% scribe: Matias Damian Cattaneo
% lastupdate: 12 November 2005
% lecture: 17
% references: Durrett, Section 2.6
% title: Poisson Processes -- Part I
% keywords: Poisson distribution, Poisson process, counting process, point process, convolution semigroups, Levy, Khinchine, Levy-Khinchine formula, Levy-Khinchine theorem, infinite divisible law, Brownian motion, compound Poisson process
% end

\documentclass[12pt, letterpaper]{article}

\include{macros}

\begin{document}

\lecture{17}{Poisson Processes -- Part I}{Matias Damian Cattaneo}
{cattaneo@econ.berkeley.edu}

\section{The Poisson Distribution}
% keywords: Poisson distribution
% end

Define $S_{n}=X_{1}+X_{2}+...+X_{n}$.  In a very simple setup, 
the $X_{i}$ are independent indicators with%
\begin{eqnarray*}
\P \!\left[ X_{i}=1\right] &=&p \\
\P \!\left[ X_{i}=0\right] &=&1-p
\end{eqnarray*}

We know that the distribution of $S_{n}$ is $\mbox{Binomial}\left(n,p\right)$. 
So we have
\begin{eqnarray*}
\P \!\left[ S_{n}=k\right] &=&\binom{n}{k}p^{k}\left( 1-p\right) ^{n-k}, \\
\E\!\left[ S_{n}\right] &=&np, \mbox{ and} \\
\var\!\left[ S_{n}\right] &=&np\left( 1-p\right).
\end{eqnarray*}

For fixed $p$, we have that%
\begin{equation*}
\frac{S_{n}-np}{\sqrt{np\left( 1-p\right) }}\dcv\mathcal{N}\left( 0,1\right)
\end{equation*}%
as $n\rightarrow \infty $.

Now we let $n\rightarrow \infty$ and choose $p_n$ small so that $np_n=\lambda$.
We see that%
\begin{eqnarray*}
\P \!\left[ S_{n}=0\right] &=&\!\left( 1-p\right) ^{n}=\left( 1-\frac{\lambda }{%
n}\right) ^{n}\longrightarrow e^{-\lambda }, \\
\P \!\left[ S_{n}=1\right] &=&np\left( 1-p\right) ^{n-1}=\lambda \left( 1-%
\frac{\lambda }{n}\right) ^{n-1}\longrightarrow \lambda e^{-\lambda },
\end{eqnarray*}%
and in general%
\begin{eqnarray*}
\P \!\left[ S_{n}=k\right] &=&\binom{n}{k}p^{k}\left( 1-p\right) ^{n-k} \\
&\approx &\frac{n^{k}}{k!}p^{k}\left( 1-p\right) ^{n}\longrightarrow \frac{%
\lambda ^{k}}{k!}e^{-\lambda},
\end{eqnarray*}%
which is the mass function of a Poisson distribution.

\begin{definition} The \emph{Poisson distribution} with
parameter $\lambda $ is given by mass function%
\begin{equation*}
P_{\lambda }\!\left( k\right) =\frac{\lambda ^{k}}{k!}e^{-\lambda}.
\end{equation*}%
\end{definition}

Observe that%
\begin{equation*}
\sum_{k=0}^{\infty }P_{\lambda }\!\left( k\right) =1
\end{equation*}

Summing up, we see that as $n\rightarrow \infty$, if we let  
$p\rightarrow 0$ such that $np = \lambda \in \left[ 0,\infty \right)$,%
\begin{equation*}
S_{n}\dcv \mbox{Poisson}\!\left( \lambda \right)
\end{equation*}

Observe that we can relax the assumption that the indicators are identically 
distributed and extend this result to the triangular array setup.  Let the 
$X_{n,i}$s be taken such that
%
\begin{equation*}
\P \!\left[ X_{n,i}=1\right] =p_{n,i}
\end{equation*}
%
Note we have $\E\!\left[ S_{n}\right] =\sum_{i=1}^{n}p_{ni}$
and assuming that as $n\rightarrow \infty $ we have 
$\sum_{i=1}^{n}p_{n,i} \longrightarrow \lambda$ and 
$\max_{i} p_{n,i} \longrightarrow 0$,
we obtain the result:%
\begin{equation*}
S_{n}\dcv \mbox{Poisson}\!\left( \lambda \right).
\end{equation*}

The proof of this result is formalized in \cite{durrett}.

Observe the following facts about the Poisson distribution:

\begin{enumerate}
\item As shown above, it is the limit of properly chosen binomial 
distributions.
\item The sum of independent Poisson random variables with parameters 
$\lambda$ and $\upsilon$ is another Poisson random variable. This result can 
be written as follows:
%
\begin{equation*}
\mbox{Poisson}\!\left( \lambda \right) \ast \mbox{Poisson}\!\left( \upsilon \right)
=\mbox{Poisson}\!\left( \lambda +\upsilon \right),
\end{equation*}
%
where we use the fact that for discrete distributions $P$ and $Q$ on $%
\left\{ 0,1,2,...\right\}$,%
\begin{equation*}
\left( P+Q\right) \!\left( n\right) =\sum_{k=0}^{n}P\!\left( k\right)
Q\!\left( n-k\right)
\end{equation*}%
for $n=0,1,2,...$, the convolution formula for the distribution of sums of 
independent random variables.
\end{enumerate}

\section{Basics of Poisson Processes}
% keywords: Poisson process, point process
% end

We discuss the basics of Poisson processes here; for details, see 
\cite{durrett}.

Consider the positive numbers divided up into intervals of length $10^{-6}$.  
We can think of this as time broken up into very short intervals.
Consider a process $(X_1, X_2, \ldots)$ where $X_i$ is $1$ if a certain
thing happens in the $i$th such time interval and $0$ otherwise.  
Suppose further that these $X_i$s are 
independent and are $1$ with probability $\lambda \cdot 10^{-6}$.

The waiting time until the first occurrence of this thing is defined as%
\begin{equation*}
T_{1}=\left\{ first\text{ }n:\text{ }X_{i}=1\right\}
\end{equation*}%
here $X_{i}=1$ with probability $\frac{\lambda }{n}$.

We see that%
\begin{equation*}
\P \!\left[ T_{1}>m\right] =\left( 1-p\right) ^{m}=\left[ \left( 1-p\right)
^{n}\right] ^{\frac{m}{n}}\longrightarrow e^{-\lambda \frac{m}{n}}
\end{equation*}%
as $m\rightarrow \infty $, and thus%
\begin{equation*}
\P \!\left[ \frac{T_{1}}{n}>t\right] =\P \!\left[ T_{1}>nt\right]
\longrightarrow e^{-\lambda t}
\end{equation*}%
as $m\rightarrow \infty $. Notice that this is an exponential distribution.

When $n=10^{6}\rightarrow \infty $ we obtain a \emph{point process} $(T_i)$, 
an increasing sequence of random variables where%
\begin{eqnarray*}
T_{1} &\thicksim &\mbox{Exponential}\!\left( \lambda \right) \\
T_{2}-T_{1} &\thicksim &\mbox{Exponential}\!\left( \lambda \right) \mbox{,
independent of }T_{1} \\
&&\vdots
\end{eqnarray*}%
so we get $T_{1},\left( T_{2}-T_{1}\right) ,\left( T_{3}-T_{2}\right)$
are iid $\mbox{Exponential}\!\left( \lambda \right)$.

Alternatively, the \emph{counting process} is defined by%
%
\begin{equation*}
N_{t}=\#\left\{ i:T_{i}\leq t\right\} =\sum_{k=1}^{n}\1\left\{
T_{k}\leq t\right\}.
\end{equation*}%
%
In general $\left\{ N_{t}\right\} _{t\geq 0}$ is called a 
\emph{Poisson process} with rate $\lambda$ on $\left( 0,\infty \right)$.
It is not difficult to show that 
$N_{t}\thicksim \mbox{Poisson}\!\left( \lambda t\right)$.

It is important to note that almost by construction this process $\left(
N_{t},t\geq 0\right) $ has \emph{stationary independent increments}: that
is, for $0<t_{1}<t_{2}<...<t_{n}$, we have that
$N\!\left( t_{1}\right) ,N\!\left( t_{2}\right) -N\!\left( t_{1}\right)
,...,N\!\left( t_{n}\right) -N\!\left( t_{n-1}\right)$
are independent Poisson variables with means
$\mu t_{1},\left( t_{2}-t_{1}\right) \mu ,...,\left( t_{n}-t_{n-1}\right) \mu$.

\section{Details of Poisson Processes}
% keywords: Poisson process, counting process, point process
% end

In this section we present details on Poisson processes.

\subsection{The Poisson Process and its counting process}

\begin{definition}
A real valued process $N_t$, $t\geq 0$, is a
\emph{Poisson Process} with rate $\lambda$ (or $PP(\lambda)$) if
$N_t$ has stationary independent increments, and for all $s,t \ge
0$, the increment $X_{t+s} - X_t$ is a Poisson random variable
with parameter $\lambda s$.
\end{definition}

Interpretation: Jumps of $N_t$ are ``arrivals'' or ``points''.

Let $T_k = \inf\{t:\mbox{ } N_t=k\}$. We say that $0<T_1<T_2<\cdots$ are
these arrival times.  Note that $N_{T_k} =k$ and $N_{T_k^-} =k-1$.

As a convention, we will always work with the right continuous version of
$(N_t$, $t\geq 0)$ which is increasing and hence has a left limit a.s.

\begin{theorem} A counting process $N_t$, $t\geq 0$
(increasing, right continuous, left limits exists, jumps of 1
only) is a $PP(\lambda)$ if and only if the distribution of
its jumps $(T_1, T_2, T_3, \ldots)$ satisfies that $(T_k - T_{k-1},
k=1,2,\ldots)$ is a sequence of i.i.d.\ $\mbox{Exponential}(\lambda)$ random 
variables. ($T_0 =0$.)
\end{theorem}

\begin{proof}
For the ``if'' direction, see \cite{durrett}, Section 2.6 (c. Poisson 
process). 
The idea is given i.i.d.\ $(W_k)$ such that 
$\P(W_k >t)=e^{-\lambda t}$, $t\geq 0$,
we construct $(N_t, t\geq 0)$ by the formula:%
\begin{eqnarray*}
  T_k &=& W_1 + \cdots + W_k \\
  N_t &=& {\rm \ number\ of\ } \{ k: T_k\leq t \} \\
      &=& \sum_{k=1}^{\infty} \1\{T_k\leq t\}
\end{eqnarray*}
Informally, the $\{ T_k \}$ are the points of the
$PP(\lambda)$ denoted by $N$.\footnote{Friendly reference:
"Probability", J.Pitman, Springer 1992.}

The ``only if'' direction is easier.  Suppose $N_t$ $\sim $ \mbox{Poisson}
$(\lambda )$. By the Strong Markov Property (see \cite[section 5.2]{durrett}), we can deduce
that $T_k -T_{k-1}$ are independent. We also can deduce:
%
$\P(T_k >t)=\P(N_t <k) = 
\sum_{j=0}^{k-1} e^{-\lambda t} \frac{(\lambda t)^j}{j !}.$ 
%
Hence,
by applying $\frac{d}{dt}$, we get $\P(T_k \in dt)= e^{-\lambda t}
\frac{(\lambda t)^{k-1}}{(k-1) !}dt$.

This shows $T_k\sim gamma(k, \lambda)$. Since $\phi _{T_k
-T_{k-1}} = (\phi _{T_k})^{1/n} $, and the characteristic
function uniquely determines the distribution, this determines the distributions
of $T_k -T_{k-1}$. Since
$T_k -T_{k-1} \sim exponential(\lambda)$ will ensure $T_k \sim
gamma(k,\lambda)$ and vice versa, $T_k -T_{k-1} \sim exponential(\lambda)$.
\end{proof}

\subsection{The Poisson Point Process}

\begin{definition}
A \emph{Poisson Point Process} (P.P.P.) with intensity measure $\mu$
on $(S, \cal{S})$ is a collection of random variables $N(B,\omega)$, $B \in {\cal
S}$, $\omega \in \Omega$ defined on a probability space $(\Omega,
\mathbb{F}, \mathbb{P})$ such that:
\begin{enumerate}
\item $N(B) = N(B,\omega)$, $B\in \cal{S}$, $\omega\in
\Omega$;
\item $N(\cdot,\omega)$ is a non-negative integer or
$\infty$ -valued measure on $(S, {\cal S})$ for each $\omega \in
\Omega$; 
\item $N(B, \cdot)$ is a r.v. with $\mbox{Poisson}(\mu (B))$
distribution:
$\P(N(B)=k) = \frac{e^{-\mu (B)}(\mu(B))^k}{k!}\  \ {\rm for\ all}\ B \in {\cal
S};$ and 
\item If $B_1$, $B_2$, ... are disjoint sets then
$N(B_1, \cdot)$, $N(B_2, \cdot)$, ... are independent random
variables.
\end{enumerate}
\end{definition}

\begin{example}
Let $S= \R _+$, $\mathcal{S}$ = Borel$(\R _+)$, $\mu = \lambda
 \cdot Lebesgue$.
Let $N(B, \omega)$ be the measure of $B$, whose cumulative
distribution function is defined as the counting process $(N_t, t\geq 0)$ for 
a $PP(\lambda)$ as described before.  That is,
$$N([0,t]):= N_t = \sum_{k=1}^{\infty} 1\{T_k\leq t\}$$
So,%
\begin{eqnarray*}
\ N(B) &=& \sum_{k=1}^{\infty} \1_{(T_k\in B)} \\
   &=& \ {\rm the \ number \ of \ }T_k\ {\rm which \ fall\ in\ }B
\end{eqnarray*}
\end{example}

\begin{example}
$S= \R $, $\mathcal{S}$ = Borel$(\R )$, $\mu = \lambda
\cdot Lebesgue$. Stick together independent $PP(\lambda )$ on $\R _+$
and $\R _-$.
\end{example}

\begin{theorem}
Such a P.P.P. exists for any $\sigma$-finite measure space.
\end{theorem}

\begin{proof}
The only convincing argument is to give an explicit construction
from sequences of independent random variables.  We begin by
considering the case $\mu(S)<\infty$.
\begin{enumerate}
\item  Take $X_1$, $X_2$, ... to be i.i.d. random variables
with form $\mu (\cdot |S)$ so that $P(X_i \in B) = \frac{\mu(B)}{\mu(S)}$. 
\item Take $N(S)$ to be a Poisson
random variable with mean $\mu(S)$, independent of the $X_i$'s.
Assume all random variables are defined on the same probability
space $(\Omega, \mathcal{F}, \mathbb{P})$. 
\item Define $N(B) = \sum_{i = 1}^{N(S)} \1_{(X_i \in B)}$, for all $B \in \mathcal{S}$.
\end{enumerate}
\end{proof}

\begin{exercise}
Verify that this $N(B,\omega)$ is a P.P.P. with
intensity $\mu$. (Use thinning property of Poisson distributions.)
\end{exercise}

\begin{example} \emph{(Poissonization of multinomial)}
If $B_1$, $B_2$ are disjoint events and $B_1 \cup B_2 = S$
\begin{eqnarray*}
  P(N(B_1)=n_1,\ N(B_2)=n_2) &=& P(N=n_1 +n_2)P(N(B_1)=n_1,\ N(B_2)=n_2 | N=n_1 +n_2) \\
    &=& e^{-\lambda}\frac{\lambda ^{n_1 +n_2}}{(n_1 +n_2)!} (\frac{n_1
    +n_2}{n_1}) p^{n_1} q^{n_2},
\end{eqnarray*}
where $N = N(S)$, $p=\frac{\mu (B_1)}{\mu (S)}$, and 
$q=\frac{\mu (B_2)}{\mu(S)}$.
\end{example}

\section{General Theory Behind Poisson Processes}
% keywords:convolution semigroups, Levy, Khinchine, Levy-Khinchine formula, Levy-Khinchine theorem, infinite divisible law, Brownian motion, compound Poisson process
% end

It is important to note that we have presented two different examples of \emph{convolution semigroups}
(convolution 1/2groups) of probability distributions on $\R$. In
general, we have a family of probability distributions $\left( F_{t},t\geq
0\right) $ such that:%
\begin{equation*}
F_{s}\ast F_{t}=F_{s+t}
\end{equation*}%
where $\ast $ is convolution. Examples are:
\begin{enumerate}
\item $F_{t}=\mathcal{N}\!\left( 0,\sigma ^{2}t\right) $
\item $F_{t}=\mbox{Poisson}\!\left( \lambda t\right) $
\item $F_{t}=\delta _{ct}$ for some real $c\in \R$
\end{enumerate}

For any such family we can create a \emph{stochastic process} $\left(
X_{t},t\geq 0\right) $ so that $X$ has stationary independent increments,
that is:%
\begin{eqnarray*}
X_{t} &\thicksim &F_{t} \\
X_{t}-X_{s} &\thicksim &F_{t-s}\text{, for }0<s<t
\end{eqnarray*}%
and so on. Notice that we did this by explicit construction for $%
F_{t}=\mbox{Poisson}\!\left( \lambda t\right) $. For the case of $F_{t}=\mathcal{N}%
\!\left( 0,\sigma ^{2}t\right) $ we need to be clever, and in general
we will need to
appeal to \emph{Kolmogorov's Extension Theorem} (see \cite{durrett} for
details).  In particular,
%
\begin{enumerate}
\item the Poisson case gives a Poisson Process; and
\item the Normal case gives a \emph{Brownian motion} (also known
as a \emph{Wiener Process}). 
\end{enumerate}

In particular, Brownian motion has the special
feature that it is possible to define the process to have 
\emph{continuous paths}; that is, for almost every 
$\omega$, $t\rightarrow X_{t}\!\left( \omega \right)$ is continuous.

\begin{definition}
A probability distribution $F$ on $\R$ is called \emph{infinitely
divisible} if for every $n$, there exist i.i.d.\ random variables 
$X_{n,1},X_{n,2},...,X_{n,n}$ such that%
\begin{equation*}
S_{n}=X_{n,1}+X_{n.2}+...+X_{n,n}\thicksim F;
\end{equation*}%
that is, there exists a distribution $F_{\frac{1}{n}}$ so that%
\begin{equation*}
F_{\frac{1}{n}}\ast F_{\frac{1}{n}}\ast ...\ast F_{\frac{1}{n}}=F,
\end{equation*}%
so $F$ has a convolution "$n$th root" for every $n$.
\end{definition}

\begin{definition}
Let $\left(X_{t},t\geq 0\right)$ be a process with stationary independent increments and distribution function $F_{t}$ at time $t$. $F_{t}$ is said to be \emph{weakly continuous in $t$} as $t \downarrow 0$ if%
\begin{equation*}
\lim_{t\downarrow 0}\P\left[ \left\vert X_{t}\right\vert >\varepsilon \right]=0
\end{equation*}%
for all $\varepsilon >0$

\end{definition}

\begin{theorem}
[L\'evy-Khinchine] There is a 1-1 correspondence between infinitely divisible
distributions $F$ and weakly continuous convolution semigroups 
$\left( F_{t},t\geq 0\right)$ with $F_{1}=F$.
\end{theorem}

\begin{corollary}
Every infinitely divisible law is associated with a process with stationary
independent increments which is continuous in P as t varies.
\end{corollary}
For a proof see \cite{feller1} or \cite{kall}

Now we present another example. 
Consider accidents that occur at times of Poisson Process with rate $\lambda $.
At the time of the $k$th accident let there be some variable $X_{k}$ like the 
damage costs covered by insurance companies.  We have%
\begin{equation*}
Y_{t}=\sum_{k=1}^{N_{t}}X_{k}
\end{equation*}%
where $Y_{t}$ is the total cost/magnitude up to time $t$. It is not
difficult to show that $\left( Y_{t},t\geq 0\right) $ has stationary
independent increments.

We look at the distribution of $Y_{t}$. This is called a \emph{compound
Poisson process}. It is hard to describe explicitly, but we can compute its
characteristic function.

\begin{exercise}
Find a formula for the characteristic function of $Y_{t}$ in terms of the 
rate $\lambda$ of $N_{t}$, and the distribution $F$ of $X_{t}$, that is,%
\begin{equation*}
F\!\left( B\right) =\P \!\left[ X_{k}\in B\right].
\end{equation*}
\end{exercise}

Observe we have%
\begin{eqnarray*}
\E\!\left[ \exp \left\{ i\theta Y_{t}\right\} \right]
&=&\sum_{n=1}^{N_{t}}\E\!\left[ \exp \left\{ i\theta
Y_{t}\right\} \cdot \1\left\{ N_{t}=n\right\} \right] \\
&=&\sum_{n=1}^{N_{t}}\E\!\left[ \exp \left\{ i\theta \left(
X_{1}+X_{2}+...+X_{n}\right) \right\} \cdot \1\left\{
N_{t}=n\right\} \right] \\
&=&\sum_{n=1}^{N_{t}}\E\!\left[ \exp \left\{ i\theta \left(
X_{1}+X_{2}+...+X_{n}\right) \right\} \right] \cdot \E\!\left[ \mathbf{%
1}\left\{ N_{t}=n\right\} \right] \\
&=&\sum_{n=1}^{N_{t}}\left( \E\!\left[ \exp \left\{ i\theta
X\right\} \right] \right) ^{n}\cdot \frac{e^{-\lambda t}\left( \lambda
t\right) ^{n}}{n!} \\
&=&e^{-\lambda t}\exp \left\{ \lambda t\E\!\left[ e^{i\theta X}\right]
\right\} \\
&=&\exp \left\{ \lambda t\int_{\R}\left( e^{i\theta
x}-1\right) \P \!\left[ X\in dx\right] \right\} \\
&=&\exp \left\{ t\int_{\R}\left( e^{i\theta x}-1\right)
L\!\left( dx\right) \right\}
\end{eqnarray*}%
where we let $L\!\left( \cdot \right) =\lambda \P \!\left[ X\in \cdot \right] $%
, which is called the \emph{L\'evy Measure} associated with the process.

It is important to note that this is an instance of the \emph{%
L\'evy-Khinchine} formula which gives the characteristic function of the most
general $\infty $-divisible law.

In the next subsection we include more details on this.

\subsection{Computation of an instance of LK Formula}

Given a P.P.P. $N$, say $N(B) = \sum_i \1\{X_i \in B\}$. For some
sequence of random variables $X_i$. Consider positive and
measurable $f$:
\begin{equation}
\int f dN\ =\ \sum_i f(X_i)
\end{equation}
has clear intuitive meanings and many applications.

\begin{example} $X_i$ could be the arrival time, location, and magnitude of
an earthquake:
$$X_i\ =\ (T_i, M_i, Y_i).$$
$f(t,m,y)$ represents the cost to the insurance company incurred by an
earthquake at time t with magnitude m in place y.
\end{example}

How do we describe its distribution?
Consider the case where $f$ is a simple function. Say $f
= \sum_{i= 1}^m x_i \1\{B_i\}$ where the $B_i$'s are disjoint
events and cover the space.  Then
\begin{equation}
\int f dN\ =\ \sum_i x_i N(B_i)
\end{equation}
where $N(B_i)$ are independent r.v.s with $\mbox{Poisson}(\mu (B_i))$
distribution.\\
This is some new infinitely divisible distribution.

Now we need a transformation. Because $f\geq 0$, it is natural to
look first at the Laplace transformation. Take $\theta >0$:
\begin{eqnarray}
\E\!\left[e^{-\theta \int f dN}\right] &=& \E\!\left[e^{-\theta \sum_i x_i N(B_i)}\right] \\
    &=& \prod_i E\!\left[e^{-\theta x_i N(B_i)}\right] \label{lap4}\\
    &=& \prod_i \exp \!\left[ -\mu(B_i)(1-e^{-\theta x_i}) \right] 
\label{lap5} \\
    &=& \exp \!\left[ -\sum_i \mu(B_i)(1-e^{-\theta x_i}) \right]\\
    &=& \exp \!\left[ -\int (1-e^{-\theta f(s)})\mu (ds) \right]
\end{eqnarray}
$\ref{lap4} implies \ref{lap5}$ because $N(B_i)\sim \mbox{Poisson} (\mu
(B_i))$, and if $N\sim \mbox{Poisson} (\lambda)$, then
\begin{eqnarray*}
\E(e^{-\theta N}) &=& \sum_{n=0}^{\infty} (e^{-\theta})^n
\frac{\lambda ^n}{n !} e^{-\lambda}\\
&=& e^{-\lambda} e^{(e^{-\theta}\lambda)}\\
&=& \exp \!\left[ -\lambda (1-e^{-\theta}) \right]
\end{eqnarray*}

\begin{theorem} For every non-negative measurable function
$f$, we have, writing $e^{-\infty}:=0$:
\begin{equation}
\E\!\left[e^{-\theta \int f dN}\right] = \exp \!\left[ \int (e^{-\theta
f(s)}-1)\mu (ds) \right]
\end{equation}
This formula is an instance of the L\'evy-Khinchine equation.
\end{theorem}

\begin{proof}
We have shown that the result holds when $f$ is a simple function. In
general, there exist sequence of simple functions $f_n$ such that
$f_n \uparrow f$. Then apply the Monotone Convergence Theorem and
Dominated Convergence Theorem.
\end{proof}


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