{R}}^{n}\) is said to be Lebesgue-measurable when, given any \(\epsilon >0\), there exists an open set \(O\) in \(\bar{\mathbb{R}}^{n}\) satisfying
\[E\subseteq O\mbox{ and }\nu^{*}(O\setminus E)<\epsilon .\]
If \(E\) is Lebesgue measurable, then its Lebesgue outer measure is Lebesgue measure and is denoted by \(\nu (E)\). \(\sharp\)
We see that if \(E\) is Lebesgue measurable then \(\nu (E)=\nu^{*}(E)\). The Lebesgue measurability of set \(E\) should not be confused with part (i) of Proposition \ref{rap18}, which states that there exists an open set \(O\) in \(\bar{\mathbb{R}}^{n}\) satisfying
\[E\subseteq O\mbox{ and }\nu^{*}(O)\leq\nu^{*}(E)+\epsilon.\]
In general, since \(O=E\cup (O\setminus E)\) when \(E\subseteq O\), we only have
\[\nu^{*}(O)\leq\nu^{*}(E)+\nu^{*}(O\setminus E).\]
From \(\nu^{*}(O)\leq\nu^{*}(E)+\epsilon\), we cannot conclude \(\nu^{*}(O\setminus E)<\epsilon\).
\begin{equation}{\label{rar61}}\tag{10}\mbox{}\end{equation}
Proposition \ref{rar61}. We have the following properties.
(i) Suppose that \(E\) is an open set in \(\bar{\mathbb{R}}^{n}\). Then, it is Lebesgue-measurable.
(ii) Suppose that \(\nu^{*}(E)=0\). Then, it is Lebesgue-measurable.
Proof. Part (i) can be realized immediately from the definition by taking \(O=E\). To prove part (ii), given any \(\epsilon >0\), part (i) of Proposition \ref{rap18} says that there exists an open set \(O\) in \(\bar{\mathbb{R}}^{n}\) satisfying
\[E\subseteq O\mbox{ and }\nu^{*}(O)\leq\nu^{*}(E)+\epsilon =\epsilon.\]
Therefore, we have
\[\nu^{*}(O\setminus E)\leq\nu^{*}(O)<\epsilon,\]
and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{rap65}}\tag{11}\mbox{}\end{equation}
Proposition \ref{rap65}. The countable union \(E=\bigcup_{k=1}^{\infty}E_{k}\) of Lebesgue-measurable sets is Lebesgue-measurable. We also have
\begin{equation}{\label{raeq63}}\tag{12}
\nu (E)\leq\sum_{k=1}^{\infty}\nu (E_{k}).
\end{equation}
Especially, an \(n\)-dimensional interval \({\bf I}\) is Lebesgue-measurable and \(v({\bf I})=\nu ({\bf I})\).
Proof. Given any \(\epsilon >0\), since each \(E_{k}\) is Lebesgue-measurable, the definition says that there exists an open set \(O_{k}\) satisfying
\begin{equation}{\label{ma55}}\tag{13}
E_{k}\subseteq O_{k}\mbox{ and }\nu^{*}(O_{k}\setminus E_{k})<\frac{\epsilon}{2}\cdot 2^{-k}.
\end{equation}
We take \(O=\bigcup_{k=1}^{\infty}O_{k}\). Then \(O\) is an open set satisfying \(E\subseteq O\). It follows
\begin{equation}{\label{ma56}}\tag{14}
O\setminus E\subseteq\bigcup_{k=1}^{\infty}(O_{k}\setminus E_{k}).
\end{equation}
Therefore, using (\ref{ma56}), part (iii) of Proposition\ref{rap62} and (\ref{ma55}) (in order), we obtain
\begin{align*}
\nu^{*}(O\setminus E) & \leq\nu^{*}\left (\bigcup_{k=1}^{\infty}(O_{k}\setminus E_{k})\right )\\
& \leq\sum_{k=1}^{\infty}\nu^{*}(O_{k}\setminus E_{k})\\
& <\frac{\epsilon}{2}\sum_{k=1}^{\infty}2^{-k}\\
& =\frac{\epsilon}{2}<\epsilon.
\end{align*}
This proves that \(E\) is Lebesgue-measurable. The inequality (\ref{raeq63}) follows from part (iii) of Proposition \ref{rap62} again. On the other hand, the \(n\)-dimensional interval \({\bf I}\) is the union of its interior and its boundary, i.e.,
\[{\bf I}=\mbox{int}({\bf I})\cup\partial ({\bf I}).\]
Since \(\mbox{int}({\bf I})\) is open and \(\partial ({\bf I})\) has Lebesgue outer measure zero, we see that \(\mbox{int}({\bf I})\) and \(\partial ({\bf I})\) are Lebesgue-measurable by Proposition \ref{rar61}. This also says that \({\bf I}\) is Lebesgue-measurable. By part (i) of Proposition \ref{rap62}, we have \(v({\bf I})=\nu ({\bf I})\), and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{rap64}}\tag{15}\mbox{}\end{equation}
Proposition \ref{rap64}. Every closed set in \(\bar{\mathbb{R}}^{n}\) is Lebesgue-measurable.
\begin{equation}{\label{rap19}}\tag{16}\mbox{}\end{equation}
Proposition \ref{rap19}. Let \(E\) be Lebesgue measurable. Then, the complement set \(E^{c}\) is also Lebesgue-measurable.
Proof. By definition, for any positive integer \(k\), there exists an open set \(O_{k}\) satisfying
\begin{equation}{\label{ma57}}\tag{17}
E\subseteq O_{k}\mbox{ and }\nu^{*}(O_{k}\setminus E)<1/k.
\end{equation}
Since the complement set \(O_{k}^{c}\) is closed, it is Lebesgue-measurable by Proposition \ref{rap64}. Let \(H=\bigcup_{k=1}^{\infty}O_{k}^{c}\). Then \(H\) is also Lebesgue-measurable by Proposition \ref{rap65}. We also have \(H\subseteq E^{c}\). Let \(Z=E^{c}\setminus H\). Then, we have
\[Z\subseteq E^{c}\setminus O_{k}^{c}=O_{k}\setminus E.\]
Using (\ref{ma57}), we also have \(\nu^{*}(Z)<1/k\) for each \(k\), which implies \(\nu^{*}(Z)=0\) by taking \(k\rightarrow\infty\). It follows that \(Z\) is Lebesgue-measurable by part (ii) of Proposition \ref{rar61}. Since \(E^{c}=H\cup Z\), we conclude that \(E^{c}\) is Lebesgue-measurable by Proposition \ref{rap65}. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{rap20}}\tag{18}\mbox{}\end{equation}
Proposition \ref{rap20}. We have the following properties.
(i) The countable intersection \(E=\bigcap_{k=1}^{\infty}E_{k}\) of Lebesgue-measurable sets is Lebesgue-measurable.
(ii) Suppose that \(E_{1}\) and \(E_{2}\) are Lebesgue-measurable. Then \(E_{1}\setminus E_{2}\) is Lebesgue-measurable.
Proof. To prove part (i), since \(E^{c}=\bigcup_{k=1}^{\infty}E_{k}^{c}\), we see that \(E^{c}\) is Lebesgue-measurable from Propositions \ref{rap65} and \ref{rap19}, which implies that \(E=(E^{c})^{c}\) is also Lebesgue-measurable using Propositions \ref{rap19} again. To prove part (ii), since \(E_{1}\setminus E_{2}=E_{1}\cap E_{2}^{c}\), the desired result follows immediately from part (i) and Proposition \ref{rap19}.
This completes the proof. \(\blacksquare\)
Corresponding to Definition \ref{ma58}, we can consider the measurability based on closed sets, which is presented below.
Proposition. A subset \(E\) of \(\bar{\mathbb{R}}^{n}\) is Lebesgue-measurable if and only if, given \(\epsilon >0\), there exists a closed set \(F\subseteq E\) satisfying \(\nu^{*}(E\setminus F)<\epsilon\)
Proof. Using Propositions \ref{rap19}, the set \(E\) is Lebesgue-measurable if and only if \(E^{c}\) is Lebesgue-measurable. By Definition \ref{ma58}, given any \(\epsilon >0\), there exists an open set \(O\) satisfying
\[E^{c}\subseteq O\mbox{ and }\nu^{*}(O\setminus E^{c})<\epsilon.\]
Let \(F=O^{c}\). Then \(F\) is a closed set and
\[O\setminus E^{c}=E\setminus O^{c}=E\setminus F.\]
Therefore, we obtain \(F\subseteq E\) and \(\nu^{*}(E\setminus F)<\epsilon\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{rap67}}\tag{19}\mbox{}\end{equation}
Proposition \ref{rap67}. Suppose that \(\{E_{k}\}_{k=1}^{\infty}\) is a sequence of disjoint Lebesgue measurable sets. Then, we have
\[\nu\left (\bigcup_{k=1}^{\infty}E_{k}\right )=\sum_{k=1}^{\infty}\nu (E_{k}).\]
\begin{equation}{\label{rac68}}\tag{20}\mbox{}\end{equation}
Corollary \ref{rac68}. Suppose that \(E_{1}\) and \(E_{2}\) are Lebesgue-measurable with \(E_{2}\subseteq E_{1}\) and \(\nu (E_{2})<+\infty\). Then, we have \(\nu (E_{1}\setminus E_{2})=\nu (E_{1})-\nu (E_{2})\).
Proof. Since \(E_{1}=E_{2}\cup (E_{1}\setminus E_{2})\), we have
\[\nu (E_{1})=\nu (E_{2})+\nu (E_{1}\setminus E_{2})\]
by Proposition \ref{rap67}. Since \(\nu (E_{2})<+\infty\), the result follows immediately. \(\blacksquare\)
\begin{equation}{\label{rap107}}\tag{21}\mbox{}\end{equation}
Proposition \ref{rap107}. Let \(\{E_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue measurable sets. Then, we have the following properties.
(i) Suppose that \(E_{k}\subseteq E_{k+1}\) for all \(k\). Let \(E=\bigcup_{k=1}^{\infty}E_{k}\). Then, we have
\[\lim_{k\rightarrow\infty}\nu (E_{k})=\nu (E).\]
(ii) Suppose that \(E_{k+1}\subseteq E_{k}\) for all \(k\). Let \(E=\bigcap_{k=1}^{\infty}E_{k}\). If \(\nu (E_{k})<+\infty\) for some \(k\), then we have
\[\lim_{k\rightarrow\infty}\nu (E_{k})=\nu (E).\]
Proof. To prove part (i), suppose that \(\nu (E_{k})=+\infty\) for some \(k\). Then both of \(\nu (E)\) and \(\lim_{k}\nu (E_{k})\) are infinite. Now, we assume \(\nu (E_{k})<+\infty\) for all \(k\). We write
\[E=E_{1}\cup (E_{2}\setminus E_{1})\cup\cdots\cup (E_{k}\setminus E_{k-1})\cup\cdots ,\]
which is a disjoint union of Lebesgue-measurable sets. Therefore, using Proposition \ref{rap67} and Corollary \ref{rac68} (in order), we have
\begin{align*}
\nu (E) & =\nu (E_{1})+\nu (E_{2}\setminus E_{1})+\cdots+\nu (E_{k}\setminus E_{k-1})+\cdots\\
& =\nu (E_{1})+(\nu (E_{2})-\nu (E_{1}))+\cdots +(\nu (E_{k})-\nu (E_{k-1}))+\cdots\\
& =\lim_{k\rightarrow\infty}\nu (E_{k}).
\end{align*}
To prove part (ii), we may just assume that \(\nu (E_{1})<+\infty\). Then, we have \(\nu (E_{k})<+\infty\) for all \(k=2,3,\cdots\). We write
\[E_{1}=E\cup (E_{1}\setminus E_{2})\cup\cdots\cup (E_{k}\setminus E_{k+1})\cup\cdots ,\]
which is a disjoint union of Lebesgue measurable subsets. Therefore, we obtain
\begin{align*}
\nu (E_{1}) & =\nu (E)+(\nu (E_{1})-\nu (E_{2}))+\cdots +(\nu (E_{k})-\nu (E_{k+1}))+\cdots\\
& =\nu (E)+\nu (E_{1})-\lim_{k\rightarrow\infty}\nu (E_{k}).
\end{align*}
Since \(\nu (E_{1})<+\infty\), this completes the proof. \(\blacksquare\)
Proposition. We have the following properties.
(i) The set \(E\) is Lebesgue measurable if and only if \(E=H\setminus Z\), where \(\nu (Z)=0\) and \(H\) is a countable intersection of open sets.
(ii) The set \(E\) is Lebesgue measurable if and only if \(E=K\cup Z\), where \(\nu (Z)=0\) and \(K\) is a countable union of closed sets.
Proof. To prove part (i), suppose that \(E\) is Lebesgue measurable. By definition, given any positive integer \(k\), there exists an open set \(O_{k}\) satisfying
\[E\subseteq O_{k}\mbox{ and }\nu (O_{k}\setminus E)<1/k.\]
Let \(H=\bigcap_{k=1}^{\infty}O_{k}\). Then, we have \(H\setminus E\subseteq O_{k}\setminus E\) for every \(k\). We also have
\[\nu (H\setminus E)\leq\mu (O_{k}\setminus E)<1/k\]
for each \(k\), which implies \(\nu (H\setminus E)=0\) by taking \(k\rightarrow\infty\). Let \(Z=H\setminus E\). We obtain \(E=H\setminus Z\) with \(\nu (Z)=0\). The converse follows immediately from part (ii) of Proposition \ref{rar61} and part (iii) of Proposition \ref{rap20}.
To prove part (ii), suppose that \(E\) is Lebesgue measurable. Then \(E^{c}\) is also Lebesgue-measurable. Using the arguments of part (i), we also have
\[E^{c}=H\setminus Z=\left (\bigcap_{k=1}^{\infty}O_{k}\right )\setminus Z,\]
where each \(O_{k}\) is open and \(\nu (Z)=0\). Therefore, we obtain
\[E=\left (\bigcup_{k=1}^{\infty}O_{k}^{c}\right )\cup Z.\]
Let \(K=\bigcup_{k=1}^{\infty}O_{k}^{c}\). Then, we have \(E=K\cup Z\). The converse follows immediately from part (ii) of Proposition \ref{rar61}
and part (iii) of Proposition \ref{rap20}. This completes the proof. \(\blacksquare\)
Proposition. Suppose that \(\nu^{*}(E)<+\infty\). The set \(E\) is Lebesgue-measurable if and only if, given any \(\epsilon >0\), there exist \(N_{1}\) and \(N_{2}\) with \(\nu^{*}(N_{1})<\epsilon\) and \(\nu^{*}(N_{2})<\epsilon\) satisfying \(E=(S\cup N_{1})\setminus N_{2}\), where \(S\) is a finite union of \(n\)-dimensional disjoint intervals. \(\sharp\)
\begin{equation}{\label{rat45}}\tag{22}\mbox{}\end{equation}
Theorem \ref{rat45}. (Caratheodory). The set \(E\) is Lebesgue measurable if and only if, for any subset \(A\) of \(\bar{\mathbb{R}}^{n}\), we have
\begin{align}
\nu^{*}(A) & =\nu^{*}(A\cap E)+\nu^{*}(A\cap E^{c})\\\ =\nu^{*}(A\cap E)+\nu^{*}(A\setminus E).\label{raeq20}\tag{23}
\end{align}
Proof. Suppose that \(E\) is Lebesgue-measurable. For any subset \(A\), since
\[A=(A\cap E)\cup (A\setminus E),\]
we have
\[\nu^{*}(A)\leq\nu^{*}(A\cap E)+\nu^{*}(A\setminus E).\]
By part (ii) of Proposition \ref{rap18}, there exists a Lebesgue-measurable set \(H\) satisfying
\begin{equation}{\label{ma59}}\tag{24}
A\subseteq H\mbox{ and }\nu^{*}(A)=\nu^{*}(H)=\nu (H).
\end{equation}
Since
\[H=(H\cap E)\cup (H\setminus E)\]
is the disjoint union of Lebesgue measurable sets \(H\cap E\) and \(H\setminus E\), Proposition \ref{rap67} says
\begin{equation}{\label{ma60}}\tag{25}
\nu (H)=\nu (H\cap E)+\nu (H\setminus E).
\end{equation}
Using using (\ref{ma59}) and (\ref{ma60}) , we obtain
\begin{align*}
\nu^{*}(A) & =\nu (H)\\\
& =\nu (H\cap E)+\nu (H\setminus E)\\ & =\nu^{*}(H\cap E)+\nu^{*}(H\setminus E)\\
& \geq\nu^{*}(A\cap E)+\nu^{*}(A\setminus E)\mbox{ (since \(A\subseteq H\))},
\end{align*}
which shows the desired equality. Conversely, suppose that \(E\) satisfies the stated equality for every \(A\). For the case of \(\nu^{*}(E)<+\infty\), by part (ii) of Proposition \ref{rap18}, we can choose a Lebesgue-measurable set \(H\) satisfying
\begin{equation}{\label{ma61}}\tag{26}
E\subseteq H\mbox{ and }\nu^{*}(H)=\nu^{*}(E)<+\infty.
\end{equation}
Therefore, we have
\[H=E\cup (H\setminus E)=(H\cap E)\cup (H\setminus E).\]
Using (\ref{ma61}), we also obtain
\begin{align*}
\nu^{*}(H) & =\nu^{*}(H\cap E)+\nu^{*}(H\setminus E)\mbox{ (by the stated equality)}\\
& =\nu^{*}(E)+\nu^{*}(H\setminus E)=\nu^{*}(H)+\nu^{*}(H\setminus E),
\end{align*}
which implies \(\nu^{*}(H\setminus E)=0\), since \(\nu^{*}(H)<+\infty\). This says that \(H\setminus E\) is Lebesgue-measurable by part (ii) of Proposition\ref{rar61}. We conclude that \(E=H\setminus (H\setminus E)\) is Lebesgue measurable by Proposition \ref{rap20}. For the case of \(\nu^{*}(E)=+\infty\), let \(B_{k}\) be an open ball with center \({\bf 0}\) and radius \(k\) for \(k=1,2,\cdots\), and let \(E_{k}=E\cap B_{k}\). Then, each \(E_{k}\) has finite outer measure and \(E=\bigcup_{k=1}^{\infty}E_{k}\). Using part (ii) of Proposition \ref{rap18}, let \(H_{k}\) be a Lebesgue-measurable set satisfying
\begin{equation}{\label{ma62}}\tag{27}
E_{k}\subseteq H_{k}\mbox{ and }\nu^{*}(E_{k})=\nu^{*}(H_{k})<+\infty.
\end{equation}
Using (\ref{ma62}), we have
\begin{align*}
\nu^{*}(H_{k}) & =\nu^{*}(H_{k}\cap E)+\nu^{*}(H_{k}\setminus E)\mbox{ (by the stated equality)}\\
& \geq\nu^{*}(H_{k}\cap E_{k})+\nu^{*}(H_{k}\setminus E)\mbox{ (since \(E_{k}\subseteq E\))}\\
& =\nu^{*}(E_{k})+\nu^{*}(H_{k}\setminus E)=\nu^{*}(H_{k})+\nu^{*}(H_{k}\setminus E)
\end{align*}
which implies \(\nu^{*}(H_{k}\setminus E)=0\), since \(\nu^{*}(H_{k})<+\infty\). Let \(H=\bigcup_{k=1}^{\infty}H_{k}\). Then \(H\) is Lebesgue-measurable and \(E\subseteq H\). Moreover, we have
\[H\setminus E=\bigcup_{k=1}^{\infty}(H_{k}\setminus E),\]
which implies \(\nu^{*}(H\setminus E)=0\), i.e., \(H\setminus E\) is Lebesgue-measurable. Therefore, we conclude that \(E=H\setminus (H\setminus E)\) is Lebesgue-measurable. This completes the proof. \(\blacksquare\)
Inspired by the equivalence presented in Theorem \ref{rat45}, we can also define the concept of Lebesgue measurability according to (\ref{raeq20}).
\begin{equation}{\label{rap21}}\tag{28}\mbox{}\end{equation}
Proposition \ref{rap21}. Given a subset \(E\) of \(\bar{\mathbb{R}}^{n}\), there exists a set \(H\) that is a countable intersection of open sets satisfying \(E\subseteq H\) and \(\nu^{*}(E\cap M)=\nu (H\cap M)\) for any Lebesgue-measurable set \(M\). \(\sharp\)
We see that Proposition \ref{rap21} reduces to part (ii) of Proposition \ref{rap18} when \(M=\bar{\mathbb{R}}^{n}\).
Definition. A collection \(\Sigma\) of subsets of \(\bar{\mathbb{R}}^{n}\) is said to be a \(\sigma\)-field (or \(\sigma\)-algebra) when the following conditions are satisfied.
- \(E\in\Sigma\) implies \(E^{c}\in\Sigma\).
- If \(E_{k}\in\Sigma\) for \(k=1,2,\cdots\), then \(\bigcup_{k=1}^{\infty}E_{k}\in\Sigma\). \(\sharp\)
We have the following observations.
- If \(E_{k}\in\Sigma\) for \(k=1,2,\cdots\), then \(\bigcap_{k=1}^{\infty}E_{k}\in\Sigma\), since
\[\left (\bigcap_{k=1}^{\infty}E_{k}\right )^{c}=\bigcup_{k=1}^{\infty}E_{k}^{c}.\] - The empty set \(\emptyset\) belongs to any \(\sigma\)-field, since \(\emptyset =E\cap E^{c}\) for any \(E\in\Sigma\).
- The entire space \(\bar{\mathbb{R}}^{n}\) belongs to any \(\sigma\)-field, since it is the complement set of empty set.
\begin{equation}{\label{rap241}}\tag{29}\mbox{}\end{equation}
Theorem \ref{rap241}. The collection of all Lebesgue-measurable subsets of \(\bar{\mathbb{R}}^{n}\) is a \(\sigma\)-field.
Proof. The result follows immediately from Propositions\ref{rap65} and \ref{rap19}. \(\blacksquare\)
Let \({\cal F}\) be a family of \(\sigma\)-fields \(\Sigma\). We define \(\bigcap_{\Sigma\in {\cal F}}\Sigma\) to be the collection of all sets that belongs to every \(\sigma\)-field \(\Sigma\in {\cal F}\).
\begin{equation}{\label{rap66}}\tag{30}\mbox{}\end{equation}
Proposition \ref{rap66}. The collection \(\bigcap_{\Sigma\in {\cal F}}\Sigma\) of sets in \(\bar{\mathbb{R}}^{n}\) is a \(\sigma\)-field.
Proof. Given any \(E\in\bigcap_{\Sigma\in {\cal F}}\Sigma\), it means that \(E\in\Sigma\) for each \(\sigma\)-field \(\Sigma\) in the collection \({\cal F}\). This also says that \(E^{c}\in\Sigma\) for each \(\sigma\)-field \(\Sigma\) in the collection \({\cal F}\), which implies \(E^{c}\in\bigcap_{\Sigma\in {\cal F}}\Sigma\). Suppose that \(E_{k}\in\bigcap_{\Sigma\in {\cal F}}\Sigma\) for \(k=1,2,\cdots\). It means \(E_{k}\in\Sigma\) for \(k=1,2,\cdots\) and for each \(\Sigma\) in the collection \({\cal F}\). It also says \(\bigcup_{k=1}^{\infty}E_{k}\in\Sigma\) for each \(\Sigma\) in the collection \({\cal F}\), i.e., \(\bigcup_{k=1}^{\infty}E_{k}\in\bigcap_{\Sigma\in {\cal F}}\Sigma\). This completes the proof. \(\blacksquare\)
Let \({\cal C}_{1}\) and \({\cal C}_{2}\) be two collections of sets in \(\bar{\mathbb{R}}^{n}\). We say that \({\cal C}_{1}\) is contained in \({\cal C}_{2}\) when every set in \({\cal C}_{1}\) is also in \({\cal C}_{2}\). Given a collection \({\cal C}\) of subsets of \(\bar{\mathbb{R}}^{n}\), it is not necessarily a \(\sigma\)-field. We are going to enlarge it to be a \(\sigma\)-field in the smallest sense. We consider the family \({\cal F}\) of all \(\sigma\)-fields which contain \({\cal C}\). Let \({\cal E}=\bigcap_{\Sigma\in {\cal F}}\Sigma\). Then, according to Proposition \ref{rap66}, \({\cal E}\) is the smallest \(\sigma\)-field containing \({\cal C }\). In other words, \({\cal E}\) contains \({\cal C}\) and if \(\bar{\cal E}\) is another \(\sigma\)-field containing \({\cal C}\), then \(\bar{\cal E}\) contains \({\cal E}\).
Definition. The smallest \(\sigma\)-field of subsets of \(\bar{\mathbb{R}}^{n}\) containing all the open subsets of \(\bar{\mathbb{R}}^{n}\) is called the Borel \(\sigma\)-field of \(\bar{\mathbb{R}}^{n}\) and is denoted by \(\mathfrak{B}\). The sets in \(\mathfrak{B}\) are called the Borel sets in \(\bar{\mathbb{R}}^{n}\). \(\sharp\)
Proposition. Every Borel set in \(\bar{\mathbb{R}}^{n}\) is Lebesgue-measurable.
Proof. Let \(\mathfrak{M}\) be the collection of all Lebesgue-measurable subsets of \(\bar{\mathbb{R}}^{n}\). Theorem \ref{rap241} says that the family \(\mathfrak{M}\) is a \(\sigma\)-field. Since each open set belongs to \(\mathfrak{M}\), and \(\mathfrak{B}\) is the smallest \(\sigma\)-field containing the open sets, it follows that \(\mathfrak{B}\) is contained in \(\mathfrak{M}\). This completes the proof. \(\blacksquare\)
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
Lebesgue-Measurable Functions.
We are going to introduce the concept of measurable functions.
Definition. Let \(f\) be an extended real-valued function defined on a subset \(E\) of \(\bar{\mathbb{R}}^{n}\), i.e., \(-\infty\leq f({\bf x})\leq +\infty\). We say that \(f\) is a Lebesgue-measurable function on \(E\) when, for every finite real number \(a\), the set
\[\left\{{\bf x}\in E:f({\bf x})>a\right\}\]
is a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n}\). \(\sharp\)
The domain \(E\) of \(f\) can be written as
\begin{equation}{\label{ma63}}\tag{31}
E=\left\{{\bf x}\in E:f({\bf x})=-\infty\right\}\bigcup\left (\bigcup_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})>-k\right\}\right ).
\end{equation}
When the function \(f\) is Lebesgue-measurable, the following set
\[\bigcup_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})>-k\right\}\]
is Lebesgue-measurable. Therefore, according to (\ref{ma63}), the measurability of \(E\) is equivalent to the measurability of the set \(\{{\bf x}\in E:f({\bf x})=-\infty\}\) when we assume that \(f\) is Lebesgue-measurable. For convenience, we shall always assume that the set \(\{{\bf x}\in E:f({\bf x})=-\infty\}\) is Lebesgue-measurable. In this case, we can consider the Lebesgue-measurable function defined on Lebesgue-measurable sets.
Definition. Let \(f\) be an extended real-valued function defined on a Borel subset \(E\) of \(\bar{\mathbb{R}}^{n}\). We say that \(f\) is a Borel-measurable function on \(E\) when, for every finite real number \(a\), the set \(\{{\bf x}\in E:f({\bf x})>a\}\) is a Borel-measurable subset of \(\bar{\mathbb{R}}^{n}\). \(\sharp\)
It is clear to see that every Borel-measurable function is Lebesgue-measurable.
\begin{equation}{\label{ma74}}\tag{32}\mbox{}\end{equation}
Theorem \ref{ma74}. Let \(f\) be an extended real-valued function defined on a Lebesgue measurable set \(E\). Then \(f\) is Lebesgue measurable if and only if the following statements are equivalent.
(a) The set \(\{{\bf x}\in E:f({\bf x})\geq a\}\) is Lebesgue-measurable for every finite real number \(a\).
(b) The set \(\{{\bf x}\in E:f({\bf x})<a\}\) is Lebesgue-measurable for every finite real number \(a\).
(c) The set \(\{{\bf x}\in E:f({\bf x})\leq a\}\) is Lebesgue-measurable for every finite real number \(a\).
Proof. The Lebesgue measurability of \(f\) implies that the set
\[\left\{{\bf x}\in E:f({\bf x})\geq a\right\}=\bigcap_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})>a-\frac{1}{k}\right\}\]
is Lebesgue-measurable. This says that the Lebesgue measurability of \(f\) implies statement (a). Since \(\{{\bf x}\in E:f({\bf x})<a\}\) is the complement set of \(\{{\bf x}\in E:f({\bf x})\geq a\}\), it shows that statement (a) implies statement (b). Since
\[\left\{{\bf x}\in E:f({\bf x})\leq a\right\}=\bigcap_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})<a+\frac{1}{k}\right\},\]
it says that statement (b) implies statement (c). Finally, since \(\{{\bf x}\in E:f({\bf x})>a\}\) is the complement set of \(\{{\bf x}\in E:f({\bf x})\leq a\}\), it shows that statement (c) implies the Lebesgue measurability of \(f\). Therefore, we have created the equivalence of the above statements, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{rac242}}\tag{33}\mbox{}\end{equation}
Corollary \ref{rac242}. Let \(f\) be an extended real-valued function defined on a Lebesgue-measurable set \(E\). Then, the following sets are all Lebesgue-measurable:
\[\{{\bf x}\in E:f({\bf x})>-\infty\},\quad\{{\bf x}\in E:f({\bf x})<+\infty\},\quad\{{\bf x}\in E:f({\bf x})=-\infty\},\]
\[\{{\bf x}\in E:f({\bf x})=+\infty\},\quad\{{\bf x}\in E:f({\bf x})=a\},\quad\{{\bf x}\in E:a<f({\bf x})<b\},\]
\[\{{\bf x}\in E:a\leq f({\bf x})<b\},\quad\{{\bf x}\in E:a<f({\bf x})\leq b\},\quad\{{\bf x}\in E:a\leq f({\bf x})\leq b\}.\]
Let \(f\) be an extended real-valued function defined on \(E\), and let \(S\) be a subset of \(\mathbb{R}\). The inverse image of \(S\) under \(f\) is defined by
\[f^{-1}(S)=\left\{{\bf x}\in E:f({\bf x})\in S\right\}.\]
\begin{equation}{\label{ral70}}\tag{34}\mbox{}\end{equation}
Lemma \ref{ral70}. Every open set in \(\mathbb{R}\) can be displayed as a countable union of disjoint open intervals. \(\sharp\)
\begin{equation}{\label{rat72}}\tag{35}\mbox{}\end{equation}
Theorem \ref{rat72}. We have the following properties.
(i) The extended real-valued function \(f\) is Lebesgue measurable if and only if, for every open set \(O\) in \(\mathbb{R}\), its inverse image \(f^{-1}(O)\) is a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n}\).
(ii) The extended real-valued function \(f\) is Borel-measurable if and only if, for every open set \(O\) in \(\mathbb{R}\), its inverse image \(f^{-1}(O)\) is a Borel-measurable subset of \(\bar{\mathbb{R}}^{n}\).
Proof. To prove part (i), suppose that, for every open set \(O\) in \(\mathbb{R}\), the inverse image \(f^{-1}(O)\) is a Lebesgue-measurable subset of \(\bar{\mathbb{R}}^{n}\). If \(O=(a,+\infty )\), then
\[f^{-1}(O)=\{{\bf x}\in E:a<f({\bf x})<+\infty\}.\]
This shows that \(f\) is Lebesgue-measurable. For the converse, suppose that \(f\) is Lebesgue-measurable. Let \(O\) be an open set in \(\mathbb{R}\). Using Lemma \ref{ral70}, we have \(O=\bigcup_{k=1}^{\infty}(a_{k},b_{k})\). Since each of the sets
\[f^{-1}((a_{k},b_{k}))=\left\{{\bf x}\in E:a_{k}<f({\bf x})<b_{k}\right\}\]
is Lebesgue measurable by Corollary \ref{rac242}, it follows that the set
\[f^{-1}(O)=\bigcup_{k=1}^{\infty}f^{-1}((a_{k},b_{k}))\]
is also Lebesgue-measurable. Part (ii) can be similarly obtained, and the proof is complete. \(\blacksquare\)
Proposition. Suppose that the set \(\{{\bf x}\in E:f({\bf x})>a\}\) is Lebesgue measurable for each \(a\in\mathbb{Q}\). Then \(f\) is Lebesgue measurable.
Proof. For any real number \(a\), there is a sequence \(\{a_{k}\}_{k=1}^{\infty}\) in \(\mathbb{Q}\) satisfying \(a_{k}\geq a\) for all \(k\) and \(\lim_{k\rightarrow\infty}a_{k}=a\). Therefore, we have
\[\left\{{\bf x}\in E:f({\bf x})>a\right\}=\bigcup_{k=1}^{\infty}\left\{{\bf x}\in E:f({\bf x})>a_{k}\right\}.\]
This completes the proof. \(\blacksquare\)
Definition. The statement “Property P holds true almost everywhere on \(E\)” means that there exists a Lebesgue-measurable set \(N\) such that \(N\subseteq E\) with \(\nu (N)=0\) and the property P holds true at every point of \(E\setminus N\). In this case, we also write “Property P holds a.e. on \(E\)”. \(\sharp\)
For example, given any two Lebesgue measurable functions \(f\) and \(g\), we say that \(f=g\) a.e. on \(E\) when
\[\nu\left (\left\{{\bf x}\in E:f({\bf x})\neq g({\bf x})\right\}\right )=0.\]
Similarly, when we write \(f\leq g\) a.e. on \(E\), it means
\[\nu\left (\left\{{\bf x}\in E:f({\bf x})>g({\bf x})\right\}\right )=0.\]
\begin{equation}{\label{rap81}}\tag{36}\mbox{}\end{equation}
Proposition \ref{rap81}. Suppose that \(f\) is Lebesgue-measurable satisfying \(f=g\) a.e. on \(E\). Then \(g\) is Lebesgue-measurable and
\[\nu\left (\{{\bf x}\in E:f({\bf x})>a\}\right )=\nu\left (\{{\bf x}\in E:g({\bf x})>a\}\right ).\]
Proof. We define \(N=\{{\bf x}\in E:f({\bf x})\neq g({\bf x})\}\). Then \(\nu (N)=0\) and \(N\) is Lebesgue-measurable by Proposition \ref{rar61}. Since
\[\left\{{\bf x}\in E:g({\bf x})>a\right\}\cup N=\left\{{\bf x}\in E:f({\bf x})>a\right\}\cup N,\]
the Lebesgue measurability of \(f\) implies that the set
\[\{{\bf x}\in E:g({\bf x})>a\}\cup N\]
is Lebesgue-measurable. Therefore, the set \(\{{\bf x}\in E:g({\bf x})>a\}\) is also Lebegsue-measurable, since \(N\) is Lebesgue measurable. This shows that the function \(g\) is Lebegsue measurable. Moreover, we have
\begin{align*}
\nu\left (\{{\bf x}\in E:g({\bf x})>a\}\right ) & =\nu\left (\{{\bf x}\in E:g({\bf x})>a\}\cup N\right )\\ & =\nu\left (\{{\bf x}\in E:f({\bf x})>a\}\cup N\right )\\ & =\nu\left (\{{\bf x}\in E:f({\bf x})>a\}\right ).
\end{align*}
This completes the proof. \(\blacksquare\)
In view of the above proposition, it is natural to extend the definition of measurability to include functions which are defined only a.e. on \(E\). We also note that if \(f\) is Lebesgue-measurable on \(E\), then it is Lebesgue-measurable on any Lebesgue-measurable subset \(E_{1}\subseteq E\), since we have
\[\left\{{\bf x}\in E_{1}:f({\bf x})>a\right\}=E_{1}\cap\left\{{\bf x}\in E:f({\bf x})>a\right\}.\]
Suppose that \(\phi\) and \(f\) are extended real-valued functions defined on \(\bar{\mathbb{R}}\) and \(\bar{\mathbb{R}}^{n}\), respectively. Then, the composition \(\phi (f({\bf x}))\) may not be Lebesgue-measurable. When \(\phi\) is continuous, we have the following interesting result.
\begin{equation}{\label{rap73}}\tag{37}\mbox{}\end{equation}
Proposition \ref{rap73}. Suppose that the extended real-valued function \(\phi\) is continuous on \(\bar{\mathbb{R}}\), and that the extended real-valued function \(f\) is finite a.e. on \(E\), i.e., the composition \(\phi\circ f\) is defined a.e. on \(E\). If \(f\) is Lebesgue-measurable, then \(\phi\circ f\) is Lebesgue-measurable.
Proof. We may assume that \(f\) is finite everywhere on \(E\) by Proposition~\ref{rap81}. Given any open set \(O\) in \(\mathbb{R}\), we see that the inverse image \(\phi^{-1}(O)\) is open by the continuity of \(\phi\). Since
\[(\phi\circ f)^{-1}(O)=\left\{{\bf x}\in E:(\phi\circ f)({\bf x})\in O\right\}=f^{-1}(\phi^{-1}(O)),\]
using the Lebesgue measurability of \(f\) and Theorem \ref{rat72}, the set \((\phi\circ f)^{-1}(O)\) is Lebesgue-measurable. Finally, using Theorem\ref{rat72} again, we conclude that \(\phi\circ f\) is Lebesgue-measurable. This completes the proof. \(\blacksquare\)
\begin{equation}{\label{rar74}}\tag{38}\mbox{}\end{equation}
Remark \ref{rar74}. Suppose that we take the function \(\phi (t)=|t|\), \(\phi (t)=|t|^{p}\) for \(p>0\) or \(\phi (t)=e^{ct}\). If \(f\) is Lebesgue-measurable, then we can directly show that the functions \(|f|\), \(|f|^{p}\) for \(p>0\) and \(e^{cf}\) are Lebesgue-measurable without considering the composition, even if we do not assume that \(f\) is finite a.e. on \(E\). \(\sharp\)
Proposition. We have the following properties.
(i) Suppose that the extended real-valued functions \(f\) and \(g\) are Lebesgue-measurable. Then, the set
\[\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:f({\bf x})>g({\bf x})\right\}\]
is Lebesgue measurable.
(ii) Suppose that the extended real-valued function \(f\) is Lebesgue-measurable, and that \(\lambda\) is any real number. Then \(f+\lambda\) and \(\lambda f\) are Lebesgue-measurable.
(iii) Suppose that the extended real-valued functions \(f\) and \(g\) are Lebesgue-measurable. Then \(f+g\) is Lebesgue-measurable.
(iv) Suppose that the extended real-valued functions \(f\) and \(g\) are Lebesgue-measurable. Then \(f\cdot g\) is Lebesgue-measurable. If we further assume that \(g\neq 0\) a.e., then \(f/g\) is Lebesgue-measurable.
Proof. There exists a sequence \(\{r_{k}\}_{k=1}^{\infty}\) in \(\mathbb{Q}\)such that the following equalities are satisfied
\begin{align*}
\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:f({\bf x})>g({\bf x})\right\}
& =\bigcup_{k=1}^{\infty}\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:f({\bf x})>r_{k}>g({\bf x})\right\}\\
& =\bigcup_{k=1}^{\infty}\left [\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:f({\bf x})>r_{k}\right\}
\cap\left\{{\bf x}\in\bar{\mathbb{R}}^{n}:g({\bf x})<r_{k}\right\}\right ].
\end{align*}
This proves part (i). Part (ii) is obvious by definition. Suppose that \(g\) is Lebesgue-measurable. Then \(a-g\) is also Lebegsue-measurable for any real number \(a\) by part (ii). Since
\[\left\{{\bf x}\in E:f({\bf x})+g({\bf x})>a\right\}=\left\{{\bf x}\in E:f({\bf x})>a-g({\bf x})\right\},\]
part (iii) follows from part (i) immediately. Uisng Remark \ref{rar74}, if \(f\) is Lebesgue-measurable, then \(f^{2}=|f|^{2}\) is Lebesgue-measurable. Therefore, the following formula
\[fg=\frac{1}{4}\left [(f+g)^{2}-(f-g)^{2}\right ]\]
implies that \(fg\) is Lebesgue-measurable. The Lebesgue measurability of \(f/g\) with \(g\neq 0\) a.e. is left as an exercise. \(\blacksquare\)
Given an extended real-valued function \(f:E\rightarrow\bar{\mathbb{R}}\) defined on \(E\subseteq\bar{\mathbb{R}}^{n}\), the positive part and negative part of \(f\) are defined by
\[f^{+}=\max\left\{f,0\right\}\mbox{ and }f^{-}=\max\left\{-f,0\right\},\]
respectively. Then, the functions \(f^{+}\) and \(f^{-}\) are nonnegative. Moreover, we have
\[f=f^{+}-f^{-}\mbox{ and }|f|=f^{+}+f^{-}.\]
\begin{equation}{\label{rap80}}\tag{39}\mbox{}\end{equation}
Proposition \ref{rap80}. Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of Lebesgue measurable functions. Then, we have the following properties.
(i) The following functions
\[\sup_{k\geq 1}f_{k}\mbox{ and }\inf_{k\geq 1}f_{k}\]
are Lebesgue-measurable. In particular, if \(f\) is Lebesgue-measurable, then the functions
\[f^{+}=\max\{f,0\}\mbox{ and }f^{-}=\max\{-f,0\}=-\min\{f,0\}\]
are also Lebesgue-measurable.
(ii) The following functions
\[\liminf_{k\rightarrow\infty}f_{k}\mbox{ and }\limsup_{k\rightarrow\infty}f_{k}\]
are Lebesgue-measurable. In particular, if the limit function \(\lim_{k\rightarrow\infty}f_{k}\) exists a.e., then it is Lebesgue-measurable.
Proof. The following fact
\[\left\{{\bf x}\in E:\sup_{k\geq 1}f_{k}({\bf x})>a\right\}=\bigcup_{k=1}^{\infty}\left\{{\bf x}\in E:f_{k}({\bf x})>a\right\}\]
shows the Lebesgue measurability of the function \(\sup_{k\geq 1}f_{k}\). Since
\[\inf_{k\geq 1}f_{k}=-\sup_{k\geq 1}(-f_{k}),\]
this proves part (i). Part (ii) can be realized by the definitions of limit superior and inferior. This completes the proof. \(\blacksquare\)
The characteristic function of a set \(E\) is defined by
\[\chi_{E}({\bf x})=\left\{\begin{array}{ll}
1 & \mbox{if \({\bf x}\in E\)}\\
0 & \mbox{if \({\bf x}\not\in E\)}.
\end{array}\right .\]
It is obvious that the function \(\chi_{E}\) is Lebesgue-measurable if and only if the set \(E\) is Lebesgue-measurable.
Definition. Let \(\phi :\bar{\mathbb{R}}^{n}\rightarrow\bar{\mathbb{R}}\) be an extended real-valued function. We say that \(\phi\) is a simple function when
\begin{equation}{\label{raeq10}}\tag{40}
\phi =\sum_{i=1}^{n}\alpha_{i}\chi_{E_{i}},
\end{equation}
where \(E_{i}=\{x:\phi ({\bf x})=\alpha_{i}\}\). \(\sharp\)
We see that the simple function \(\phi\) is Lebesgue-measurable if and only if each of the sets \(E_{i}\) is Lebesgue-measurable. Note that the step function is also a simple function defined in (\ref{raeq10}). However, the converse is not necessary true.
\begin{equation}{\label{rat127}}\tag{41}\mbox{}\end{equation}
Theorem \ref{rat127}. We have the following properties.
(i) Suppose that the extended real-valued function \(f\) is nonnegative. Then, there exists a sequence \(\{\phi_{k}\}_{k=1}^{\infty}\) of nonnegative simple functions such that \(f\) is the limit function of \(\{\phi_{k}\}_{k=1}^{\infty}\) satisfying \(\phi_{k}\leq\phi_{k+1}\) for all \(k\).
(ii) Every extended real-valued function \(f\) can be written as the limit function of a sequence \(\{\phi_{k}\}_{k=1}^{\infty}\) of simple functions.
(iii) Suppose that the extended real-valued function \(f\) in either (i) or (ii) is Lebesgue-measurable. Then, the sequence \(\{\phi_{k}\}_{k=1}^{\infty}\) of simple functions can be taken to be Lebesgue-measurable.
Proof. To prove part (i), for \(f\geq 0\), we define
\[\phi_{k}({\bf x})=\left\{\begin{array}{ll}
{\displaystyle \frac{j-1}{2^{k}},} & {\displaystyle \mbox{if }\frac{j-1}{2^{k}}\leq
f({\bf x})<\frac{j}{2^{k}}\mbox{ for }j=1,\cdots ,k2^{k}} \\
k, & \mbox{if \(f({\bf x})\geq k\).}
\end{array}\right .\]
It is clear that each \(\phi_{k}\) is a nonnegative simple function defined everywhere on the domain of \(f\). We also have \(\phi_{k}\leq\phi_{k+1}\) for all \(k\). We consider the following cases.
- Suppose that \(f({\bf x})=+\infty\). Then \(\phi_{k}({\bf x})=k\). In this case, we have
\[\lim_{k\rightarrow\infty}\phi_{k}({\bf x})=\lim_{k\rightarrow\infty} k=+\infty=f({\bf x}).\] - Suppose that \(f({\bf x})<+\infty\). Since
\[0\leq f({\bf x})-\phi_{k}({\bf x})\leq 2^{-k},\]
we have
\[\lim_{k\rightarrow\infty}\phi_{k}({\bf x})=f({\bf x}).\]
To prove part (ii), since \(f=f^{+}-f^{-}\), where \(f^{+}\) and \(f^{-}\) are nonnegative, using part (i), there exist two sequences \(\{\phi^{\prime}_{k}\}_{k=1}^{\infty}\) and \(\{\phi^{\prime\prime}_{k}\}_{k=1}^{\infty}\) satisfying
\[\lim_{k\rightarrow\infty}\phi^{\prime}_{k}=f^{+}\mbox{ and }\lim_{k\rightarrow\infty}\phi^{\prime\prime}_{k}=f^{-}.\]
Then \(\phi_{k}\equiv\phi^{\prime}_{k}-\phi^{\prime\prime}_{k}\) are simple functions for all \(k\) and
\[\lim_{k\rightarrow\infty}\phi_{k}=\lim_{k\rightarrow\infty}\phi^{\prime}_{k}-
\lim_{k\rightarrow\infty}\phi^{\prime\prime}_{k}=f^{+}-f^{-}=f.\]
To prove part (iii), for \(f\geq 0\), let
\[E_{j}({\bf x})=\left\{{\bf x}\in E:\frac{j-1}{2^{k}}\leq f({\bf x})\leq\frac{j}{2^{k}}\right\}\mbox{ for }j=1,\cdots ,k2^{k}.\]
Then, we have
\begin{equation}{\label{raeq74}}\tag{42}
\phi_{k}(x)=\sum_{j=1}^{k2^{k}}\frac{j-1}{2^{k}}\cdot\chi_{E_{j}}+k\cdot\chi_{\{{\bf x}\in E:f({\bf x})\geq k\}}.
\end{equation}
Suppose that \(f\) is Lebesgue-measurable. Then, all the sets \(E_{j}\) and \(\{{\bf x}\in E:f({\bf x})\geq k\}\) in (\ref{raeq74}) are Lebesgue-measurable. This shows that the simple function \(\phi_{k}\) is Lebesgue-measurable. In general, we have \(f=f^{+}-f^{-}\), where \(f^{+}\) and \(f^{-}\) are nonnegative. Therefore, we can similarly obtain the desired results, and the proof is complete. \(\blacksquare\)
\begin{equation}{\label{rat88}}\tag{43}\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 \(F\). \(\sharp\)
Theorem. (Lusin’s Theorem) Suppose that the extended real-valued function \(f\) is defined and finite on a Lebesgue-measurable set \(E\). Then, the function \(f\) is Lebesgue-measurable if and only if, given \(\epsilon >0\), there exists a closed set \(F\subseteq E\) such that \(\nu (E\setminus F)<\epsilon\) and \(f\) is continuous on \(F\). \(\sharp\)
