Jacques Carabain (1834-1933) was a Dutch-Belgian painter.
We have sections
- Integration of Simple Functions
- Integration of Nonnegative Functions
- Integration of Extended Real-Valued Function
- The Radon-Nikodym Theorem
- Integration on Product Spaces
We are going to generalize the Lebesgue integral by considering the extended real-valued functions defined on an abstract measure space instead of Lebesgue measure space.
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Integration of Simple Functions.
Suppose that the simple function \(\phi :(X,\mathfrak{M},\mu)\rightarrow [0,+\infty ]\) defined on a measure space \((X,\mathfrak{M},\mu)\) is measurable and has the form of
\[\phi=\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}},\]
where \(\alpha_{1},\cdots ,\alpha_{n}\) are the distinct values of \(\phi\). Given any measurable set \(E\in\mathfrak{M}\), we define the integral
\begin{equation}{\label{raeq2}}\tag{1}
\int_{E}\phi d\mu =\sum_{i=1}^{n}\alpha_{i}\mu\left (A_{i}\cap E\right ).
\end{equation}
The convention \(0\cdot\infty=0\) is used here. It may happen that \(\alpha_{i}=0\) and \(\mu (A_{i}\cap E)=\infty\) for some \(i\).
\begin{equation}{\label{rap89}}\tag{2}\mbox{}\end{equation}
Proposition \ref{rap89}. We have the following properties.
(i) Let the simple function \(\phi\) defined on a measure space \((X,\mathfrak{M},\mu)\) be nonnegative and measurable. Given any \(E\in\mathfrak{M}\), we define
\[\zeta (E)=\int_{E}\phi d\mu .\]
Then \(\zeta\) is a measure on \(\mathfrak{M}\). In particular, given a sequence \(\{E_{k}\}_{k=1}^{\infty}\) of disjoint measurable sets with \(E=\bigcup_{k=1}^{\infty}E_{k}\), we have
\[\int_{E}\phi d\mu=\sum_{k=1}^{\infty}\int_{E_{k}}\phi d\mu.\]
(ii) Let the simple functions \(\phi\) and \(\psi\) defined on a measure space \((X,\mathfrak{M},\mu)\) be nonnegative and measurable. Then, we have
\[\int_{X}(\phi +\psi )d\mu =\int_{X}\phi d\mu +\int_{X}\psi d\mu .\]
Proof. To prove part (i), let \(\{E_{k}\}_{k=1}^{\infty}\) be a sequence of disjoint measurable sets with \(E=\bigcup_{k=1}^{\infty}E_{k}\). By the countable additivity of \(\mu\), we have
\begin{align*}
\zeta (E) & =\int_{E}\phi d\mu =\sum_{i=1}^{n}\alpha_{i}\mu\left (A_{i}\cap E\right )\\
& =\sum_{i=1}^{n}\alpha_{i}\left [\sum_{k=1}^{\infty}\mu\left (A_{i}\cap E_{k}\right )\right ]
\\ & =\sum_{k=1}^{\infty}\sum_{i=1}^{n}\alpha_{i}\mu\left (A_{i}\cap E_{k}\right )\\
& =\sum_{k=1}^{\infty}\int_{E_{k}}\phi d\mu =\sum_{k=1}^{\infty}\zeta (E_{k}).
\end{align*}
We also have \(\zeta (\emptyset )=0\). Therefore, \(\zeta\) is not identically \(+\infty\). This shows that \(\zeta\) is a measure. To prove part (ii), suppose that
\[\phi=\sum_{i=1}^{n}\alpha_{i}\chi_{A_{i}}\mbox{ and }\psi=\sum_{j=1}^{m}\beta_{j}\chi_{B_{j}}.\]
Let \(E_{ij}=A_{i}\cap B_{j}\). Then, we have
\begin{align*} \int_{E_{ij}}\left (\phi +\psi\right )d\mu & =\left (\alpha_{i}+\beta_{j}\right )\mu (E_{ij})\\ & =
\alpha_{i}\mu (E_{ij})+\beta_{j}\mu (E_{ij})\\ & =\int_{E_{ij}}\phi d\mu +\int_{E_{ij}}\psi d\mu .\end{align*}
Since \(X\) is the disjoint union of the sets \(E_{ij}\) for \(i=1,\cdots,n\) and \(j=1,\cdots,m\), using part (i), we have
\begin{align*} \int_{X}\left (\phi +\psi\right )d\mu & =\sum_{i,j}\int_{E_{ij}}\left (\phi +\psi\right )d\mu
\\ & =\sum_{i,j}\int_{E_{ij}}\phi d\mu +\sum_{i,j}\int_{E_{ij}}\psi d\mu
\\ & =\int_{X}\phi d\mu +\int_{X}\psi d\mu .\end{align*}
This completes the proof. \(\blacksquare\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Integration of Nonnegative Functions.
We shall present two approaches to define the integral of nonnegative functions.
First Approach.
Suppose that the nonnegative function \(f:(X,\mathfrak{M},\mu)\rightarrow [0,+\infty ]\) defined on a measure space \((X,\mathfrak{M},\mu)\) is measurable. We can define the integral by mimicing the lower Riemann integral. Given any measurable set \(E\in\mathfrak{M}\), the integral of \(f\) on \(E\) with respect to the measure \(\mu\) is defined by
\[\int_{E}fd\mu =\sup_{0\leq\phi\leq f}\int_{E}\phi d\mu ,\]
where the supremum is taken over all measurable simple functions \(\phi\) satisfying \(0\leq\phi\leq f\). We can also observe that if \(f\) happens to be a simple function, then the definition coincides with (\ref{raeq2}).
Proposition. We have the following properties.
(i) If \(0\leq f\leq g\), then
\[\int_{E}fd\mu\leq\int_{E}gd\mu .\]
(ii) For any two measurable sets \(A\) and \(B\), if \(f\geq 0\), then
\[\int_{A}fd\mu\leq\int_{B}fd\mu .\]
(iii) Given a nonnegative constant \(c\), if \(f\geq 0\), then
\[\int_{E}cfd\mu =c\int_{E}fd\mu .\]
(iv) If \(f(x)=0\) for all \(x\in E\), then
\[\int_{E}fd\mu =0\]
even if \(\mu (E)=\infty\).
(v) If \(\mu (E)=0\), then
\[\int_{E}fd\mu =0\]
even if \(f(x)=0\) for every \(x\in E\).
(vi) If \(f\geq 0\), then
\[\int_{E}fd\mu =\int_{X}\chi_{E}fd\mu .\]
\begin{equation}{\label{rat90}}\tag{3}\mbox{}\end{equation}
Lemma \ref{rat90}. Suppose that the nonnegative and extended real-valued function \(f:(X,\mathfrak{M})\rightarrow [0,+\infty ]\) defined on a measurable space \((X,\mathfrak{M})\) is measurable. Then, there exists a sequence of simple functions \(\{s_{n}\}_{n=1}^{\infty}\) satisfying
\[\phi_{n+1}\geq\phi_{n}\mbox{ for all }n\mbox{ and }\lim_{n\rightarrow\infty}\phi_{n}(x)=f(x)\mbox{ for each }x\in X.\]
\begin{equation}{\label{rap92}}\tag{4}\mbox{}\end{equation}
Lemma \ref{rap92}. Let \(\mu\) be a nonnegative measure defined on a measurable space \((X,\mathfrak{M})\). Then, we have the following properties.
(i) We have \(\mu (\emptyset )=0\).
(ii) Let \(A_{1},\cdots ,A_{n}\) be pairwise disjoint measurable sets. Then, we have
\[\mu\left (\bigcup_{i=1}^{n}A_{i}\right )=\sum_{i=1}^{n}\mu (A_{i}).\]
(iii) Given any measurable sets \(A\) and \(B\) satisfying \(A\subseteq B\), we have \(\mu (A)\leq\mu (B)\).
(iv) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\) satisfying
\[A_{n}\subseteq A_{n+1}\mbox{ for all }n\mbox{ and }A=\bigcup_{n=1}^{\infty}A_{n}.\]
Then, we have
\[\lim_{n\rightarrow\infty}\mu (A_{n})=\mu (A).\]
(v) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\) satisfying
\[A_{n+1}\subseteq A_{n}\mbox{ for all }n,\quad\mu (A_{1})<\infty\mbox{ and }A=\bigcap_{n=1}^{\infty}A_{n}.\]
Then, we have
\[\lim_{n\rightarrow\infty}\mu (A_{n})=\mu (A).\]
(vi) Given any sequence \(\{A_{i}\}_{i=1}^{\infty}\) in \(\mathfrak{M}\), we have
\[\mu\left (\bigcup_{n=1}^{\infty}A_{n}\right )\leq\sum_{n=1}^{\infty}\mu (A_{n}).\]
The above Lemmas \ref{rat90} and \ref{rap92} can refer to the page General Measures.
\begin{equation}{\label{rat95}}\tag{5}\mbox{}\end{equation}
Theorem \ref{rat95}. (Monotone Convergence Theorem).
Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative and measurable functions defined on \(X\) satisfying
\[f_{k}(x)\leq f_{k+1}(x)\mbox{ for each }x\in X\mbox{ and for all }k\]
and
\[\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on }X.\]
Then \(f\) is measurable and
\[\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu =\int_{X}fd\mu .\]
Proof. The limit function \(f\) is obviously measurable. Suppose that
\[N=\{x\in X:f_{k}(x)\not\rightarrow f(x)\}.\]
Then \(\mu (N)=0\). Therefore, we have
\[\int_{X\setminus N}f_{k}d\mu =\int_{X}f_{k}d\mu\mbox{ and }\int_{X\setminus N}fd\mu =\int_{X}fd\mu .\]
Therefore, it is without lose of generality to assume
\[\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ for each }x\in X.\]
Since
\[\int_{X}f_{k}d\mu\leq\int_{X}f_{k+1}d\mu ,\]
there exists \(\alpha\in\bar{\mathbb{R}}_{+}\) satisfying
\[\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu=\alpha,\]
where \(\alpha\) can be \(+\infty\). Since \(f_{k}\leq f\), we also have
\begin{equation}{\label{raeq94}}\tag{6}
\alpha =\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu\leq\int_{X}fd\mu .
\end{equation}
Using Lemma \ref{rat90}, there exists a simple measurable function \(\phi\) satisfying \(0\leq\phi\leq f\). Let \(c\) be a constant satisfying \(0<c<1\). We define
\[E_{k}=\left\{x\in X:f_{k}(x)\geq c\phi (x)\right\}.\]
Then, each \(E_{k}\) is measurable. Since \(f_{k}\leq f_{k+1}\), it follows \(E_{k}\subseteq E_{k+1}\). We are going to claim
\begin{equation}{\label{raeq91}}\tag{7}
X=\bigcup_{k=1}^{\infty}E_{k}.
\end{equation}
For any \(x\in X\), suppose that \(f(x)=0\). Then \(\phi (x)=0\), since \(0\leq\phi\leq f\). This shows \(x\in E_{1}\). Now, we assume \(f(x)>0\). Since \(0<c<1\), we have \(c\phi (x)<f(x)\). Suppose that \(x\not\in E_{k}\) for each \(k\). Then, we have \(c\phi (x)>f_{k}(x)\) for each \(k\), which implies \(c\phi (x)\geq f(x)\) by taking \(k\rightarrow\infty\). A contradiction occurs for \(0\leq\phi\leq f\). Therefore, we obtain \(x\in E_{k}\) for some \(k\), i.e., \(x\in\bigcup_{k=1}^{\infty}E_{k}\). This shows the equality (\ref{raeq91}). Using part (i) of Proposition \ref{rap89} and part (iv) of Lemma \ref{rap92}, we have
\begin{equation}{\label{raeq93}}\tag{8}
\lim_{k\rightarrow\infty}\int_{E_{k}}\phi d\mu =\lim_{k\rightarrow\infty}\zeta(E_{k})=\zeta(X)=\int_{X}\phi d\mu .
\end{equation}
On the other hand, since \(f_{k}\geq 0\), we also have
\[\int_{X}f_{k}d\mu\geq\int_{E_{k}}f_{k}d\mu\geq c\int_{E_{k}}\phi d\mu .\]
By taking the limit and using (\ref{raeq93}), we obtain
\begin{align*} \alpha & =\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu\\ & \geq c\lim_{k\rightarrow\infty}\int_{E_{k}}\phi d\mu \\ & =c\int_{X}\phi d\mu .\end{align*}
Since the above inequality holds true for every \(0<c<1\), we obtain
\[\alpha\geq\int_{X}\phi d\mu\]
for every simple function \(\phi\) satisfying \(0\leq\phi\leq f\). By taking the supremum, we obtain
\[\alpha\geq\sup_{0\leq\phi\leq f}\int_{X}\phi d\mu=\int_{X}fd\mu .\]
Combining the above inequality with (\ref{raeq94}), we complete the proof. \(\blacksquare\)
Theorem. (Fatou’s Lemma). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative measurable functions defined on \(X\). Then, we have
\[\int_{X}\left (\liminf_{k\rightarrow\infty} f_{k}\right )d\mu\leq\liminf_{k\rightarrow\infty}\int_{X}f_{k}d\mu .\]
In particular, if
\[\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on } E,\]
we have
\[\int_{E}fd\mu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\mu .\]
Proof. We define
\[g_{k}(x)=\inf_{i\geq k}f_{i}(x).\]
Then, we see that each \(g_{k}\) is measurable. Therefore, we have
\[\int_{X}g_{k}d\mu\leq\int_{X}f_{k}d\mu\mbox{ for all }k.\]
Since \(0\leq g_{k}\leq g_{k+1}\) for all \(k\), by the definition of limit inferior, we have
\[\lim_{k\rightarrow\infty}g_{k}=\sup_{k}g_{k}=\sup_{k}\inf_{i\geq k}f_{i}=\liminf_{k\rightarrow\infty}f_{k}.\]
Therefore, using the monotone convergence Theorem \ref{rat95}, we obtain
\begin{align*}
\int_{X}\left (\liminf_{k\rightarrow\infty} f_{k}\right )d\mu & =\int_{X}\left (\lim_{k\rightarrow\infty}g_{k}\right )d\mu\\
& =\lim_{k\rightarrow\infty}\int_{X}g_{k}d\mu \\
& =\liminf_{k\rightarrow\infty}\int_{X}g_{k}d\mu\\ & \leq\liminf_{k\rightarrow\infty}\int_{X}f_{k}d\mu\mbox{ (since \(g_{k}\leq f_{k}\))}.
\end{align*}
This completes the proof. \(\blacksquare\)
Using the Fatou’s Lemma, we can prove another version of monotone convergence theorem as shown below.
Theorem. (Monotone Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative and measurable functions defined on \(X\) satisfying \(f_{k}\leq f\) for all \(k\) and
\[\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on }X.\]
Then, we have
\[\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu =\int_{X}d\mu .\]
Proposition. Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative and measurable functions defined on \(X\). Then, we have
\[\int_{X}\sum_{k=1}^{\infty}f_{k}(x)d\mu =\sum_{k=1}^{\infty}\int_{X}f_{k}d\mu .\]
\begin{equation}{\label{rap41}}\tag{9}\mbox{}\end{equation}
Proposition \ref{rap41}. Suppose that the nonnegative function \(f:(X,\mathfrak{M})\rightarrow [0,+\infty ]\) is measurable. Given a measure \(\mu_{1}\) on \((X,\mathfrak{M})\), we define
\[\mu_{2}(E)=\int_{E}fd\mu_{1}\mbox{ for any }E\in\mathfrak{M}.\]
Then \(\mu_{2}\) is a measure on \(\mathfrak{M}\) and
\[\int_{X}gd\mu_{2}=\int_{X}gfd\mu_{1}\]
for each nonnegative measurable function \(g:(X,\mathfrak{M})\rightarrow [0,+\infty ]\).
Second Approach.
Let \(f\) be nonnegative on a measurable set \(E\). We can mimic the lower Riemann sum by considering
\[{\cal L}({\cal P})=\sum_{k=1}^{n}\left [\mu (E_{k})\cdot\inf_{{\bf x}\in E_{k}}f({\bf x})\right ],\]
where \({\cal P}=\{E_{1},\cdots,E_{n}\}\) is a decomposition of \(E\) satisfying that \(E=\bigcup_{k=1}^{n}E_{k}\) is a finite number of disjoint union of measurable sets \(E_{k}\). By mimicing the lower Riemann integral, the integral of \(f\) on \(E\) with respect to \(\mu\) is defined by
\begin{equation}{\label{raeq43}}\tag{10}
\int_{E}fd\mu =\sup_{{\cal P}}{\cal L}({\cal P})=\sup_{{\cal P}}\sum_{k=1}^{n}\left [\mu (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 measurable sets \(E_{k}\). The convention \(0\cdot\infty=\infty\cdot 0=0\) is adopted. Although (\ref{raeq43}) does not require the measurability of \(f\), many of the familiar properties of the integral are valid only for measurable functions. Therefore, all the functions are assumed to be measurable. The definition reduces to the Lebesgue integral when \(X=\bar{\mathbb{R}}^{n}\), the \(\sigma\)-field \(\mathfrak{M}\) is taken to be the class of Lebesgue measurable sets, the measure \(\mu\) is taken to be the Lebesgue measure and \(f\) is assumed to be Lebesgue measurable.
Proposition. Let \((X,\mathfrak{M},\mu )\) be a measure space, and let \(\phi\) be a nonnegative simple measurable function defined on a measurable set \(E\). Suppose that \(\phi\) takes values \(\alpha_{1},\cdots ,\alpha_{n}\) on disjoint sets \(E_{1},\cdots ,E_{n}\). Then, we have
\[\int_{E}\phi d\mu =\sum_{k=1}^{n}\alpha_{k}\mu (E_{k}).\]
Proposition. Let \(\{\phi_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative and simple measurable functions defined on a measurable set \(E\). Suppose that
\[\phi_{k}\leq\phi_{k+1}\mbox{ for all }k\mbox{ and }\lim_{k\rightarrow\infty}\phi_{k}(x)=f(x)\mbox{ a.e. on }E,\]
Then, we have
\[\lim_{k\rightarrow\infty}\int_{E}\phi_{k}d\mu =\int_{E}fd\mu .\]
The properties shown in the first approach are also valid in the second approach.
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Integration of Extended Real-Valued Function.
A nonnegative function \(f\) is called integrable on \(E\) with respect to \(\mu\) when it is measurable and
\[\int_{E}fd\mu <+\infty\]
for any one of the above two approaches. An arbitrary function \(f\) is said to be integrable on \(E\) with respect to \(\mu\) when \(f^{+}\) and \(f^{-}\) are integrable on \(E\)with respect to \(\mu\). In this case, we define
\[\int_{E}fd\mu =\int_{E}f^{+}d\mu -\int_{E}f^{-}d\mu .\]
This definition is valid to the above two approaches.
Proposition. Given any two constants \(\alpha\) and \(\beta\), suppose that \(f\) and \(g\) are integrable with respect to \(\mu\). Then \(\alpha f+\beta g\) is integrable with respect to \(\mu\) satisfying
\[\int_{X}\left (\alpha f+\beta g\right )d\mu =\alpha\int_{X}fd\mu+\beta\int_{X}gd\mu .\]
Proposition. Suppose that \(f\) is integrable with respect to \(\mu\). Then, we have
\[\left |\int_{X}fd\mu\right |\leq\int_{X}|f|d\mu .\]
We write \(f\sim g\) when \(f=g\) a.e. \([\mu ]\) on \(X\). It is easy to see that “$\sim$” is an equivalence relation. The transitivity is a consequence of the fact that the union of two sets of measure zero has measure zero.
Proposition. Suppose that \(f=g\) a.e. \([\mu ]\) on \(X\). Then, we have
\begin{equation}{\label{raeq4}}\tag{11}
\int_{E}fd\mu =\int_{E}gd\mu
\end{equation}
for any \(E\in\mathfrak{M}\).
Proof. Let \(N=\{x:f(x)\neq g(x)\}\). Then \(\mu(N)=0\). We also see that \(E\) is the union of the disjoint set \(E\setminus N\) and \(E\cap N\) satisfying \(\mu (E\cap N)=0\) and \(f=g\) on \(E\setminus N\), which implies (\ref{raeq4}). \(\blacksquare\)
The above proposition says that the sets of measure zero are negligible in integration.
Proposition. We have the following properties.
(i) Suppose that the nonnegative function \(f:(X,\mathfrak{M})\rightarrow [0,+\infty ]\) is measurable. Suppose that
\[\int_{E}fd\mu =0\mbox{ for any }E\in\mathfrak{M}.\]
Then \(f=0\) a.e. \([\mu]\) on \(E\).
(ii) Suppose that \(f\) is integrable with respect to \(\mu\) and
\[\left |\int_{X}fd\mu\right |=\int_{X}|f|d\mu .\]
Then, there is a constant \(\alpha\) satisfying \(\alpha f=|f|\) a.e. \([\mu ]\) on \(X\). \(\sharp\)
Proposition. Let \(\{E_{n}\}_{n=1}^{\infty}\) be a sequence of measurable sets in \(X\) satisfying
\[\sum_{n=1}^{\infty}\mu (E_{n})<\infty .\]
Then, almost all \(x\in X\) lie in at most finitely many of the sets \(E_{n}\). \(\sharp\)
Theorem. (Monotone Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions defined on a measurable set \(E\). We have the following properties.
(i) Suppose that
\[f_{k}\leq f_{k+1}\mbox{ for all }\mbox{ and }\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on }E,\]
and that there exists an integrable function \(\phi\) on \(E\) satisfying \(f_{k}(x)\geq\phi(x)\) a.e. on \(E\) for all \(k\). Then, we have
\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\mu =\int_{E}fd\mu .\]
(ii) Suppose that
\[f_{k+1}\leq f_{k}\mbox{ for all }k\mbox{ and }\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on }E,\]
and that there exists an integrable function \(\phi\) on \(E\) satisfying \(f_{k}(x)\leq\phi(x)\) a.e. on \(E\) for all \(k\). Then, we have
\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\mu =\int_{E}fd\mu .\]
Theorem. (Uniform Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of integrable functions defined on a measurable set \(E\) with \(\mu (E)<+\infty\), and let \(\{f_{k}\}_{k=1}^{\infty}\) converges uniformly to \(f\) on \(E\). Then \(f\) is integrable on \(E\) and
\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\mu =\int_{E}fd\mu .\]
Theorem. (Fatou’s Lemma). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions defined on a measurable set \(E\). Suppose that there exists an integrable function \(\phi\) on \(E\) satisfying \(f_{k}(x)\geq\phi(x)\) a.e. on \(E\) for all \(k\). Then, we have
\[\int_{E}\left (\liminf_{k\rightarrow\infty}f_{k}\right )d\mu\leq\liminf_{k\rightarrow\infty}\int_{E}f_{k}d\mu .\]
Theorem. (Fatou’s Lemma). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions defined on a measurable set \(E\). Suppose that there exists an integrable function \(\phi\) on \(E\) satisfying \(f_{k}(x)\leq\phi(x)\) a.e. on \(E\) for all \(k\). Then, we have
\[\int_{E}\left (\limsup_{k\rightarrow\infty}f_{k}\right )d\mu\geq\limsup_{k\rightarrow\infty}\int_{E}f_{k}d\mu .\]
Theorem. (Dominated Convergence Theorem). Let \(\{f_{n}\}_{n=1}^{\infty}\) be a sequence of measurable functions defined on \(X\) satisfying
\[f(x)=\lim_{n\rightarrow\infty}f_{n}(x)\mbox{ a.e. }[\mu ].\]
Suppose that there exists an integrable function \(g\) with respect to \(\mu\) satisfying \(|f_{n}(x)|\leq g(x)\) a.e. \([\mu ]\) on \(X\). Then \(f\) is integrable with respect to \(\mu\) satisfying
\[\lim_{n\rightarrow\infty}\int_{X}\left |f_{n}-f\right |d\mu =0\]
and
\[\lim_{n\rightarrow\infty}\int_{X}f_{n}d\mu =\int_{X}fd\mu .\]
Theorem. (Bounded Convergence Theorem). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of measurable functions defined on a measurable set \(E\) satisfying
\[\lim_{k\rightarrow\infty}f_{k}(x)=f(x)\mbox{ a.e. on }E.\]
Suppose that \(\mu (E)<+\infty\), and that there is a finite constant \(k\) satisfying \(|f_{k}(x)|\leq k\) a.e. on \(E\). Then, we have
\[\lim_{k\rightarrow\infty}\int_{E}f_{k}d\mu =\int_{E}fd\mu .\]
Proposition. Suppose that \(\{f_{n}\}_{n=1}^{\infty}\) is a sequence of measurable functions defined a.e. \([\mu ]\) on \(X\) satisfying
\[\sum_{n=1}^{\infty}\int_{X}|f_{n}|d\mu <\infty .\]
Then, the series
\[f(x)=\sum_{n=1}^{\infty}f_{n}(x)\mbox{ converges a.e. \([\mu ]\) on \(X\)}\]
such that \(f\) is integral with respect to \(\mu\) and
\[\int_{X}fd\mu =\sum_{n=1}^{\infty}\int_{X}f_{n}d\mu .\]
Definition. Let \((X,\mathfrak{M})\) be a measurable space, and let \(\{\mu_{k}\}_{k=1}^{\infty}\) be a sequence of set functions defined on \((X,\mathfrak{M})\). We say that \(\{\mu_{k}\}_{k=1}^{\infty}\) converges setwise to the set function \(\mu\) when, for each \(E\in\mathfrak{M}\), we have
\[\mu (E)=\lim_{k\rightarrow\infty}\mu_{k}(E).\]
With the above notion, we can generalize the Fatou’s Lemma and dominated convergence theorem.
Theorem. Let \((X,\mathfrak{M})\) be a measurable space, and let \(\{\mu_{k}\}_{k=1}^{\infty}\) be a sequence of measures defined on \((X,\mathfrak{M})\) such that \(\{\mu_{k}\}_{k=1}^{\infty}\) converges setwise to a measure \(\mu\). Let \(\{f_{k}\}_{k=1}^{\infty}\) be a sequence of nonnegative measurable functions which converge pointwise to the function \(f\). Then, we have
\[\int_{X}fd\mu\leq\liminf_{k\rightarrow\infty}\int_{X}f_{k}d\mu_{k}.\]
Theorem. Let \((X,\mathfrak{M})\) be a measurable space, and let \(\{\mu_{k}\}_{k=1}^{\infty}\) be a sequence of measures defined on \((X,\mathfrak{M})\) such that \(\{\mu_{k}\}_{k=1}^{\infty}\) converges setwise to a measure \(\mu\). Let \(\{f_{k}\}_{k=1}^{\infty}\) and \(\{g_{k}\}_{k=1}^{\infty}\) be two sequences of measurable functions which converge pointwise to the function \(f\) and \(g\), respectively. Suppose that \(|f_{k}|\leq g_{k}\) and
\[\lim_{k\rightarrow\infty}\int_{X}g_{k}d\mu_{k}=\int_{X}gd\mu <+\infty\]
Then, we have
\[\lim_{k\rightarrow\infty}\int_{X}f_{k}d\mu_{k}=\int_{X}fd\mu .\]
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
The Radon-Nikodym Theorem.
Suppose that the nonnegative function \(f:(X,\mathfrak{M})\rightarrow [0,+\infty ]\) is measurable. Given a measure \(\mu_{1}\) on \((X,\mathfrak{M})\), we define
\[\mu_{2}(E)=\int_{E}fd\mu_{1}\mbox{ for any }E\in\mathfrak{M}.\]
Then, Proposition \ref{rap41} says that \(\mu_{2}\) is a measure on \(\mathfrak{M}\). Suppose that \(\mu_{1}(E)=0\). Then, we have
\[0=\int_{E}fd\mu_{1}=\mu_{2}(E).\]
This says that \(\mu_{2}\ll\mu_{1}\). The converse is shown below.
\begin{equation}{\label{rat43}}\tag{12}\mbox{}\end{equation}
Theorem \ref{rat43}. (Radon-Nikodym Theorem). Let \((X,\mathfrak{M},\mu_{1})\) be a \(\sigma\)-finite measure space, and let \(\mu_{2}\) be a measure defined on \((X,\mathfrak{M})\) satisfying \(\mu_{2}\ll\mu_{1}\). Then, there exists a nonnegative measurable function \(f\) satisfying
\[\mu_{2}(E)=\int_{E}fd\mu_{1}\mbox{ for each }E\in\mathfrak{M}.\]
The function \(f\) is unique in the sense that if \(g\) is any measurable function with this property, then \(g=f\) a.e. \([\mu_{1}]\). \(\sharp\)
The function \(f\) given in Theorem \ref{rat43} is called the Radon-Nokodym derivative of \(\mu_{2}\) with respect to \(\mu_{1}\). It is also denoted by
\[f=\left [\frac{d\mu_{2}}{d\mu_{1}}\right ].\]
Proposition. The Radon-Nikodym derivative has the following properties.
(i) Suppose that \(\mu_{2}\ll\mu_{1}\), and that \(f\) is a nonnegative measurable function. Then, we have
\[\int_{X}fd\mu_{2}=\int_{X}f\left [\frac{d\mu_{2}}{d\mu_{1}}\right ]d\mu_{1}.\]
(ii) We have
\[\left [\frac{d(\mu_{2}+\mu_{3})}{d\mu_{1}}\right ]=\left [\frac{d\mu_{2}}{d\mu_{1}}\right ]+\left [\frac{d\mu_{3}}{d\mu_{1}}\right ].\]
(iii) Suppose that \(\mu_{3}\ll\mu_{2}\ll\mu_{1}\). Then, we have
\[\left [\frac{d\mu_{3}}{d\mu_{1}}\right ]=\left [\frac{d\mu_{3}}{d\mu_{2}}\right ]\left [\frac{d\mu_{2}}{d\mu_{1}}\right ].\]
(iv) Suppose that \(\mu_{2}\ll\mu_{1}\) and \(\mu_{1}\ll\mu_{2}\). Then, we have
\[\left [\frac{d\mu_{2}}{d\mu_{1}}\right ]=\left [\frac{d\mu_{1}}{d\mu_{2}}\right ]^{-1}.\]
Using the Radon-Nikodym theorem, we can prove the following decomposition theorem.
\begin{equation}{\label{rat42}}\tag{13}\mbox{}\end{equation}
Theorem \ref{rat42}. (Lebesgue Decomposition). Let \((X,\mathfrak{M},\mu_{1})\) be a \(\sigma\)-finite measure space. Given any \(\sigma\)-finite measure \(\mu_{2}\) defined on \((X,\mathfrak{M})\), there exist a measure \(\mu_{s}\) that is mutually singular with respect to \(\mu_{1}\) and a measure \(\mu_{a}\) that is absolutely continuous with respect to \(\mu_{1}\) satisfying \(\mu_{2}=\mu_{s}+\mu_{a}\). The measures \(\mu_{s}\) and \(\mu_{a}\) are unique. \(\sharp\)
For \(0<p<+\infty\), the space \(L^{p}(\mu )\) denotes the collection of all measurable functions \(f\) defined on \((X,\mathfrak{M})\) satisfying
\[\int_{X}|f|^{p}d\mu <+\infty.\]
We also define the norm of \(f\) in \(L^{p}(\mu)\) by
\[\parallel f\parallel_{p}=\left (\int_{X}|f|^{p}d\mu\right )^{1/p}\mbox{ for }0<p<+\infty\]
and
\begin{align*} \parallel f\parallel_{\infty} & =\mbox{ess sup}|f|\\ & =\inf\left\{\alpha :\mu (\{{\bf x}\in E:f({\bf x})>\alpha\})=0\right\}.\end{align*}
A measurable function \(f\) is said to be essentially bounded when \(\mbox{ess sup}|f|<+\infty\). The space of all measurable functions that are essentially bounded is denoted by \(L^{\infty}(\mu )\). We can similarly obtain the H\”{o}lder’s inequality
\[\parallel fg\parallel_{1}\leq\parallel f\parallel_{p}\cdot\parallel g\parallel_{q}\mbox{ for }1\leq p\leq +\infty\mbox{ and }1/p+1/q=1,\]
and the Minkiwski’s inequality
\[\parallel f+g\parallel_{p}\leq\parallel f\parallel_{p}+\parallel g\parallel_{p}
\mbox{ for }1\leq p\leq +\infty.\]
From the H\”{o}lder’s inequality, for each \(g\in L^{q}(\mu )\), we can define a linear functional \(l\) on \(L^{p}(\mu )\) by
\[l(f)=\int_{X}fgd\mu .\]
It is not difficult to show that \(\parallel l\parallel =\parallel g\parallel_{q}\). The converse is the so-called Riesz representation theorem.
Theorem. (Riesz Representation Theorem). Let \(l\) be a bounded linear functional on \(L^{p}(\mu )\) for \(1\leq p<+\infty\), and let \(\mu\) be a \(\sigma\)-finite measure. Then, there exists a unique element \(g\in L_{q}(\mu )\) with \(1/q+1/p=1\) satisfying
\[l(f)=\int_{X}fgd\mu\mbox{ and }\parallel l\parallel =\parallel g\parallel_{q}.\]
If \(p=1\), it is necessary that \(\mu\) should be \(\sigma\)-finite. If \(p>1\), then the \(\sigma\)-finiteness is not required.
Theorem. (Riesz Representation Theorem). Let \(l\) be a bounded linear functional on \(L^{p}(\mu )\) for \(1<p<+\infty\). Then, there exists a unique element \(g\in L_{q}(\mu )\) with \(1/q+1/p=1\) satisfying
\[l(f)=\int_{X}fgd\mu\mbox{ and }\parallel l\parallel =\parallel g\parallel_{q}.\]
\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}
Integration on Product Spaces.
Let \((X,\mathfrak{M}_{X},\mu_{X})\) and \((Y,\mathfrak{M}_{Y},\mu_{Y})\) be two complete measure spaces. We consider the product \(X\times Y\). For \(A\subseteq X\) and \(B\subseteq Y\), we call \(A\times B\) a rectangle. If \(A\in\mathfrak{M}_{X}\) and \(B\in\mathfrak{M}_{Y}\), we call \(A\times B\) a measurable rectangle. The collection \(\mathfrak{R}\) of measurable rectangles is a semialgebra, since
\[(A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)\]
and
\[(A\times B)^{c}=(A^{c}\times B)\cup (A\times B^{c})\cup (A^{c}\times B^{c}).\]
In this case, we define
\[\tilde{\mu}(A\times B)=\mu_{X}(A)\cdot\mu_{Y}(B).\]
\begin{equation}{\label{rap47}}\tag{14}\mbox{}\end{equation}
Lemma \ref{rap47}. Let \(\mathfrak{C}\) be a semialgebra, and let \(\tilde{\mu}\) be a nonnegative set function defined on \(\mathfrak{C}\) satisfying \(\tilde{\mu}(\emptyset )=0\) if \(\emptyset\in\mathfrak{C}\). Then \(\tilde{\mu}\) has a unique extension to a measure \(\hat{\mu}\) on the algebra \(\mathfrak{A}\) generated by \(\mathfrak{C}\) when the following conditions are satisfied.
- If a set \(C\in\mathfrak{C}\) is the union of a finite disjoint collection \(\{C_{i}\}_{i=1}^{n}\) of sets in \(\mathfrak{C}\), then
\[\tilde{\mu}(C)=\sum_{i=1}^{n}\tilde{\mu}(C_{i}).\] - If a set \(C\in\mathfrak{C}\) is the union of a countable disjoint collection \(\{C_{k}\}_{k=1}^{\infty}\) of sets in \(C\), then
\[\tilde{\mu}(C)\leq\sum_{k=1}^{\infty}\tilde{\mu}(C_{k}).\]
\begin{equation}{\label{rat49}}\tag{15}\mbox{}\end{equation}
Lemma \ref{rat49}. (Caratheodory). Let \(\hat{\mu}\) be a measure on an algebra \(\mathfrak{A}\), and let \(\mu^{*}\) be an outer measures induced by \(\hat{\mu}\). Then, the restriction \(\bar{\mu}\) of \(\mu^{*}\) to the \(\mu^{*}\)-measurable sets is an extension of \(\hat{\mu}\) to a \(\sigma\)-field containing \(\mathfrak{A}\). We also have the following properties.
(i) Suppose that \(\hat{\mu}\) is finite \((\)resp. \(\sigma\)-finite$)$. Then \(\bar{\mu}\) is also finite \((\)resp. \(\sigma\)-finite$)$.
(ii) Suppose that \(\hat{\mu}\) is \(\sigma\)-finite. Then \(\bar{\mu}\) is the only measure on the smallest \(\sigma\)-field containing \(\mathfrak{A}\). \(\sharp\)
The above Lemmas \ref{rap47} and \ref{rap49} can refer to the page General Measures.
\begin{equation}{\label{rap48}}\tag{16}\mbox{}\end{equation}
Proposition \ref{rap48}. Let \(\{A_{k}\times B_{k}\}_{k=1}^{\infty}\) be a countable disjoint collection of measurable rectangles whose union is measurable rectangle \(A\times B\). Then, we have
\begin{align*} \tilde{\mu}(A\times B) & =\sum_{k=1}^{\infty}\tilde{\mu}(A_{k}\times B_{k})\\ & =\sum_{k=1}^{\infty}\mu_{X}(A_{k})\cdot\mu_{Y}(B_{k}).\end{align*}
Proposition \ref{rap48} says that \(\tilde{\mu}\) satisfies the conditions of Lemma \ref{rap47}. Therefore, there exists a unique extension to a measure on the algebra \(\mathfrak{A}\) consisting of all finite disjoint union of sets in \(\mathfrak{R}\). Moreover, Lemma \ref{rat49} allows us to extend \(\tilde{\mu}\) to be a complete measure \(\mu\) on a \(\sigma\)-field containing \(\mathfrak{R}\). This extended measure \(\mu\) is called the product measure of \(\mu_{X}\) and \(\mu_{Y}\) and is also denoted by \(\mu_{X}\times\mu_{Y}\). Suppose that \(\mu_{X}\) and \(\mu_{Y}\) are finite (resp. \(\sigma\)-finite). Then, we see that \(\mu_{X}\times\mu_{Y}\) is also finite (resp. \(\sigma\)-finite). In particular, when \(X=Y=\bar{\mathbb{R}}\) and \(\mu_{X}=\mu_{Y}=\nu\) the Lebesgue measure, the product measure \(\nu\times\nu\) is called the two-dimensional Lebesgue measure for the plane \(\bar{\mathbb{R}}^{2}\). Let \(E\) be any subset of \(X\times Y\). Given a point \(x\in X\) and a point \(y\in Y\), we define the cross sections $E_{x}$ and \(E_{y}\) by
\[E_{x}=\left\{y\in Y:(x,y)\in E\right\}\mbox{ and }E_{y}=\left\{x\in X:(x,y)\in E\right\}.\]
Proposition. We have the following properties.
(i) Let \(x\) be a point of \(X\), and let \(E\in\mathfrak{R}_{\sigma\delta}\). Then \(E_{x}\) is a measurable subset of \(Y\).
(ii) Let \(E\) be a set in \(\mathfrak{R}_{\sigma\delta}\) satisfying \((\mu_{X}\times\mu_{Y})(E)<+\infty\). Then, the function \(g\) defined by \(g(x)=\mu_{Y}(E_{x})\) is measurable and
\[\int_{X}gd\mu_{X}=(\mu_{X}\times\mu_{Y})(E).\]
(iii) Let \(E\) be a set satisfying \((\mu_{X}\times\mu_{Y})(E)=0\). Then, for almost all \(x\in X\), we have \(\mu_{Y}(E_{x})=0\).
(iv) Let \(E\) be a measurable subset of \(X\times Y\) satisfying \((\mu_{X}\times\mu_{Y})(E)<+\infty\). Then, for almost all \(x\in X\), the set \(E_{x}\) is a measurable subset of \(Y\). The function \(g\) defined by \(g(x)=\mu_{Y}(E_{x})\) is measurable for almost all \(x\in X\) and
\[\int_{X}gd\mu_{X}=(\mu_{X}\times\mu_{Y})(E).\]
The following two theorems enable us to interchange the order of integration and to calculate integrals with respect to product measures by iteration.
Theorem. (Fubini’s Theorem). Let \((X,\mathfrak{M}_{X},\mu_{X})\) and \((Y,\mathfrak{M}_{Y},\mu_{Y})\) be two complete measure spaces, and let \(f\) be integrable function on \(X\times Y\). Then, we have the following properties.
(i) For almost all \(x\in X\), the function \(f_{x}(y)=f(x,y)\) is an integrable function on \(Y\). For almost all \(y\in Y\), the function \(f_{y}(x)=f(x,y)\) is an integrable function on \(X\).
(ii) The function
\[\int_{Y}f(x,y)d\mu_{Y}(y)\]
is integrable on \(X\), and the function
\[\int_{X}f(x,y)d\mu_{X}(x)\]
is integrable on \(Y\).
(iii) We have
\begin{align*}
\int_{X\times Y}f(x,y)d(\mu_{X}\times\mu_{Y}) & =\int_{X}\left [\int_{Y}f(x,y)d\mu_{Y}(y)\right ]d\mu_{X}(x)\\
& =\int_{Y}\left [\int_{X}f(x,y)d\mu_{X}(x)\right ]d\mu_{Y}(y).
\end{align*}
In order to apply the Fubini’s theorem, one must first verify that \(f\) is integrable with respect to \(\mu_{X}\times\mu_{Y}\); that is, one must show that \(f\) is a measurable function on \(X\times Y\) and that
\[\int_{X\times Y}|f(x,y)|d(\mu_{X}\times\mu_{Y})<+\infty .\]
The measurability of \(f\) on \(X\times Y\) is sometimes difficult to check. If \(\mu_{X}\) and \(\mu_{Y}\) are \(\sigma\)-finite, the integrability of \(f\) can be determined by iterated integration, which will be shown below.
Theorem. (Tonelli’s Theorem). Let \((X,\mathfrak{M}_{X},\mu_{X})\) and \((Y,\mathfrak{M}_{Y},\mu_{Y})\) be two \(\sigma\)-finite measure spaces, and let \(f\) be nonnegative measurable function on \(X\times Y\). Then, we have the following properties.
(i) For almost all \(x\in X\), the function \(f_{x}(y)=f(x,y)\) is a measurable function on \(Y\). For almost all \(y\in Y\), the function \(f_{y}(x)=f(x,y)\) is a measurable function on \(X\).
(ii) The function
\[\int_{Y}f(x,y)d\mu_{Y}(y)\]
is measurable on \(X\), and the function
\[\int_{X}f(x,y)d\mu_{X}(x)\]
is measurable on \(Y\).
(iii) We have
\begin{align*}
\int_{X\times Y}f(x,y)d(\mu_{X}\times\mu_{Y}) & =\int_{X}\left [\int_{Y}f(x,y)d\mu_{Y}(y)\right ]d\mu_{X}(x)\\
& =\int_{Y}\left [\int_{X}f(x,y)d\mu_{X}(x)\right ]d\mu_{Y}(y).
\end{align*}
