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\@@toccaption{{\lx@tag[ ]{{7}}{\small Fuzzy attributes "$low$", "$medium$" and
"$high$" for $T=\\{0,1,\ldots,9\\}$.}}}\@@caption{{\lx@tag[. ]{{Figure
7}}{\small Fuzzy attributes "$low$", "$medium$" and "$high$" for
$T=\\{0,1,\ldots,9\\}$.}}} \end{figure} \end{Example}
\par\par\par\par\par\@@numbered@section{section}{toc}{Measure Theoretical
Approach} Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space, where
$\Omega\subseteq\mathbb{R}$ is the set of possible outcomes, $\mathscr{F}$ is
a $\sigma$-algebra over $\Omega$ consisting of all events, and $\mathbb{P}$ is
a measure on $(\Omega,\mathscr{F})$ with $\mathbb{P}(\Omega)=1$. We assume
that $\mathscr{F}$ is the Borel $\sigma$-algebra on $\Omega$. Now, let
$X:\Omega\to\mathbb{R}$ be the identity random variable, and $A$ be a fuzzy
attribute of $\Omega$. Hence, the probability measure induced by $X$, denoted
by $\mathbb{P}_{X}$, coincides with $P$. For the sake of simplicity, in the
sequel, we denote all probability measures and their derived notations (e.g.,
$\mathbb{P}_{X}$) by $\mathbb{P}$, when the underlying measurable space
$(\Omega,\mathscr{F})$ is known. For any $x\in\Omega$, assume that
$\mathbb{P}(x\text{ is }A)$ is the chance of selecting $x$ as $A$ in the
random experiment whose sample space is $\\{\text{Selecting $x$ as
$A$},\text{Not selecting $x$ as $A$}\\}$, and it is measurable as a function
of $x$. Considering Experiment ($\star$), We have the following assumption for
any $x\in\Omega$: $$\mathbb{P}(\Omega\text{ is }A|X=x)=\mathbb{P}(x\text{ is
}A).$$ It follows that $$\mathbb{P}(\Omega\text{ is
}A)=\int_{\mathbb{R}}\mathbb{P}(\Omega\text{ is
}A|X=x)\,\mathrm{d}\mathbb{P}_{X}(x)=\int_{\mathbb{R}}\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}(x).$$ where, $\mathbb{P}_{X}$ is the probability
measure on $\mathbb{R}$ induced by $X$. Let $\xi_{X,A}$ be the random variable
associated to Experiment ($\star$). Similar to the above, we consider the
following assumption for any $x\in\Omega$ and $E\in\mathscr{F}$:
$$\mathbb{P}(\xi_{X,A}\in E|X=x)=\mathbb{P}(\xi_{x,A}\in E).$$ Let
$E\in\mathscr{F}$. Then, by the total law of probability, we have that:
\@@amsalign\mathbb{P}(\xi_{X,A}\in
E)&=\int_{\mathbb{R}}\mathbb{P}(\xi_{X,A}\in
E|X=x)\,\mathrm{d}\mathbb{P}_{X}(x)=\int_{\mathbb{R}}\mathbb{P}(\xi_{x,A}\in
E)\,\mathrm{d}\mathbb{P}_{X}(x)\\\ &=\int_{E}\mathbb{P}(\xi_{x,A}\in
E)\,\mathrm{d}\mathbb{P}_{X}(x)+\int_{{\mathbb{R}}\backslash
E}\mathbb{P}(\xi_{x,A}\in E)\,\mathrm{d}\mathbb{P}_{X}(x). If $x_{A}\in E$,
then: \@@amsalign\int_{E}\mathbb{P}(\xi_{x,A}\in
E)\,\mathrm{d}\mathbb{P}_{X}(x)&=\int_{E}\left(\mathbb{P}(\xi_{x,A}=x)+\mathbb{P}(\xi_{x,A}=x_{A})\right)\,\mathrm{d}\mathbb{P}_{X}(x)=\int_{E}\,\mathrm{d}\mathbb{P}_{X}(x),\\\
\int_{\mathbb{R}\backslash E}\mathbb{P}(\xi_{x,A}\in
E)\,\mathrm{d}\mathbb{P}_{X}(x)&=\int_{\mathbb{R}\backslash
E}\mathbb{P}(\xi_{x,A}=x_{A})\,\mathrm{d}\mathbb{P}_{X}(x)\\\
&=\int_{\mathbb{R}\backslash E}\left(1-\mathbb{P}(x\text{ is
}A)\right)\,\mathrm{d}\mathbb{P}_{X}(x). Further, if $x_{A}\notin E$, then:
\@@amsalign&\int_{E}\mathbb{P}(\xi_{x,A}\in
E)\,\mathrm{d}\mathbb{P}_{X}(x)=\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}(x)=\int_{E}\mathbb{P}(x\text{
is }A)\,\mathrm{d}\mathbb{P}_{X}(x),\\\ &\int_{\mathbb{R}\backslash
E}\mathbb{P}(\xi_{x,A}\in E)\,\mathrm{d}\mathbb{P}_{X}(x)=0. It follows from
$\int_{\mathbb{R}}\,\mathrm{d}\mathbb{P}_{X}(x)=1$ that:
$$\mathbb{P}(\xi_{X,A}\in
E)=\left\\{\begin{array}[]{ll}1-\displaystyle\int_{\mathbb{R}\backslash
E}\mathbb{P}(x\text{ is }A)\,\mathrm{d}\mathbb{P}_{X}(x)&,x_{A}\in E\\\
\displaystyle\int_{E}\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}(x)&,x_{A}\notin E\end{array}\right..$$ We can
rewrite the above formula as follows: \begin{equation*}\mathbb{P}(\xi_{X,A}\in
E)=\int_{E}\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}(x)+\alpha\delta_{x_{A}}(E),\quad\alpha=1-\int_{\mathbb{R}}\mathbb{P}(x\text{
is }A)\,\mathrm{d}\mathbb{P}_{X}(x),\end{equation*} where $\delta_{x_{A}}$ is
the Dirac measure with respect to $x_{A}$. The above formula equals the
following as well: \begin{equation*}\mathbb{P}(\xi_{X,A}\in
E)=\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}(x)+\beta\delta_{x_{A}}(E),\quad\beta=1-\int_{\mathbb{R}}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}(x),\end{equation*}
\par\par Now, we determine the expected value of $\xi_{X,A}$ in the following
theorem. \begin{Theorem}
$$\mathbb{E}(\xi_{X,A})=x_{A}+\int_{\mathbb{R}}(x-x_{A})\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}(x).$$ \end{Theorem} \@proof Let
$E\in\mathscr{F}$. Then, we have that
\@@amsalign\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}(\mathbb{P}_{X}+\beta\delta_{x_{A}})&=\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}+\beta\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\delta_{x_{A}}\\\
&=\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}+\beta\underbrace{\mathbb{P}(\xi_{x_{A},A}=x_{A})}_{=1}\delta_{x_{A}}(E)\\\
&=\mathbb{P}(\xi_{X,A}\in E). It follows that
\@@amsalign\mathbb{E}(\xi_{X,A})&=\int_{\mathbb{R}}x\,\mathrm{d}\mathbb{P}_{\xi_{X,A}}=\int_{\mathbb{R}}x\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}(\mathbb{P}_{X}+\beta\delta_{x_{A}})\\\
&=\int_{\mathbb{R}}x\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}+\beta\int_{\mathbb{R}}x\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\delta_{x_{A}}\\\
&=\int_{\mathbb{R}}x\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}+\beta
x_{A}\mathbb{P}(\xi_{x_{A},A}=x_{A})\\\
&=\int_{\mathbb{R}}x\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}+x_{A}\left(1-\int_{\mathbb{R}}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}\right)\\\
&=x_{A}+\int_{\mathbb{R}}(x-x_{A})\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}\mathbb{P}_{X}\\\
&=x_{A}+\int_{\mathbb{R}}(x-x_{A})\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}. \qed\end@proof In the following theorem, we
provide the classic continuous distributions of $\xi_{X,A}$. To do so, we
initially assume that $\mathbb{P}(x\text{ is }A)$ and the probability density
function (pdf) of $X$ are continuous functions of the variable $x$, and hence
they are Riemann integrable. \par\begin{Theorem} Let $X$ be a classic
continuous distribution whose pdf is $f_{X}(x)$. Then, $\xi_{X,A}$ has a
generalized pdf, denoted by $f_{\xi_{X,A}}$, and it is given by the following:
$$f_{\xi_{X,A}}(t)=\mathbb{P}(t\text{ is }A)f_{X}(t)+\alpha\delta(t-x_{A}),$$
where $\delta(x-x_{A})$ is the Dirac delta function, and we have that
$$\alpha=1-\int_{-\infty}^{\infty}\mathbb{P}(x\text{ is
}A)f_{X}(x)\,\mathrm{d}x.$$ \end{Theorem} \@proof We have that
$\mathrm{d}\mathbb{P}_{X}=f(x)\,\mathrm{d}x$. First, we note that
$$\alpha=1-\int_{\mathbb{R}}\mathbb{P}(x\text{ is
}A)\,\mathrm{d}\mathbb{P}_{X}(x)=1-\int_{-\infty}^{\infty}\mathbb{P}(x\text{
is }A)f_{X}(x)\,\mathrm{d}x.$$ Now, we show that $\xi_{X,A}$ has a generalized
pdf (see Appendix \ref{gpdf}). To do so, it follows from the proof of Theorem
\ref{expectedvalueingeneral} that $$\mathbb{P}(\xi_{X,A}\in
E)=\int_{E}\mathbb{P}(\xi_{x,A}=x)\,\mathrm{d}(\mathbb{P}_{X}+\beta\delta_{x_{A}}),$$
while $\mathbb{P}_{X}$ is absolutely continuous (see Appendix
\ref{continuousrandomvariables}). Hence, $\mathbb{P}_{X}+\beta\delta_{x_{A}}$
is absolutely continuous with respect to $\lambda+\beta\delta_{x_{A}}$, where
$\lambda$ is the Lebesgue measure on $\mathbb{R}$. It follows that $\xi_{X,A}$
is absolutely continuous with respect to $\lambda+\beta\delta_{x_{A}}$, and
hence it has a generalized pdf. \par Denote the distribution function of
$\xi_{X,A}$ by $F$. Then, we have that \@@amsalign
F(t)=\mathbb{P}(\xi_{X,A}\leq t)&=\int_{-\infty}^{t}\mathbb{P}(x\text{ is
}A)f_{X}(x)\,\mathrm{d}x+\alpha\delta_{x_{A}}(-\infty,t]\\\
&=\int_{-\infty}^{t}\mathbb{P}(x\text{ is }A)f_{X}(x)\,\mathrm{d}x+\alpha
U_{x_{A}}(t), \par\par where $U_{x_{A}}$ is the Heaviside step function. It
follows that
$$f_{\xi_{X,A}}(t)=\frac{\mathrm{d}F}{\mathrm{d}x}(t)=\mathbb{P}(t\text{ is
}A)f_{X}(t)+\alpha\delta(t-x_{A}).$$ \qed\end@proof \par By a similar process
(i.e., taking the total law of probability over the values of $X$ and using
the Bayes rule), we can define the conditional probabilities
$\mathbb{P}(\xi_{X,A}\in E^{\prime}|E)$ and $\mathbb{P}(E|\xi_{X,A}\in
E^{\prime})$ that we described in the discrete case in Section \ref{frame}.
For instance, we have that \@@amsalign&\mathbb{P}(\xi_{X,A}\in
E^{\prime}|E)=\left\\{\begin{array}[]{ll}1-\displaystyle\int_{\mathbb{R}\backslash
E^{\prime}}\mathbb{P}(x\text{ is
}A|X=x,E)\,\mathrm{d}\mathbb{P}^{E}_{X}&,x_{A}\in E^{\prime}\\\
\displaystyle\int_{E^{\prime}}\mathbb{P}(x\text{ is
}A|X=x,E)\,\mathrm{d}\mathbb{P}^{E}_{X}&,x_{A}\notin
E^{\prime}\end{array}\right.,\\\
&\mathbb{E}(\xi_{X,A}|E)=x_{A}+\int_{\mathbb{R}}(x-x_{A})\mathbb{P}(x\text{ is
}A|X=x,E)\,\mathrm{d}\mathbb{P}^{E}_{X}, where $\mathbb{P}^{E}$ is the
probability measure on $(\Omega,\mathscr{F})$ given $E$, and
$\mathbb{P}_{X}^{E}$ is the probability measure induced by $X$ on $\mathbb{R}$
with respect to $\mathbb{P}^{E}$. \par Now, we have the following theorem,
which is similar to Theorem \ref{theoremcont} but in conditional case. In this
theorem, we assume that $f_{X|E}(t|E)$ and $\mathbb{P}(t\text{ is }A|X=t,E)$
are continuous functions of $t$. \par\begin{Theorem} Let $X$ be a classic
continuous distribution whose probability density function (pdf) is
$f_{X}(x)$. Then, the generalized pdf of $\xi_{X,A}$ given $E$, denoted by
$f_{\xi_{X,A}|E}$, is as follows: $$f_{\xi_{X,A}|E}(t|E)=\mathbb{P}(t\text{ is
}A|X=t,E)f_{X|E}(t|E)+\alpha^{E}\delta(t-x_{A}),$$ where,
$\alpha^{E}=1-\int_{-\infty}^{\infty}\mathbb{P}(x\text{ is
}A|X=x,E)f_{X|E}(x|E)\,\mathrm{d}x$. \end{Theorem} The above theorem follows
the following corollaries: \par\begin{Corollary}Considering the assumption of
Theorem \ref{cond}, if $\xi_{x,A}$ and $X$ are independent given $E$, then we
have that $$f_{\xi_{X,A}|E}(t|E)=\mathbb{P}(t\text{ is
}A|E)f_{X|E}(t|E)+\alpha^{E}\delta(t-x_{A}),$$ where,
$\alpha^{E}=1-\int_{-\infty}^{\infty}\mathbb{P}(x\text{ is
}A|E)f_{X|E}(x|E)\,\mathrm{d}x$. \end{Corollary}
\par\begin{Corollary}Considering the assumption of Theorem \ref{cond}, if $E$
is the event of "$\xi_{X,B}=x_{B}$", and $B$ is a fuzzy attribute of $X$, then
we have that $$f_{\xi_{X,A}|E}(t|E)=\mathbb{P}(t\text{ is }A|\neg(t\text{ is
}B))f_{X|E}(t|E)+\alpha\delta(t-x_{A}),$$ where,
$\alpha^{E}=1-\int_{-\infty}^{\infty}\mathbb{P}(x\text{ is }A|\neg(x\text{ is
}B))f_{X|E}(x|E)\,\mathrm{d}x$. \end{Corollary} \par Let's finish this section
by the following example: \par\begin{Example}Let $\Omega$ denotes the rainfall
in a specific region and $\Omega^{\prime}$ be collected water on the ground at
a certain place located in the region. Consider the fuzzy attributes $"high"$
and $"wet"$ for $\Omega$ and $\Omega^{\prime}$, respectively. Also, assume
that $X$ and $Y$ are the inclusion random variables corresponding to $\Omega$
and $\Omega^{\prime}$, respectively. One could think of being $"wet"$ for a
specific $y_{0}$ not depending on the joint distribution of the following
events: \begin{itemize}\par\itemize@item@$X=x$, \par\itemize@item@$Y=y$, and
\par\itemize@item@selecting $x^{\prime}$ as $"high"$ \par\end{itemize} for any
$x,x^{\prime}\in\mathrm{Supp}(X)$ and $y_{0},y\in\mathrm{Supp}(Y)$. Here,
since all conditions are removed, the above formulas for conditional
probabilities are not applicable. Hence, consider the case that we drop the
third one (i.e. selecting $x^{\prime}$ as "high"). However, for greater values
of $X=x$ and $Y=y$, since grater values of $Y$ are more likely available, we
might select $y_{0}$ with a lower probability as $"wet"$. Also, being $"high"$
depends on $X$, since for instance what is considered high rainfall in London
is significantly different from what is considered high rainfall in Lut desert
(located in Iran). Hence, dependencies of the above random variables come from
the subjective decisions that might be made by criteria such as the required
precision. Therefore, by considering $x_{high}=y_{wet}=0$ and the
aforementioned independence in selecting $y_{0}$ as $"wet"$ (without the third
one), we have the followings: \@@amsalign&\mathbb{P}(y_{0}\text{ is
}wet|Y=y,X=x)=\mathbb{P}(y_{0}\text{ is }wet),\\\
&f_{\xi_{Y,wet}}(y_{0})=\mathbb{P}(y_{0}\text{ is }wet)f_{Y}(y_{0}),\\\
&\mathbb{E}(\xi_{Y,wet})=\int_{-\infty}^{\infty}y\mathbb{P}(y\text{ is
}wet)f_{Y}(y)\,\mathrm{d}y,\\\
&f_{\xi_{Y,wet}|X}(y_{0}|x)=\mathbb{P}(y_{0}\text{ is
}wet)f_{Y|X}(y_{0}|x),\\\ &f_{Y|\xi_{X,high}}(y_{0}|x)=f_{Y|E}(y_{0}|E),\quad
E=(x\text{ is }high)\&(X=x), where $y_{0}\in\mathrm{Supp}(Y)\backslash\\{0\\}$
and $x\in\mathrm{Supp}(X)\backslash\\{0\\}$. Now, if we assume that selecting
$x$ as "high" is independent from the joint distribution of $X$ and $Y$ as
well, then the last formula can be written as follows:
\@@amsalign&f_{Y|\xi_{X,high}}(y_{0}|x)=f_{Y|X}(y_{0}|x). \end{Example}
\par\par\@@numbered@section{section}{toc}{Relationship Between Zadeh's PFL and
our Theory} Zadeh's PFL is different from our theory in nature. Indeed,
Zadeh's theory is a probability theory for fuzzy events, while our theory is
about the probability of selecting crisp elements with a certain fuzzy
attribute. Anyway, several aspects of Zadeh's theory could be interpreted in
our theory. Consider a probability space $(\Omega,\mathscr{F},P)$, where
$\Omega\subseteq\mathbb{R}$ and $\mathscr{F}$ is the Borel $\sigma$-algebra on
$\Omega$. Let $X:\Omega\to\mathbb{R}$ be the inclusion random variable, and
let $A$ be a fuzzy attribute of $\Omega$. Zadeh defined the probability of
$\widetilde{A}$ (i.e. the fuzzy subset of $\Omega$ associated to $A$) as
follows: $$P(\widetilde{A})=\int_{\mathbb{R}}\mu_{A}(x)\,\mathrm{d}P_{X},$$
where $P_{X}$ is the probability function induced by $X$. The above
probability in our Simple Fuzzy Model is equal to $\mathbb{P}(\Omega\text{ is
}A)$ (see Section \ref{Measure Theoretical Approach}). Zadeh also defined a
mean for $P(\widetilde{A})$ as follows:
$$m(\widetilde{A})=\frac{1}{P(\widetilde{A})}\int_{\mathbb{R}}x\mu_{A}(x)\,\mathrm{d}P_{X}.$$
Note that $m(\widetilde{A})$ in our Simple Fuzzy Model could be obtained as
follows:
$$m(\widetilde{A})=\frac{\mathbb{E}\left(\xi_{X,A}\right)}{\mathbb{P}(\Omega\text{
is }A)},$$ where $x_{A}=0$.
\par\par\par\@@numbered@section{section}{toc}{Conclusion} \par One major
problem with Deep Learning Algorithms is that they fail when it comes to
reasoning (see \cite[cite]{[\@@bibref{}{faghihi2020association}{}{}]},
\cite[cite]{[\@@bibref{}{madan2021fast}{}{}]}, and
\cite[cite]{[\@@bibref{}{pearl2018book}{}{}]}). Hence, it has been suggested
in \cite[cite]{[\@@bibref{}{faghihi2020cog}{}{}]} and
\cite[cite]{[\@@bibref{}{faghihi2020association}{}{}]} to use Non-Classical
Logics to equip the machine with reasoning. In this paper, we constructed a
PFL framework that can solve basic problems but fundamental in some different
types of reasoning such as Causal Reasoning. Our PFL suggests a probabilistic
screening criterion to feed a machine with datapoints satisfying a specific
fuzzy attribute at a time (see Section \ref{Problem Setting and Some
Motivations}). Further, it provides an assignment mechanism to calculate the
fuzzy average treatment effect for Causal Inference. Our Fuzzy Causal
Inference framework is flexible enough to integrate the vagueness of the
selection criteria into different Causal Inference concepts such as
ignorability condition. Note that as we mentioned in Section \ref{Related
Works}, the previous PFL frameworks have at least one of the following
drawbacks: \begin{itemize}\par\itemize@item@The lack of a precise mathematical
setting. \par\itemize@item@Providing a framework for which events are just
fuzzy, and hence it is not possible to have crisp events while the criteria of
the random selection are fuzzy. \par\itemize@item@Computationally expensive.
\par\end{itemize} Our PFL framework solves all of the above drawbacks. \par In
the near future, we intend to create an integration theory of Fuzzy Logic and
Causal Inference by using our PFL framework. Although we very briefly touched
the concepts of Causal Inference in this paper, we prepared the ground for the
fully integration of PFL and Causal Inference in general. To demonstrate this,
we showed an application of our PFL theory in Causal Inference. \par We
recommend other researchers to take part in developing the PFL framework
suggested in this paper. For instance, the followings are recommended:
\begin{itemize}\par\itemize@item@Working on the probabilistic aspects of our
PFL theory such as studying the independence, dependence and correlation of
random variables, and studying famous inequalities and theorems in Probability
Theory. \par\itemize@item@Constructing new models of our PFL capable of
processing specific needs for specific applications in the areas such as
Medicine, Market Research, Social Studies and Control Systems.
\par\itemize@item@Working on the philosophical aspects of our PFL theory. A
notable work in our research was to interpret "selecting {\it nothing }" as
"selecting a specific element". This could be controversial for some
researchers and philosophers. \par\end{itemize}
\par\par\@@unnumbered@section{section}{}{Acknowledgments} \par The research of
Amir Saki has been supported by the Institute for Research in Fundamental
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\emph{Intuitive topology}, Universities Press, 1998\.
\par\lx@bibitem{causalinference} Rationality, \emph{Encyclopædia britannica},
Encyclopædia Britannica, Inc., 2014\. \par\lx@bibitem{zadeh1996fuzzy}
Lotfi~{}A Zadeh, \emph{Fuzzy probabilities}, Fuzzy Sets, Fuzzy Logic, And
Fuzzy Systems: Selected Papers by Lotfi A Zadeh, World Scientific, 1996,
pp.~{}643--652. \par\lx@bibitem{zadeh1968probability} Lotfi~{}Asker Zadeh,
\emph{Probability measures of fuzzy events}, Journal of mathematical analysis
and applications {23} (1968), no.~{}2, 421--427. \par\endthebibliography
\par\par\@@numbered@section{appendix}{toc}{Topological Spaces} Let
$f:\mathbb{R}\to\mathbb{R}$ be a function. Then, $f$ is continuous if and only
if $f^{-1}(I)$ is a union of open intervals for any open interval
$I\subseteq\mathbb{R}$. Continuous functions are important for many
mathematical purposes such as integration. To generalize continuity on
arbitrary spaces (not only $\mathbb{R}$), one common and popular way is to
define the so called open sets.
\par\par\@@numbered@section{subsection}{toc}{Definition of a Topology} \par
Let $\Omega$ be a set. A collection $\tau$ of subsets of $\Omega$ is called a
topology on $\Omega$, whenever
\begin{itemize}\par\itemize@item@$\emptyset,\Omega\in\tau$,
\par\itemize@item@$U,V\in\tau$ implies that $U\cap V\in\tau$, and
\par\itemize@item@for any family $\\{U_{i}\\}_{i\in I}$ of elements $\tau$, we
have that $\bigcup_{i\in I}U_{i}\in\tau$. \par\end{itemize} Then,
$(\Omega,\tau)$ (or simply $\Omega$) is called a topological space. Moreover,
each element of $\tau$ is called an open subset of $\Omega$. The complement of
each open subset of $\Omega$ is called a closed subset of $\Omega$.
\par\par\@@numbered@section{subsection}{toc}{Base for a Topology} \par Let
$(\Omega,\tau)$ be a topological space. A subcollection
$\mathcal{B}=\\{B_{i}\\}_{i\in I}$ of $\tau$ is called a base for $\tau$, if
\begin{itemize}\par\itemize@item@$\bigcup_{i\in I}B_{i}=X$, and
\par\itemize@item@for any $\omega\in B_{i}\cap B_{j}$, there exists $l\in I$
with $\omega\in B_{l}\subseteq B_{i}\cap B_{j}$ for any $i,j\in I$.
\par\end{itemize} A topology could be uniquely determined by its base. Indeed,
if $\mathcal{B}$ is a base for the topology $\tau$, then $\tau$ is the
collection of arbitrary unions of the elements of $\mathcal{B}$. Conversely,
any collection $\mathcal{B}$ of subsets of a set $\Omega$ satisfying the above
conditions uniquely defines a topology on $\Omega$. \par Let $\Omega$ be a
set. Then, $\tau_{1}=\\{\emptyset,\Omega\\}$ and
$\tau_{2}=\mathcal{P}(\Omega)$ are the smallest and the largest topologies
defined on $\Omega$, respectively. The standard topology on $\mathbb{R}$ is
the collection of arbitrary unions of open intervals.
\par\@@numbered@section{subsection}{toc}{Continuous Functions} Let
$(\Omega,\tau)$ and $(\Omega^{\prime},\tau^{\prime})$ be two topological
spaces. A function $f:(\Omega,\tau)\to(\Omega^{\prime},\tau^{\prime})$ is
called continuous, whenever $f^{-1}(U)\in\tau$ for any $U\in\tau^{\prime}$.
One could see that $f:(\Omega,\tau)\to(\Omega^{\prime},\tau^{\prime})$ is
continuous if and only if $f^{-1}(B^{\prime})\in\tau$ for any
$B^{\prime}\in\mathcal{B}^{\prime}$, where $\mathcal{B}^{\prime}$ is a base
for $\tau^{\prime}$. \par\par\@@numbered@section{subsection}{toc}{Product
Topology} \par Let $(\Omega,\tau)$ and $(\Omega^{\prime},\tau^{\prime})$ be
two topological spaces. Then, the product topology on the Cartesian product
$\Omega\times\Omega^{\prime}$ is the topology whose base is $\\{U\times
U^{\prime}:U\in\tau,U^{\prime}\in\tau^{\prime}\\}$. Note that if $\mathcal{B}$
and $\mathcal{B}^{\prime}$ are bases of $\tau$ and $\tau^{\prime}$,
respectively, then $\\{B\times
B^{\prime}:B\in\mathcal{B},B^{\prime}\in\mathcal{B}^{\prime}\\}$ is the base
for the product topology on $\Omega\times\Omega^{\prime}$. Now, inductively,
one could define the product topology on the Cartesian product
$\Omega_{1}\times\cdots\times\Omega_{n}$ of topological spaces
$(\Omega_{i},\tau_{i})$ for any $i=1\ldots,n$. For instance, the product
topology of the standard topology on $\mathbb{R}$ with itself on
$\mathbb{R}^{2}$ has the following base: $$\\{I\times J:I\text{ and }J\text{
are open intervals in }\mathbb{R}\\}.$$ In general, the following set (i.e.,
the set of all hypercubes in $\mathbb{R}^{n}$) is a base for the standard
topology on $\mathbb{R}^{n}$: $$\\{I_{1}\times\cdots\times I_{n}:I_{i}\text{'s
are open intervals in }\mathbb{R}^{n}\\}.$$ In the sequel, we consider the
standard topology on $\mathbb{R}^{n}$. Note that it could be shown that any
open set in $\mathbb{R}^{n}$ could be written as a countable union of disjoint
hypercubes. \par\@@numbered@section{subsection}{toc}{Extended Real Line} \par
Let $\overline{\mathbb{R}}=\mathbb{R}\cup\\{\pm\infty\\}$. We define the
following arithmetic: \@@amsalign&a\times\infty=\infty\times
a=-((-a)\times\infty)=-(\infty\times(-a))=\infty,\\\
&b+\infty=\infty+\infty=\infty\times\infty=\infty,\qquad\frac{b}{\infty}=0.
for any $a,b\in\mathbb{R}$ with $a>0$. Indeed, there are similar identities
for $-\infty$. Note that we leave $\pm(\infty-\infty)$, $0\times(\pm\infty)$
and $(\pm\infty)\times 0$ undefined. One could see that the following set is a
base for a topology on $\overline{\mathbb{R}}$ called standard base:
$$\mathcal{B}=\\{(a,\infty]:a\in\mathbb{R}\\}\cup\\{[-\infty,b):a\in\mathbb{R}\\}\cup\\{(a,b):a,b\in\mathbb{R},\;a<b\\},$$
which is called the standard topology on $\overline{\mathbb{R}}$. We define
$\overline{\mathbb{R}^{n}}:=\overline{\mathbb{R}}^{n}$ for any positive
integer $n$, and we equip it by the product topology. It is worth mentioning
that each open set in $\overline{\mathbb{R}}^{n}$ could be written as a
countable union of disjoint elements of the form of $E_{1}\times\cdots\times
E_{n}$, where $E_{i}\in\mathcal{B}$ for any $1\leq i\leq n$.
\par\vspace*{.4cm} Note that several important concepts such as connected
spaces, compact spaces, local properties, quotient spaces and several other
concepts are discussed in Topology as a branch of mathematics. To study more
in General Topology, readers are referred to the following textbooks
\cite[cite]{[\@@bibref{}{armstrong2013basic}{}{}]},
\cite[cite]{[\@@bibref{}{munkres2000topology}{}{}]} and
\cite[cite]{[\@@bibref{}{prasolov1998intuitive}{}{}]}.
\par\par\@@numbered@section{appendix}{toc}{Measure Theory} In this appendix,
we provide some definitions and theorems required for explaining the measure
theoretical point of view of Probability Theory (see Appendix \ref{Probability
Theory (Measure Theoretical Point of View)}). To see more details, we refer
readers to study a standard textbook such as
\cite[cite]{[\@@bibref{}{bogachev2007measure}{}{}]} and
\cite[cite]{[\@@bibref{}{folland1999real}{}{}]}.
\par\@@numbered@section{subsection}{toc}{$\sigma$-Algebras} \par Let $\Omega$
be a set. A $\sigma$-algebra on $\Omega$ is a collection $\mathscr{F}$ of
subsets of $\Omega$ which satisfies the following properties:
\begin{enumerate}\par\enumerate@item@$\Omega\in\mathscr{F}$,
\par\enumerate@item@$E\in\mathscr{F}$ implies that
$\Omega\backslash\\{E\\}\in\mathscr{F}$ for any $E\in\mathscr{F}$, and
\par\enumerate@item@if $\\{E_{i}\\}_{i=1}^{\infty}$ is a family of the
elements of $\mathscr{F}$ then $\bigcup_{i=1}^{\infty}E_{i}\in\mathscr{F}$.
\par\end{enumerate} Let $\Omega$ be a set and $\mathscr{F}$ be a
$\sigma$-algebra on $\Omega$. It follows from these properties that
$\emptyset$ and the intersection of any countable family of elements of a
$\sigma$-algebra on $\Omega$ belong to the $\sigma$-algebra as well. We
sometimes refer to elements of a $\sigma$-algebra on $\Omega$ as measurable
subsets of $\Omega$. \par For any set $\Omega$, the collection
$\\{\emptyset,\Omega\\}$ and the powerset of $\Omega$ are the smallest and
largest $\sigma$-algebras defined on $\Omega$. The $\sigma$-algebra generated
by a collection $\mathcal{S}$ of subsets of $\Omega$ is defined to be the
intersection of all $\sigma$-algebras on $\Omega$ containing $\mathcal{S}$.
Note that the $\sigma$-algebra generated by all open subsets of
$\mathbb{R}^{n}$ is called the Borel $\sigma$-algebra and is denoted by
$\mathscr{B}(\mathbb{R}^{n})$. The $\sigma$-algebra generated by each of the
following sets is $\mathscr{B}(\mathbb{R})$:
\@@amsalign\mathcal{B}_{1}&=\\{(a,b):a,b\in\mathbb{R},\;a<b\\},&\mathcal{B}_{2}&=\\{(a,b]:a,b\in\mathbb{R},\;a<b\\},\\\
\mathcal{B}_{3}&=\\{[a,b):a,b\in\mathbb{R},\;a<b\\},&\mathcal{B}_{4}&=\\{[a,b]:a,b\in\mathbb{R},\;a\leq
b\\},\\\
\mathcal{B}_{5}&=\\{(a,\infty):a\in\mathbb{R}\\}\cup\\{(-\infty,b]:b\in\mathbb{R}\\},&&\\\
\mathcal{B}_{6}&=\\{[a,\infty):a\in\mathbb{R}\\}\cup\\{(-\infty,b]:b\in\mathbb{R}\\},&&\\\
\mathcal{B}_{7}&=\\{(a,\infty):a\in\mathbb{R}\\}\cup\\{(-\infty,b):b\in\mathbb{R}\\},&&\\\
\mathcal{B}_{8}&=\\{[a,\infty):a\in\mathbb{R}\\}\cup\\{(-\infty,b):b\in\mathbb{R}\\}.&&
\par\par\par Indeed, the Borel $\sigma$-algebra is defined for any topological
space. Especially, for any integer $n$, we have the Borel $\sigma$-algebra on
$\overline{\mathbb{R}}^{n}$, denoted by
$\mathscr{B}(\overline{\mathbb{R}}^{n})$(see Appendix~{}\ref{1} for the
definition of $\overline{\mathbb{R}}^{n}$ and the the standard topology on
it). \par\par\@@numbered@section{subsection}{toc}{Measurable Spaces and
Measurable Functions} \par For a set $\Omega$ and a $\sigma$-algebra
$\mathscr{F}$ on $\Omega$, the pair $(\Omega,\mathscr{F})$ is called a
measurable space. Let $(\Omega,\mathscr{F})$ and
$(\Omega^{\prime},\mathscr{F}^{\prime})$ be two measurable spaces. A function
$f:(\Omega,\mathscr{F})\to(\Omega^{\prime},\mathscr{F}^{\prime})$ is called
measurable if $f^{-1}(E^{\prime})\in\mathscr{F}$ for any
$E^{\prime}\in\mathscr{F}^{\prime}$. If $\Omega$ and $\Omega^{\prime}$ are
both topological spaces and $\mathscr{F}$ and $\mathscr{F}^{\prime}$ are the
Borel $\sigma$-algebras on $\Omega$ and $\Omega^{\prime}$, respectively, then
any continuous function from $(\Omega,\mathscr{F})$ to
$(\Omega^{\prime},\mathscr{F}^{\prime})$ is measurable (but the converse is
wrong). One could see that the composition of two measurable functions is also
measurable. As a convention, by a real valued measurable function
$f:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}$, we mean
$f:(\Omega,\mathscr{F})\to(\overline{\mathbb{R}},\mathscr{B}(\overline{\mathbb{R}}))$
is measurable. We also consider the similar convention for multi-valued
functions into $\overline{\mathbb{R}}^{n}$.
\par\par\@@numbered@section{subsection}{toc}{Indicator Functions and Simple
Functions} Let $(\Omega,\mathscr{F})$ be a measurable space and
$E\in\mathscr{F}$. Then, the indicator function of $E$, denoted by
$\mathbbm{1}_{E}$, is defined as follows:
$$\mathbbm{1}_{E}:\Omega\to\mathbb{R},\qquad\mathbbm{1}_{E}(\omega)=\left\\{\begin{array}[]{ll}1&,\omega\in
E\\\ 0&,\omega\notin E\end{array}\right..$$ Let
$E_{1},\ldots,E_{n}\in\mathscr{F}$ be pairwise disjoint and
$r_{1},\ldots,r_{n}\in\mathbb{R}$. Then,
$\sum_{i=1}^{n}r_{i}\mathbbm{1}_{E_{i}}$ is called a simple function. One
could see that the sum, product and composition of two simple functions is a
simple function as well. Note that a simple function (and hence an indicator
function) is measurable. \par A famous result states that any non-negative
measurable function $g:\Omega\to\overline{\mathbb{R}}$ is the point-wise limit
of a sequence of simple functions defined on $\Omega$. This result plays a key
role in the sequel. \par\@@numbered@section{subsection}{toc}{Measures} \par
Let $(\Omega,\mathscr{F})$ be a measurable space. A function
$\mu:\mathscr{F}\to[0,\infty]$ is called a measure, whenever it satisfies the
following conditions: \begin{enumerate}\par\enumerate@item@$\mu(\emptyset)=0$,
and
\par\enumerate@item@$\mu\left(\bigcup_{i=1}^{\infty}E_{i}\right)=\sum_{i=1}^{\infty}\mu(E_{i})$
for any countable family $\\{E_{i}\\}_{i=1}^{\infty}$ of pairwise disjoint
elements $\mathscr{F}$. \par\end{enumerate} One could see that a measure $\mu$
defined on a measurable space $(\Omega,\mathscr{F})$ is increasing (i.e.,
$\mu(E)\leq\mu(E^{\prime})$ if $E\subseteq E^{\prime}$ for any
$E,E^{\prime}\in\mathscr{F}$). If $\mu$ is a measure defined on the measurable
space $(\Omega,\mathscr{F})$, then $(\Omega,\mathscr{F},\mu)$ is called a
measure space. \par As an example, assume that $\mathscr{F}$ is the powerset
of $\Omega$ and $\omega$ is an arbitrary element of $\Omega$. Then, the Dirac
measure with respect to $\omega$ is defined as follows:
$$\delta_{\omega}(E)=\left\\{\begin{array}[]{ll}1&,\omega\in E\\\
0&,\omega\notin E\end{array}\right.$$ for any subset $E$ of $\Omega$.
\par\par\@@numbered@section{subsection}{toc}{Continuity Property of Measures}
\par Let $(\Omega,\mathscr{F},\mu)$ be a measure space and
$\\{E_{n}\\}_{n=1}^{\infty}$ be a family of elements of $\mathscr{F}$ with
$E_{1}\subseteq E_{2}\subseteq\cdots$. Let us assume that
$E=\bigcup_{n=1}^{\infty}E_{n}$ (i.e., $E_{n}\uparrow E$). Then, obviously $E$
is measurable. Now, we show that $\mu(E)=\lim_{N\to\infty}\mu(E_{N})$. To see
this, assume that $$B_{1}=E_{1},\qquad B_{n}=E_{n}\backslash E_{n-1},\quad
n=2,3,\ldots.$$ Then, $\\{B_{n}\\}_{n=1}^{\infty}$ is a family of pairwise
disjoint measurable sets with $E=\bigcup_{n=1}^{\infty}B_{n}$. Thus, we have
that
\@@amsalign\mu(E)=\mu\left(\bigcup_{n=1}^{\infty}B_{n}\right)=\sum_{n=1}^{\infty}\mu(B_{n})=\lim_{N\to\infty}\sum_{n=1}^{N}\mu(B_{n})=\lim_{N\to\infty}\mu\left(\bigcup_{n=1}^{N}B_{n}\right)=\lim_{N\to\infty}\mu(E_{N}).
Similarly, one could see that if $E_{1}\supseteq E_{2}\supseteq\cdots$,
$E=\bigcap_{n=1}^{\infty}E_{n}$ (i.e., $E_{n}\downarrow E$), and
$\mu(E_{1})<\infty$, then $\mu(E)=\lim_{N\to\infty}\mu(E_{N})$.
\par\@@numbered@section{subsection}{toc}{Almost Everywhere Properties} \par
Let $(\Omega,\mathscr{F},\mu)$ be a measure space and $\mathcal{P}$ be a
property of elements in $\Omega$. It is said that $\mathcal{P}$ holds almost
everywhere in $\Omega$ if there exists $E\in\mathscr{F}$ with $\mu(E)=0$, and
$\mathcal{P}$ holds for each $\omega\in\Omega\backslash\\{E\\}$. For instance,
assume that the measure of any singleton is zero. If a property is satisfied
for all elements of $\Omega$ except for a countable subset of $\Omega$, since
the measure of any countable set is zero, then that property holds almost
everywhere in $\Omega$. In the sequel, any equality or property involving
measures holds almost everywhere but we might no longer mention it. For
instance, when we say that "there is a unique function satisfying the property
$\mathcal{P}$", we mean that if two functions satisfy $\mathcal{P}$, then they
are almost everywhere equal.
\par\par\par\@@numbered@section{subsection}{toc}{Lebesgue Measure} \par Any
measure defined on a Borel $\sigma$-algebra is called a Borel measure. A
theorem guaranties that there is a unique Borel measure $\lambda$ on
$\overline{\mathbb{R}}$ in such a way that $\lambda([a,b])=b-a$ for any real
numbers $a$ and $b$ with $a\leq b$. Similarly, there is a unique Borel measure
$\lambda$ on $\overline{\mathbb{R}}^{n}$ such that
$\lambda(H)=\mathrm{Vol}(H)$, where $H$ is a hypercube (i.e., a Cartesian
product of $n$ intervals in $\overline{\mathbb{R}}$) and $\mathrm{Vol}(H)$ is
the volume of $H$. The aforementioned Borel measure on
$\overline{\mathbb{R}}^{n}$ is called the Lebesgue measure. For instance,
hypercubes in $\overline{\mathbb{R}}^{2}$ and $\overline{\mathbb{R}}^{3}$ are
squares and cubes, respectively. Hence, intuitively the Lebesgue measures in
$2$ and $3$ dimensions represent the area and the ordinary volume,
respectively. \par\par\par\@@numbered@section{subsection}{toc}{Integrals} \par
Let $(\Omega,\mathscr{F},\mu)$ be a measure space and
$s=\sum_{i=1}^{n}r_{i}\mathbbm{1}_{E_{i}}$ be a non-negative simple function.
The integral of $s$ on $\Omega$ is defined as follows:
$$\int_{\Omega}s\,\mathrm{d}\mu=\sum_{i=1}^{n}r_{i}\mu(E_{i}).$$ Now, assume
that $f:\Omega\to[0,\infty]$ is a measurable function. Then, the integral of
$f$ on $\Omega$ is defined as below:
$$\int_{\Omega}f\,\mathrm{d}\mu=\sup\left\\{\int_{\Omega}s\,\mathrm{d}\mu:s\text{
is a simple function with }0\leq s\leq f\right\\}.$$ If $E\in\mathscr{F}$,
then the integral of $f$ on $E$ is defined as follows:
$$\int_{E}f\,\mathrm{d}\mu=\int_{\Omega}f\mathbbm{1}_{E}\,\mathrm{d}\mu.$$
Note that if $f$ is only defined on $E\in\mathscr{F}$, then we can extend it
to a measurable function $\tilde{f}:\Omega\to\mathbb{R}$ by the following
setting: $$\tilde{f}(x)=\left\\{\begin{array}[]{ll}f(x)&,x\in E\\\ 0&,x\notin
E\end{array}\right.,$$ and we define
$$\int_{E}f\,\mathrm{d}\mu=\int_{E}\tilde{f}\,\mathrm{d}\mu.$$ Now, let
$f:\Omega\to\mathbb{R}$ be a measurable function. Assume that
$f^{+}=\max\\{f,0\\}$ and $f^{-}=\max\\{-f,0\\}$. Both $f^{+}$ and $f^{-}$ are
non-negative and measurable, and we have that $f=f^{+}-f^{-}$ and
$|f|=f^{+}+f^{-}$. We say that $f$ is integrable, if
$\int_{\Omega}|f|\,\mathrm{d}\mu<\infty$. Now, for an integrable function
$f:\Omega\to\mathbb{R}$, we define
$$\int_{E}f\,\mathrm{d}\mu=\int_{E}f^{+}\,\mathrm{d}\mu-\int_{E}f^{-}\,\mathrm{d}\mu.$$
To be more precise, sometimes we use the notation
$\int_{E}f(\omega)\,\mathrm{d}\mu(\omega)$ to clarify the variable we are
taking the integral with respect to.
\par\par\par\@@numbered@section{subsection}{toc}{Integrals of Multi-Values
Functions} \par Let $(\Omega,\mathscr{F},\mu)$ be a measure space and
$f:\Omega\to\overline{\mathbb{R}}^{n}$ be non-negative (in each component) and
measurable. Assume that $f(\omega)=(f_{1}(\omega),\ldots,f_{n}(\omega))$ for
any $\omega\in\Omega$. For any $1\leq i\leq n$, note that
$\pi_{i}:\overline{\mathbb{R}}^{n}\to\overline{\mathbb{R}}$ sending each
$\omega\in\overline{\mathbb{R}}^{n}$ to its $i^{\text{th}}$ component is
measurable, which implies that $f_{i}=\pi_{i}\circ f$ is measurable as well.
In general, $f$ is measurable if and only if $f_{i}$ is measurable for any
$1\leq i\leq n$. Now, we define the integral of $f$ as follows:
$$\int_{E}f\,\mathrm{d}\mu:=\left(\int_{E}f_{1}\,\mathrm{d}\mu,\ldots,\int_{E}f_{n}\,\mathrm{d}\mu\right).$$
Similarly, an arbitrary function $f:\Omega\to\overline{\mathbb{R}}^{n}$ (not
necessarily non-negative) is integrable if and only if $f_{i}$ is intgrable
for any $1\leq i\leq n$. Thus, the above definition is considered for the
integrable function $f$ as well. Hence, many properties of the integration of
the single valued functions (provided in the sequel) hold for multi-values
functions as well. \par\@@numbered@section{subsection}{toc}{Some Properties of
Integrals} Let $(\Omega,\mathscr{F},\mu)$ be a measure space, and let
$f,g:\Omega\to\overline{\mathbb{R}}$ be non-negative and measurable (or in
general case, one could consider the integrable functions). If
$\\{E_{i}\\}_{i=1}^{\infty}$ is a family of pairwise disjoint measurable
subsets of $\Omega$, then we have that
$$\int_{\bigcup_{i=1}^{\infty}E_{i}}f\,\mathrm{d}\mu=\sum_{i=1}^{\infty}\int_{E_{i}}f\,\mathrm{d}\mu.$$
Further, if $\lambda,\eta\in\mathbb{R}$, then for any measurable subset $E$ of
$\Omega$, we have that $$\int_{E}(\lambda f+\eta
g)\,\mathrm{d}\mu=\lambda\int_{E}f\,\mathrm{d}\mu+\eta\int_{E}g\,\mathrm{d}\mu.$$
Furthermore, if $\\{f_{n}\\}_{n=1}^{\infty}$ is a sequence of non-negative
measurable real valued functions defined on $(\Omega,\mathscr{F})$, then we
have that
$$\sum_{n=1}^{\infty}\int_{E}f_{n}(\omega)\,\mathrm{d}\mu(\omega)=\int_{E}\sum_{n=1}^{\infty}f_{n}(\omega)\,\mathrm{d}\mu(\omega)$$
for any $E\in\mathscr{F}$.
\par\par\@@numbered@section{subsection}{toc}{Measures Defined via Measures}
\par Let $(\Omega,\mathscr{F},\mu)$ be a measure space and
$f:(\Omega,\mathscr{F})\to[0,\infty]$ be a non-negative measurable function.
Define $\nu:\mathscr{F}\to[0,\infty]$ as follows:
$$\nu(E)=\int_{E}f\,\mathrm{d}\mu$$ for any $E\in\mathscr{F}$. Then, $\nu$ is
a measure on $(\Omega,\mathscr{F})$. This method to define new measures is
common to define probability measures.
\par\par\par\@@numbered@section{subsection}{toc}{Monotone Convergence Theorem}
Let $(\Omega,\mathscr{F},\mu)$ be a measure space and
$f:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}$ be measurable and non-
negative. If $\\{s_{n}\\}_{n=1}^{\infty}$ is a sequence of non-negative simple
functions defined on $(\Omega,\mathscr{F})$ which converges to $f$ almost
everywhere, then
$$\lim_{n\to\infty}\int_{E}s_{n}(\omega)\,\mathrm{d}\mu(\omega)=\int_{E}f(\omega)\,\mathrm{d}\mu(\omega)$$
for any $E\in\mathscr{F}$.
\par\par\@@numbered@section{subsection}{toc}{Dominating Convergence Theorem}
\par Let $(\Omega,\mathscr{F},\mu)$ be a measure space, and
$\\{f_{n}\\}_{n=1}^{\infty}$ be a sequence of non-negative measurable
functions from $(\Omega,\mathscr{F})$ to $\overline{\mathbb{R}}^{n}$. If
$\\{f_{n}\\}_{n=1}^{\infty}$ converges point-wise to $f$, and there exists an
integrable function $g:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ in
such a way that $|f_{n}(\omega)|\leq g(\omega)$ for any $\omega\in\Omega$,
then we have that
$$\lim_{n\to\infty}\int_{E}f_{n}(\omega)\,\mathrm{d}\mu(\omega)=\int_{E}f(\omega)\,\mathrm{d}\mu(\omega)$$
for any $E\in\mathscr{F}$.
\par\par\par\@@numbered@section{subsection}{toc}{Equality of Finite Borel
Measures} \par Let $\Omega$ be a topological space, and let $\mu_{1}$ and
$\mu_{2}$ be two finite Borel measures on $\Omega$. Then, by
\cite[cite]{[\@@bibref{}{bogachev2007measure}{}{}, Lemma 7.1.2]}, if $\mu_{1}$
and $\mu_{2}$ are equal on all open sets in $\Omega$, then they are equal on
all Borel sets in $\Omega$. Now, let's consider the case
$\Omega=\mathbb{R}^{n}$. We know that any open set in $\mathbb{R}^{n}$ is a
countable union of disjoint hypercubes, which implies that if two Borel
measures on $\mathbb{R}^{n}$ are equal on all hypercubes, then they are equal
on all open sets and hence equal on all Borel sets in $\Omega$. Similarly, if
two Borel measures are equal on a base $\mathcal{B}$ for the standard topology
on $\overline{\mathbb{R}}^{n}$ (discussed in \ref{extendedrealline}), then
they are equal on all Borel sets in $\overline{\mathbb{R}}^{n}$ .
\par\par\@@numbered@section{subsection}{toc}{Computing Lebesgue Integrals via
Riemann Integrals} Consider the measure space
$(\mathbb{R},\mathscr{B}(\mathbb{R}),\lambda)$, where $\lambda$ is the
Lebesgue measure. Then, one could see that any Riemann integrable function
$f:[a,b]\to\mathbb{R}$ is Lebesgue integrable, and we have that
$$\int_{[a,b]}f\,\mathrm{d}\mu=\int_{a}^{b}f\,\mathrm{d}x.$$ Also, if
$f:\mathbb{R}\to\mathbb{R}$ is a non-negative Riemann integrable function (in
the sense of improper integrability), then $f$ is Lebesgue integrable, and we
have that
$$\int_{\mathbb{R}}f\,\mathrm{d}\mu=\int_{-\infty}^{\infty}f\,\mathrm{d}x.$$ A
similar statement holds when we consider any interval like $[a,\infty)$ and
$(-\infty,b]$ for any real numbers $a$ and $b$.
\par\par\@@numbered@section{subsection}{toc}{Radon-Nikodym Theorem} \par Let
$(\Omega,\mathscr{F},\mu)$ be a measure space. Then, $\mu$ is called a finite
measure, if $\mu(\Omega)<\infty$ (and hence $\mu(E)<\infty$ for any
$E\in\mathscr{F}$). Further, $\mu$ is called a $\sigma$-finite measure if
there is a countable family $\\{E_{i}\\}_{i=1}^{\infty}$ of measurable subsets
of $\Omega$ with finite measure (i.e., $\mu(E_{i})<\infty$ for any $i$) such
that $\Omega=\bigcup_{i=1}^{\infty}E_{i}$. Obviously, a finite measure is a
$\sigma$-finite measure. \par Now, assume that $(\Omega,\mathscr{F})$ is a
measurable space, and $\mu$ and $\nu$ are two measures defined on this space.
Then, we say that $\nu$ is absolutely continuous with respect to $\mu$, if
$\mu(E)=0$ implies that $\nu(E)=0$ for any $E\in\mathscr{F}$. \par Now, we are
ready to state the Radon-Nikodym theorem. Let $(\Omega,\mathscr{F})$ be a
measurable space, and $\mu$ and $\nu$ be two $\sigma$-finite measures on this
space. If $\nu$ is absolutely continuous with respect to $\mu$, then there
exists a non-negative measurable function $f:\Omega\to[0,\infty]$ in such a
way that $$\nu(E)=\int_{E}f\,\mathrm{d}\mu$$ for any $E\in\mathscr{F}$.
\par\par\@@numbered@section{subsection}{toc}{Changing the Measure of
Integration} \par Let $(\Omega,\mathscr{F})$ be a measurable space. Also, let
$\mu$ and $\nu$ be two measures on $(\Omega,\mathscr{F})$ such that $\nu$ is
absolutely continuous with respect to $\mu$. Then, by Radon-Nikodym theorem,
there exists a measurable function $f:\Omega\to[0,\infty]$ satisfying the
following: $$\nu(E)=\int_{E}f\,\mathrm{d}\mu$$ for any $E\in\mathscr{F}$. Now,
if $g:\Omega\to[0,\infty]$ is a measurable function, then for any
$E\in\mathscr{F}$, we have that
$$\int_{E}g\,\mathrm{d}\nu=\int_{E}gf\,\mathrm{d}\mu.$$ To prove it, assume
that $E,E^{\prime}\in\mathscr{F}$. If $g=\mathbbm{1}_{E^{\prime}}$, then
$$\int_{E}\mathbbm{1}_{E^{\prime}}\,\mathrm{d}\nu=\nu(E\cap
E^{\prime})=\int_{E\cap
E^{\prime}}f\,\mathrm{d}\mu=\int_{E}\mathbbm{1}_{E^{\prime}}f\,\mathrm{d}\mu.$$
Hence, the desired identity holds for all indicator functions $g$. By the
linearity of integrals, this identity holds for all simple functions $g$ as
well. Now, by the monotone convergence theorem, we obtain the identity for all
non-negative measurable functions $g$.
\par\par\@@numbered@section{subsection}{toc}{Change of Variable Theorem} \par
Let $(\Omega,\mathscr{F},\mu)$ and $(\Omega^{\prime},\mathscr{F}^{\prime})$ be
a measure space and a measurable space, respectively. Let
$F:(\Omega,\mathscr{F})\to(\Omega^{\prime},\mathscr{F}^{\prime})$ be a
measurable function. Then, $F$ induces a measure on
$(\Omega^{\prime},\mathscr{F}^{\prime})$ as follows:
$$\mu_{F}:\mathscr{F}^{\prime}\to[0,\infty],\quad\mu_{F}(E^{\prime})=\mu(F^{-1}(E^{\prime}))$$
for any $E^{\prime}\in\mathscr{F}^{\prime}$. Now, if
$g:(\Omega^{\prime},\mathscr{F}^{\prime})\to[0,\infty]$ is measurable, then we
have that $$\int_{\Omega}g\circ
F(\omega)\,\mathrm{d}\mu(\omega)=\int_{\Omega^{\prime}}g(\omega^{\prime})\,\mathrm{d}\mu_{F}(\omega^{\prime}).$$
\par Naturally, the above theorem holds for integrable functions
$g:(\Omega^{\prime},\mathscr{F}^{\prime})\to\overline{\mathbb{R}}^{n}$.
\par\@@numbered@section{appendix}{toc}{Probability Theory (Measure Theoretic
Point of View)} In this appendix, we provide a very brief description of the
basics of the measure theoretical Probability Theory. To deal with conditional
probabilities, we provide two different points of view, and explain how these
two points of view are consistent with each other. To study more in the
measure theoretical Probability Theory, we recommend the following textbooks
\cite[cite]{[\@@bibref{}{ash2000probability}{}{}]},
\cite[cite]{[\@@bibref{}{capinski2004measure}{}{}]} and
\cite[cite]{[\@@bibref{}{ccinlar2011probability}{}{}]}.
\par\par\@@numbered@section{subsection}{toc}{Probability Space} \par A
probability space is a measure space $(\Omega,\mathscr{F},P)$ such that
$P(\Omega)=1$. If $(\Omega,\mathscr{F},P)$ is a probability space, then
$\Omega$, $\mathscr{F}$ and $P$ are called the sample space, the event space
and the probability measure, respectively. Further, each element of the event
space is called an event.
\par\par\@@numbered@section{subsection}{toc}{Discrete Probability Spaces} \par
Let $(\Omega,\mathscr{F},P)$ be a probability space. If $\Omega$ is countable
and $\mathscr{F}$ is the powerset of $\Omega$, then $(\Omega,\mathscr{F},P)$
is called a discrete probability space. Note that for any $E\in\mathscr{F}$,
we have that $$P(E)=P\left(\bigcup_{e\in E}\\{e\\}\right)=\sum_{e\in
E}P(\\{e\\}).$$ \par\par\@@numbered@section{subsection}{toc}{Random Variables
and Random Vectors} \par Let $(\Omega,\mathscr{F},P)$ be a probability space.
A measurable function $X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ is
called a random vector. A random variable is obtained when in the above
definition we assume that $n=1$. Let
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. We
note that $X$ induces a probability measure $P_{X}$ on
$(\overline{\mathbb{R}}^{n},\mathscr{B}(\overline{\mathbb{R}}^{n}))$ as
follows: $$P_{X}(E^{\prime})=P(X^{-1}(E^{\prime}))$$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. We define the support
of $X$, denoted by $\mathrm{Supp}(X)$, to be the smallest closed set $V$ with
$P_{X}(V)=1$. Especially, if $X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}$
is a random variable which takes a finitely number of values (i.e., a simple
random variable), we have that
$$\mathrm{Supp}(X)=\\{r\in\overline{\mathbb{R}}:P_{X}(r)>0\\}.$$ If the set of
values of a random vector $X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$
is countable, then $X$ is called a discrete random vector. We note that the
support of a discrete random variable can be obtained similar to the support
of a simple random variable. \par\@@numbered@section{subsection}{toc}{Joint
Probability Distributions} \par Let $(\Omega,\mathscr{F},P)$ be a probability
space and
$X:(\Omega,\mathscr{F})\to(\overline{\mathbb{R}}^{n},\mathscr{B}(\overline{\mathbb{R}}^{n}))$
and
$Y:(\Omega,\mathscr{F})\to(\overline{\mathbb{R}}^{m},\mathscr{B}(\overline{\mathbb{R}}^{m}))$
be two random vectors, respectively. Also, let
$\mathscr{B}(\overline{\mathbb{R}}^{n})\otimes\mathscr{B}(\overline{\mathbb{R}}^{m})$
be the $\sigma$-algebra generated by the set
$\mathscr{B}(\overline{\mathbb{R}}^{n})\times\mathscr{B}(\overline{\mathbb{R}}^{m})$
on $\overline{\mathbb{R}}^{n}\times\overline{\mathbb{R}}^{m}$. One could see
that
$\overline{\mathbb{R}}^{n}\times\overline{\mathbb{R}}^{m}=\overline{\mathbb{R}}^{n+m}$
and
$\mathscr{B}(\overline{\mathbb{R}}^{n})\otimes\mathscr{B}(\overline{\mathbb{R}}^{m})=\mathscr{B}(\overline{\mathbb{R}}^{n+m})$.
Then,
$(X,Y):(\Omega,\mathscr{F})\to(\overline{\mathbb{R}}^{n+m},\mathscr{B}(\overline{\mathbb{R}}^{n+m}))$
is a random vector defined by setting $(X,Y)(\omega)=(X(\omega),Y(\omega))$
for any $\omega\in\Omega$. The joint probability distribution of $X$ and $Y$
is determined by the probability measure induced by $(X,Y)$ (i.e.,
$P_{X,Y}(E)=P(\\{\omega\in\Omega:(X(\omega),Y(\omega))\in E\\})$ for any
$E\in\mathscr{B}(\overline{\mathbb{R}^{n+m}})$). Two Random vectors $X$ and
$Y$ defined on $(\Omega,\mathscr{F})$ are said to be independent if
$P_{X,Y}(E^{\prime}\times
E^{\prime\prime})=P_{X}(E^{\prime})P_{Y}(E^{\prime\prime})$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}^{n}})$ and
$E^{\prime\prime}\in\mathscr{B}(\overline{\mathbb{R}^{m}})$. Note that
\@@amsalign P_{X}(E^{\prime})=P(\\{\omega\in\Omega:X(\omega)\in
E^{\prime}\\})&=P(\\{\omega\in\Omega:(X(\omega),Y(\omega))\in
E^{\prime}\times\overline{\mathbb{R}}^{m}\\})\\\
&=P_{X,Y}(E^{\prime}\times\overline{\mathbb{R}}^{m}). Similarly, we have that
$P_{Y}(E^{\prime\prime})=P_{X,Y}(\overline{\mathbb{R}}^{n}\times
E^{\prime\prime})$. One could generalize the above procedure to define the
joint distribution of any finite number of random variables defined on
$(\Omega,\mathscr{F},P)$. Further, as we explained above, the joint
distribution of any non-empty subset of a finite set of random vectors defined
on $(\Omega,\mathscr{F},P)$ could be obtained from the joint distribution of
all random vectors. \par\par\@@numbered@section{subsection}{toc}{Conditional
Probability (Conditions on the Values of a Random Variable)} \par Let
$(\Omega,\mathscr{F},P)$ be a probability space, $E\in\mathscr{F}$, and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. We
define the measure
$\mu_{E}:\mathscr{B}(\overline{\mathbb{R}}^{n})\to[0,\infty]$ as follows:
$$\mu_{E}(E^{\prime})=P(X^{-1}(E^{\prime})\cap E)$$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. Note that
$P_{X}(E^{\prime})=0$ implies that $\mu_{E}(E^{\prime})=0$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. Hence, $\mu_{E}$ is
absolutely continuous with respect to $P_{X}$. Thus, it follows from the
Radon-Nikodym theorem that there exists a unique measurable function
$f_{E}:\overline{\mathbb{R}}^{n}\to[0,\infty]$ such that
$$\mu_{E}(E^{\prime})=\int_{E^{\prime}}f_{E}(x)\,\mathrm{d}P_{X}(x)$$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. It is a common notation
to denote $f_{E}(x)$ by $P(E|X=x)$ for any possible or impossible value $X=x$.
Therefore, we have that \begin{equation}P((X\in E^{\prime})\cap
E)=\int_{E^{\prime}}P(E|X=x)\,\mathrm{d}P_{X}(x)\end{equation} for any
$E\in\mathscr{F}$ and $E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$.
The function $P(\,\cdot\,|X=x):\mathscr{F}\to[0,\infty]$ is called a
conditional probability function conditioned on $X=x$. As a result, we have
the following equation for any $x_{0}\in\overline{\mathbb{R}}^{n}$:
\begin{equation}P(E\cap(X=x_{0}))=\int_{\\{x_{0}\\}}P(E|X=x)\,\mathrm{d}P_{X}(x)=P(E|X=x_{0})P_{X}(x_{0}).\end{equation}
\par\par Let's see an example as follows. Assume that $X$ is a discrete random
variable and $E\in\mathscr{F}$. Then, it follows from Equation
(\ref{conditionalformula}) that we must have $$P(E|X=x)=\dfrac{P((X=x)\cap
E)}{P(X=x)},\quad x\in\mathrm{Supp}(X).$$ For any
$x\in\overline{\mathbb{R}}\backslash\mathrm{Supp}(X)$, we can assign any value
to $P(E|X=x)$ (for instance, one could define $P(E|X=x)=0$). Now, we show that
$P(\,\cdot\,|X=x)$ with the above definition satisfies Equation
(\ref{conddef}). To do so, let
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. Then, we have that
\@@amsalign\int_{E^{\prime}}P(E|X=x)\,\mathrm{d}P_{X}(x)&=\sum_{x\in
E^{\prime}}P(E|X=x)P_{X}(x)=\sum_{x\in E^{\prime}}P(E\cap(X=x))\\\
&=P\left(E\cap\left(\bigcup_{x\in
E^{\prime}}\\{x\\}\right)\right)=P(E\cap(X\in E^{\prime})). Hence,
$P(\,\cdot\,|X=x)$ satisfies Equation (\ref{conddef}). One could see that for
any $x\in\mathrm{Supp}(X)$, $P(\,\cdot\,|X=x)$ is a probability measure as
well. \par If $X_{i}:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n_{i}}$ is
a random vector for any $1\leq i\leq l$, then we know that
$(X_{1},\ldots,X_{l})$ is a random vector. Hence, we can define
$$P(E|X_{1}=x_{1},\ldots,X_{n}=x_{n})=P(E|(X_{1},\ldots,X_{n})=(x_{1},\ldots,x_{n})).$$
Note that the concept of being independent is naturally defined here. Indeed,
let $X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{k_{1}}$,
$Y:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{k_{2}}$ and
$Z:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{k_{3}}$ be three random
vectors defined on $(\Omega,\mathscr{F})$. Then, $X$ and $Y$ are said to be
independent given $Z=z$, whenever $$P((X,Y)\in E_{1}\times E_{2}|Z=z)=P(X\in
E_{1}|Z=z)P(Y\in E_{2}|Z=z)$$ for any Borel sets $E_{1}$ and $E_{2}$ in
$\overline{\mathbb{R}}^{k_{1}}$ and $\overline{\mathbb{R}}^{k_{2}}$,
respectively. \par\@@numbered@section{subsection}{toc}{Expected Values of
Random Variables} \par Let $(\Omega,\mathscr{F},P)$ be a probability space and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}$ be a random variable. Then,
the expected value of $X$ is defined as follows:
$$\mathbb{E}(X)=\int_{\Omega}X(\omega)\,\mathrm{d}P.$$ Thus, if $X$ is a
discrete random variable, and $x_{1},\ldots,x_{n},\ldots,$ are all distinct
values of $X$, then we have that
\@@amsalign\mathbb{E}(X)&=\int_{\bigcup_{i=1}^{\infty}X^{-1}(x_{i})}X(\omega)\,\mathrm{d}P=\sum_{i=1}^{\infty}\int_{X^{-1}(x_{i})}X(\omega)\,\mathrm{d}P\\\
&=\sum_{i=1}^{\infty}x_{i}\int_{X^{-1}(x_{i})}\,\mathrm{d}P=\sum_{i=1}^{\infty}x_{i}P(X^{-1}(x_{i}))=\sum_{i=1}^{\infty}x_{i}P_{X}(x_{i}).
\par\par\@@numbered@section{subsection}{toc}{Expected Values of Random
Vectors} \par Let $(\Omega,\mathscr{F},P)$ be a probability space and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. Then,
the expected value of $X$ is defined as follows:
$$\mathbb{E}(X)=\left(\mathbb{E}(X_{1}),\ldots,\mathbb{E}(X_{n})\right),$$
where we assume that $X(\omega)=(X_{1}(\omega),\ldots,X_{n}(\omega))$ for any
$\omega\in\Omega$. Note that
$\pi_{i}:\overline{\mathbb{R}}^{n}\to\overline{\mathbb{R}}$ is measurable, and
hence it is a random variable for any $1\leq i\leq n$. It follows that
$X_{i}=\pi_{i}\circ X$ is a random variable as well for any $1\leq i\leq n$.
\par\@@numbered@section{subsection}{toc}{Conditional Expectation and
Conditional Probability (Conditions on a Sub $\sigma$-Algebra)} \par Let
$(\Omega,\mathscr{F},P)$ be a probability space and $\mathscr{G}$ be a sub
$\sigma$-algebra of $\mathscr{F}$. Also, let
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. The
expectation of $X$ with respect to $\mathscr{G}$, denoted by
$\mathbb{E}(X|\mathscr{G})$, is a random vector from $(\Omega,\mathscr{G})$ to
$\overline{\mathbb{R}}^{n}$ in such a way that
$$\int_{E}\mathbb{E}(X|\mathcal{G})(\omega)\,\mathrm{d}P(\omega)=\int_{E}X(\omega)\,\mathrm{d}P(\omega)$$
for any $E\in\mathscr{G}$. Indeed, $\mathbb{E}(X|\mathscr{G})$ exists and it
is almost every where unique. To see this, define the measure
$\nu:\mathscr{G}\to[0,1]$ by setting
$$\nu(E)=\int_{E}X(\omega)\,\mathrm{d}P(\omega)$$ for any $E\in\mathscr{G}$.
Note that $P(E)=0$ implies that $\nu(E)=0$ for any $E\in\mathscr{G}$. It
follows that $\nu$ is absolutely continuous with respect to
$P_{|_{\mathscr{G}}}:\mathscr{G}\to[0,\infty]$. Hence, there exists a unique
measurable function $g:(\Omega,\mathscr{G})\to\overline{\mathbb{R}}^{n}$ for
which we have that $\nu(E)=\int_{E}g(\omega)\,\mathrm{d}P(\omega)$ for any
$E\in\mathscr{G}$. Finally, $g$ is denoted by $\mathbb{E}(X|\mathscr{G})$.
\par\par\par Now, let $E\in\mathscr{G}$. The conditional probability of $E$
given $\mathscr{G}$ is defined as follows:
$$\mathbb{P}(E|\mathcal{G})=\mathbb{E}(\mathbbm{1}_{E}|\mathcal{G}).$$ Hence,
$\mathbb{P}(E|\mathcal{G})$ is a random variable defined on
$(\Omega,\mathscr{G})$ such that
$$\int_{E^{\prime}}\mathbb{P}(E|\mathcal{G})\,\mathrm{d}P=\int_{E^{\prime}}\mathbbm{1}_{E}\,\mathrm{d}P=P(E\cap
E^{\prime})$$ for any $E^{\prime}\in\mathscr{G}$. We also define a probability
measure $P(\,\cdot\,|E^{\prime}):\mathscr{G}\to[0,1]$ in such a way that
$$P(E|E^{\prime})P(E^{\prime})=P(E\cap E^{\prime})$$ for any
$E^{\prime}\in\mathscr{G}$. It follows that
$$P(E|E^{\prime})P(E^{\prime})=\int_{E^{\prime}}\mathbb{P}(E|\mathcal{G})\,\mathrm{d}P$$
for any $E^{\prime}\in\mathscr{G}$. \par Considering $P(\,\cdot\,|E^{\prime})$
(defined above), being independent given an event for two random variables is
defined similar to the definition given in Appendix \ref{con}.
\par\@@numbered@section{subsection}{toc}{ The Relationship Between the
Conditional Probability Points of View introduced in Appendices \ref{con} and
\ref{consigma} } Let $(\Omega,\mathscr{F},P)$ be a probability space and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. Then,
it is easy to see that
$\mathscr{G}=\\{X^{-1}(E^{\prime\prime}):E^{\prime\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})\\}$
is a sub $\sigma$-algebra of $\mathscr{F}$. We show that
$$\mathbb{P}(E|\mathscr{G})(\omega)=P(E|X=X(\omega))$$ for almost every
$\omega\in\Omega$, where $P(E|X=X(\omega))$ and $\mathbb{P}(E|\mathscr{G})$
are defined with respect to the points of view introduced in Appendices
\ref{con} and \ref{consigma}, respectively. To do so, first we note that
$P(E|X=X(\omega))=P(E|X=\,\cdot\,)\circ X$ as the composition of two
measurable functions is measurable. Hence, it follows from the Change of
Variable theorem that
$$\int_{X^{-1}(E^{\prime\prime})}P(E|X=X(\omega))\,\mathrm{d}P(\omega)=\int_{E^{\prime\prime}}P(E|X=x)\,\mathrm{d}P_{X}(x)=P(E\cap
X^{-1}(E^{\prime\prime})).$$ One could see that when $E$ is a specific value
of a random variable on $(\Omega,\mathscr{F})$, then two concepts of
independence discussed in Appendices \ref{con} and \ref{consigma} coincide
with each other. \par\@@numbered@section{subsection}{toc}{Conditional
Probabilities of Conditional Probabilities} \par Let $(\Omega,\mathscr{F},P)$
be a probability space, $E,E_{0}\in\mathscr{F}$ and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. Let
$P^{E}:=P(\,\cdot\,|E)$ be a conditional probability measure, and $P_{X}^{E}$
be the conditional probability measure induced by $X$ on
$\overline{\mathbb{R}}^{n}$ (i.e.,
$P_{X}^{E}(E^{\prime})=P(X^{-1}(E^{\prime})|E)$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$). Then, by the same
argument discussed in Appendix \ref{con}, there exists a unique measurable
function $f^{E}_{E_{0}}:\overline{\mathbb{R}}^{n}\to[0,\infty]$ that
$$P^{E}((X\in E^{\prime})\cap
E_{0})=\int_{E^{\prime}}f_{E_{0}}^{E}(x)\,\mathrm{d}P^{E}_{X}(x)$$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. Now, we denote
$f_{E_{0}}^{E}(x)$ by $P(E_{0}|E,X=x)$ for any
$x\in\overline{\mathbb{R}}^{n}$. It follows that $$P((X\in E^{\prime})\cap
E_{0}|E)=\int_{E^{\prime}}P(E_{0}|E,X=x)\,\mathrm{d}P^{E}_{X}(x)$$ for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$.
\par\par\@@numbered@section{subsection}{toc}{Total law of Probability for
Conditional Probabilities} Similar to the assumptions in Appendix
\ref{conofcon}, let $(\Omega,\mathscr{F},P)$ be a probability space,
$E,E_{0}\in\mathscr{F}$ and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector. Also,
assume that $P^{E}:=P(\,\cdot\,|E)$ be a conditional probability measure, and
$P_{X}^{E}$ be the conditional probability measure induced by $X$ on
$\overline{\mathbb{R}}^{n}$ with respect to $P^{E}$. Then, by Appendix
\ref{conofcon}, we have that
$$P(E_{0}|E)=P((X\in\overline{\mathbb{R}}^{n})\cap
E_{0}|E)=\int_{\overline{\mathbb{R}}^{n}}P(E_{0}|E,X=x)\,\mathrm{d}P_{X}^{E}(x).$$
\par\par\@@numbered@section{subsection}{toc}{Relationship Between Conditional
Probabilities of Conditional Probabilities and Conditional Probabilities Given
the Joint Distribution of two Random Vectors} Once again, let us assume that
$(\Omega,\mathscr{F},P)$ is a probability space, $E_{0}\in\mathscr{F}$, and
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ and
$Y:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{m}$ are two random vectors.
Also, assume that $P^{E}:=P(\,\cdot\,|E)$ is a conditional probability
measure, and $P_{X}^{E}$ is the conditional probability measure induced by $X$
on $\overline{\mathbb{R}}^{n}$ with respect to $P^{E}$. Note that if
$E=Y^{-1}(y)$ for some value $y\in\mathrm{Supp}(Y)$, then $P(E_{0}|Y=y,X=x)$
(defined in Appendix \ref{con}) coincides with
$f_{E_{0}}^{E}(x)=P(E_{0}|E,X=x)$ (defined in Appendix~{}\ref{conofcon}). To
see this, let $E=Y^{-1}(y)$ with $y\in\mathrm{Supp}(Y)$. For simplicity, we
denote $P_{X}^{E}$ and $f_{E_{0}}^{E}$ by $P_{X}^{y}$ and $f_{E_{0}}^{y}$,
respectively. Let
$P(E_{0}|X=\,\cdot\,\,,\,Y=\cdot\,):\overline{\mathbb{R}}^{n}\times\overline{\mathbb{R}}^{m}\to[0,1]$
be a conditional probability function in the sense of Appendix \ref{con}.
Assume that $P(E_{0}|X=\,\cdot\,\,,\,Y=\cdot\,)$ is integrable. It is enough
to show that $$P(E_{0}\cap(X\in
E^{\prime})|Y=y)=\int_{E^{\prime}}P(E_{0}|X=x,Y=y)\,\mathrm{d}P_{X}^{y}(x)$$
for any $E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$. To prove the
above equality, it is enough to show that for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$ and
$E^{\prime\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{m})$, we have that
$$P(E_{0}\cap((X,Y)\in E^{\prime}\times
E^{\prime\prime}))=\int_{E^{\prime\prime}}\int_{E^{\prime}}P(E_{0}|X=x,Y=y)\,\mathrm{d}P_{X}^{y}(x)\mathrm{d}P_{Y}(y),$$
since $P(E_{0}\cap(X\in E^{\prime})|Y=\,\cdot\,)$ is a unique function
satisfying the following equality for any
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$ and
$E^{\prime\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{m})$ (see Appendic
\ref{con}): $$P(E_{0}\cap((X,Y)\in E^{\prime}\times
E^{\prime\prime}))=\int_{E^{\prime\prime}}P(E_{0}\cap(X\in
E^{\prime})|Y=y)\,\mathrm{d}P_{Y}(y).$$ To do so, let
$E^{\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$ and
$E^{\prime\prime}\in\mathscr{B}(\overline{\mathbb{R}}^{m})$. Note that it
follows from Appendix~{}\ref{con} that $$P(E_{0}\cap((X,Y)\in E^{\prime}\times
E^{\prime\prime}))=\int_{E^{\prime}\times
E^{\prime\prime}}P(E_{0}|X=x,Y=y)\,\mathrm{d}P_{X,Y}(x,y).$$ Thus, it is
enough to show that $$\int_{E^{\prime}\times
E^{\prime\prime}}f(x,y)\,\mathrm{d}P_{X,Y}(x,y)=\int_{E^{\prime\prime}}\int_{E^{\prime}}f(x,y)\,\mathrm{d}P_{X}^{y}(x)\mathrm{d}P_{Y}(y),$$
where $f(x,y)=P(E_{0}|X=x,Y=y)$. To prove this, first assume that
$f=\mathbbm{1}_{E_{1}\times E_{2}}$ for some
$E_{1}\in\mathscr{B}(\overline{\mathbb{R}}^{n})$ and
$E_{2}\in\mathscr{B}(\overline{\mathbb{R}}^{m})$. Then, we have that
\@@amsalign\int_{E^{\prime}\times E^{\prime\prime}}\mathbbm{1}_{E_{1}\times
E_{2}}(x,y)\,\mathrm{d}P_{X,Y}(x,y)&=\int_{\overline{\mathbb{R}}^{n}\times\overline{\mathbb{R}}^{m}}\mathbbm{1}_{(E_{1}\cap
E^{\prime})\times(E_{2}\cap E^{\prime\prime})}(x,y)\,\mathrm{d}P_{X,Y}(x,y)\\\
&=P_{X,Y}((E_{1}\cap E^{\prime})\times(E_{2}\cap E^{\prime\prime})),\\\
\int_{E^{\prime\prime}}\int_{E^{\prime}}\mathbbm{1}_{E_{1}\times
E_{2}}(x,y)\,\mathrm{d}P_{X}^{y}(x)\mathrm{d}P_{Y}(y)&=\int_{E^{\prime\prime}\cap
E_{2}}\int_{\overline{\mathbb{R}}^{n}}\mathbbm{1}_{E_{1}\cap E^{\prime}\times
E_{2}}(x,y)\,\mathrm{d}P_{X}^{y}(x)\mathrm{d}P_{Y}(y)\\\
&=\int_{E^{\prime\prime}\cap
E_{2}}\int_{\overline{\mathbb{R}}^{n}}\mathbbm{1}_{E_{1}\cap
E^{\prime}}(x)\,\mathrm{d}P_{X}^{y}(x)\mathrm{d}P_{Y}(y)\\\
&=\int_{E^{\prime\prime}\cap E_{2}}P(X\in E_{1}\cap
E^{\prime}|Y=y)\,\mathrm{d}P_{Y}(y)\\\ &=P((X,Y)\in(E_{1}\cap
E^{\prime})\times(E_{2}\cap E^{\prime\prime}))\\\ &=P_{X,Y}((E_{1}\cap
E^{\prime})\times(E_{2}\cap E^{\prime\prime})). Hence, the equality holds for
indicator functions. By the linearity of integrals, one could see that the
equality holds for simple functions as well. Finally, by the monotone
convergence theorem and dominating convergence theorem, the equality holds for
any non-negative function.
\par\par\@@numbered@section{subsection}{toc}{Expected Value Given an Event}
\par Let $(\Omega,\mathscr{F},P)$ be a probability space and
$E\in\mathscr{F}$. Also, let
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ be a random vector.
Assume that $\\{E_{i}\\}_{i=1}^{\infty}$ is a family of pairwise disjoint
elements of $\mathscr{F}$ with $\Omega=\bigcup_{i=1}^{\infty}E_{i}$. Also,
assume that $\mathscr{G}$ is the sub $\sigma$-algebra of $\mathscr{F}$
generated by $\\{E_{i}\\}_{i=1}^{\infty}$. One could see that
$$\mathscr{G}=\left\\{\bigcup_{i\in I}E_{i}:I\subseteq\mathbb{N}\right\\}.$$
Thus, $\mathbb{E}(X|\mathscr{G})$ is constant on $E_{i}$, since
$\mathbb{E}(X|\mathscr{G})^{-1}(r)$ must be in $\mathscr{G}$ for any
$i\in\mathbb{N}$ and $r\in\overline{\mathbb{R}}$. It follows that
$$\mathbb{E}(X|\mathscr{G})=\sum_{i=1}^{\infty}r_{i}\mathbbm{1}_{E_{i}}$$ for
some real numbers $r_{i}$. In the aforementioned expression, $r_{i}$ is
denoted by $\mathbb{E}(X|E_{i})$ for any $i\in\mathbb{N}$. It follows that
$$\mathbb{E}(X|E_{i})P(E_{i})=\int_{E_{i}}\mathbb{E}(X|\mathscr{G})\,\mathrm{d}P(\omega)=\int_{E_{i}}X(\omega)\,\mathrm{d}P(\omega)$$
for any $i$. Hence, we showed that for any $E\in\mathscr{F}$,
$\mathbb{E}(X|E)P(E)$ is well-defined (i.e., it does not depend on a family of
pairwise disjoint elements of $\mathscr{F}$ which includes $E$). Now, let
$E\in\mathscr{F}$ and $P^{E}=P(\,\cdot\,|E)$ be a conditional probability
measure. Then, we claim that the following definition for $\mathbb{E}(X|E)$ is
acceptable:
$$\mathbb{E}(X|E):=\int_{\Omega}X(\omega)\,\mathrm{d}P^{E}(\omega).$$ To show
this, it is enough to show that
$$\left(\int_{\Omega}X(\omega)\,\mathrm{d}P^{E}(\omega)\right)P(E)=\int_{\Omega}X(\omega)\,\mathrm{d}P(\omega).$$
Let $E^{\prime}\in\mathscr{F}$ be arbitrary and $X=\mathbbm{1}_{E^{\prime}}$.
Then, we have that
\@@amsalign\left(\int_{\Omega}\mathbbm{1}_{E^{\prime}}(\omega)\,\mathrm{d}P^{E}(\omega)\right)P(E)&=P^{E}(E^{\prime})P(E)=P(E^{\prime}|E)P(E)\\\
&=P(E^{\prime}\cap
E)=\int_{E}\mathbbm{1}_{E^{\prime}}(\omega)\,\mathrm{d}P(\omega). One could
see that the equality holds when $X$ is a simple function as well. Finally, by
the monotone convergence theorem, the equality holds for any random vector
$X:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$.
\par\par\@@numbered@section{subsection}{toc}{Continuous Random Variables }
\par In this subsection, we briefly explain continuous random variables, their
conditional distributions, expectations, and conditional expectations. In the
sequel, we assume that $(\Omega,\mathscr{F},P)$ is a probability space,
$X,Y:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}$ are random variables, and
$Z:(\Omega,\mathscr{F})\to\overline{\mathbb{R}}^{n}$ is a random vector. We
also denote the Lebesgue measure on $\overline{\mathbb{R}}^{n}$ by $\lambda$.
\par\par\@@numbered@section{subsubsection}{toc}{Absolutely Continuous Random
Variables} The random vector $Z$ is called absolutely continuous if $P_{Z}$ is
absolutely continuous with respect to the Lebesgue measure.
\par\par\@@numbered@section{subsubsection}{toc}{Probability Density Functions}
\par Assume that $Z$ is absolutely continuous. It follows from the Radon-
Nicodym theorem that there exists a measurable function
$f:\overline{\mathbb{R}}\to[0,\infty]$ for which
$P_{Z}(E)=\int_{E}f(z)\,\mathrm{d}\lambda(z)$ for any Borel set $E$ in
$\overline{\mathbb{R}}$. Here, $f$ is called the probability density function
of $X$. Now, if the random variable $X$ is absolutely continuous,
$\mathrm{Supp}(X)\subseteq\mathbb{R}$, and $f$ is Riemann integrable, then
$$P_{X}((-\infty,t])=\int_{-\infty}^{t}f(x)\,\mathrm{d}x$$ for any
$t\in\mathbb{R}$. \par\par\@@numbered@section{subsubsection}{toc}{Distribution
Function of a Random Variable} \par Define
$F_{X}:\overline{\mathbb{R}}\to[0,1]$ by setting $F_{X}(t)=P_{X}([-\infty,t])$
for any $t\in\overline{\mathbb{R}}$. Then, $F_{X}$ is called the distribution
function of $X$. Now, let $X$ be absolutely continuous and
$\mathrm{Supp}(X)\subseteq\mathbb{R}$. Then
$$F_{X}(t)=\int_{-\infty}^{t}f(x)\,\mathrm{d}x$$ for any $t\in\mathbb{R}$,
where $f$ is the probability density function of $X$ which is assumed to be
Riemann integrable. Note that if $f$ is continuous, then $F_{X}$ is
differentiable and we have that
$$\frac{\mathrm{d}F_{X}}{\mathrm{d}x}(t)=f(t).$$
\par\@@numbered@section{subsubsection}{toc}{Definition of a Continuous Random
Variable} \par Let $(\Omega,\mathscr{F},P)$ be a probability space. Then, the
random variable $X$ is called continuous if $F_{X}$ is continuous. Now, we
show that $X$ is continuous if and only if $P_{X}(t)=0$ for any
$t\in\overline{\mathbb{R}}$. First, assume that $X$ is continuous and
$t_{0}\in\overline{\mathbb{R}}$. Let $\\{t_{n}\\}_{n=1}^{\infty}$ be an
increasing sequence in $\overline{\mathbb{R}}$ converging to $t_{0}$. Now, if
$E_{n}=(t_{n},t_{0}]$, then we have that $E_{n}\downarrow\\{t_{0}\\}$, and
hence we have that
$$0=\lim_{n\to\infty}(F_{X}(t_{0})-F_{X}(t_{n}))=\lim_{n\to\infty}P_{X}((t_{n},t_{0}])=\lim_{n\to\infty}P_{X}(E_{n})=P_{X}(t_{0}).$$
One could see that the converse holds as well. \par Let $X$ be absolutely
continuous. We show that $X$ is continuous. To do so, let
$t\in\overline{\mathbb{R}}$. Then, the Lebesgue measure of $\\{t\\}$ is 0, and
hence $P_{X}(t)=0$. It follows that $X$ is a continuous random variable. \par
Continuous random variables are often absolutely continuous. Hence, from now
on, by a continuous random variable, we mean an absolutely continuous random
variable. \par\par\@@numbered@section{subsubsection}{toc}{Examples of
Continuous Random Variables} \par As we discussed, a continuous random
variable is uniquely determined by its probability density function (pdf).
Thus, we introduce the pdfs of some famous random variables in the following:
\par\@@amsalign f(x)&:=\left\\{\begin{array}[]{ll}\frac{1}{b-a}&,a\leq x\leq
b\\\ 0&,\text{otherwise}\end{array}\right.&(\text{Unifrom Random Variable on
}[a,b]),\\\ g(x)&:=\left\\{\begin{array}[]{ll}\lambda e^{-\lambda x}&,x\geq
0\\\
0&,\text{otherwise}\end{array}\right.&\left(\begin{array}[]{l}\text{Exponential
Random Variable}\\\ \text{with the parameter $\lambda>0$
}\end{array}\right),\\\
h(x):&=\frac{1}{\sqrt{2\pi}\sigma}e^{-\dfrac{(x-\mu)^{2}}{2\sigma^{2}}}&\left(\begin{array}[]{l}\text{Normal
Random Variable}\\\ \text{with the variance $\sigma^{2}$ ($\sigma>0$)}\\\
\text{and the mean $\mu$ }\end{array}\right).
\par\par\@@numbered@section{subsubsection}{toc}{Continuous Joint Distributions
of Random Variables} \par Assume that the random vector $(X,Y)$ is absolutely
continuous, and $f_{X,Y}$ is the pdf of $(X,Y)$. Then, the pdfs of $X$ and $Y$
could be obtained as follows:
$$f_{X}(x)=\int_{\overline{\mathbb{R}}}f_{X,Y}(x,y)\,\mathrm{d}\lambda(y),\qquad
f_{Y}(y)=\int_{\overline{\mathbb{R}}}f_{X,Y}(x,y)\,\mathrm{d}\lambda(x),$$
since for instance we have that
\@@amsalign\int_{E}\int_{\overline{\mathbb{R}}}f_{X,Y}(x,y)\,\mathrm{d}\lambda(y)\mathrm{d}\lambda(x)=\int_{E\times\overline{\mathbb{R}}}f_{X,Y}(x,y)\,\mathrm{d}\lambda(x,y)=P_{X,Y}(E\times\overline{\mathbb{R}})=P_{X}(E)
for any Borel set $E$ in $\overline{\mathbb{R}}$. Indeed, we validate the
first equality of the above expression for simple functions $f_{X,Y}(x,y)$,
and then we generalize it for arbitrary functions $f_{X,Y}(x,y)$ using the
dominated convergence theorem (another way to justify this equality is the
Fubini theorem in measure theory). \par Now, assume that $X$ and $Y$ are
independent. Then, for any two Borel sets $E_{1}$ and $E_{2}$ in
$\overline{\mathbb{R}}$, we have that $P_{X,Y}(E_{1}\times
E_{2})=P_{X}(E_{1})P_{Y}(E_{2})$. Now, we claim that $f_{X,Y}=f_{X}f_{Y}$. To
do so, we note that \@@amsalign\int_{E_{1}\times
E_{2}}f_{X}(x)f_{Y}(y)\,\mathrm{d}\lambda(x,y)&=\int_{E_{1}}\int_{E_{2}}f_{X}(x)f_{Y}(y)\,\mathrm{d}\lambda(x)\mathrm{d}\lambda(y)\\\
&=\left(\int_{E_{1}}f_{X}(x)\,\mathrm{d}\lambda(x)\right)\left(\int_{E_{2}}f_{Y}(y)\,\mathrm{d}\lambda(y)\right)\\\
&=P_{X}(E_{1})P_{Y}(E_{2})=P_{X,Y}(E_{1}\times E_{2}). It follows that for any
Borel set $E^{\prime}$ in $\overline{\mathbb{R}}^{n}$, we have that
$$\int_{E^{\prime}}f_{X}(x)f_{Y}(y)\,\mathrm{d}\lambda(x,y)=P_{X,Y}(E^{\prime}),$$
which implies that $f_{X,Y}=f_{X}f_{Y}$.
\par\par\@@numbered@section{subsubsection}{toc}{Conditional Continuous Random
Variables} \par Let $E\in\mathscr{F}$ and $X$ be continuous (as we mentioned
before, here we mean being absolutely continuous). If $P(E)>0$ and $P^{E}$ be
a conditional probability measure (for its definition look at Appendix
\ref{conofcon}), then $P_{X}^{E}$ is absolutely continuous, since for any
Borel set $E^{\prime}$ in $\overline{\mathbb{R}}$, we have that
$$P^{E}_{X}(E^{\prime})P(E)=P((X\in E^{\prime})\cap E)\leq
P_{X}(E^{\prime}).$$ Otherwise, if $P(E)=0$, then we define $P_{X}^{E}$ in
such a way that it is absolutely continuous. In general, we can define
$P_{X}^{E}$ as follows:
$$P_{X}^{E}(E^{\prime})=\int_{E^{\prime}}f_{X|E}(x|E)\,\mathrm{d}\lambda(x),$$
where $f_{X|E}(\,\cdot\,|E):\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ is
a non-negative measurable function satisfying the following identity:
$$\int_{\overline{\mathbb{R}}}f_{X|E}(x|E)\,\mathrm{d}\lambda(x)=1.$$ In the
case that $E$ is the event $Y^{-1}(y)$, we denote $f_{X|E}$ by $f_{X}^{y}$.
Also, we assume that $P(X\in
E_{1}|Y=\,\cdot\,):\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ is the
probability function defined in Appendix \ref{con}. Indeed, $P(X\in
E_{1}|Y=\,\cdot\,)$ is measurable and satisfies the following equality:
$$P((X\in E_{1})\cap(Y\in E_{2}))=\int_{E_{2}}P(X\in
E_{1}|Y=y)\,\mathrm{d}P_{Y}(y)$$ for any Borel sets $E_{1}$ and $E_{2}$ in
$\overline{\mathbb{R}}$. Now, we show that
$$f_{X}^{y}(x|y)f_{Y}(y)=f_{X,Y}(x,y)$$ for any $x,y\in\overline{\mathbb{R}}$.
To do so, it is enough to show that for any Borel sets $E_{1}$ and $E_{2}$ in
$\overline{\mathbb{R}}$, we have that $$P_{X,Y}(E_{1}\times
E_{2})=\int_{E_{1}\times
E_{2}}f_{X}^{y}(x|y)f_{Y}(y)\,\mathrm{d}\lambda(x,y).$$ Denote the right side
of the above identity by $I$. Then, we have that \@@amsalign
I&=\int_{E_{2}}\left(\int_{E_{1}}f_{X}^{y}(x|y)\,\mathrm{d}\lambda(x)\right)f_{Y}(y)\,\mathrm{d}\lambda(y)=\int_{E_{2}}P_{X}^{y}(E_{1})f_{Y}(y)\mathrm{d}\lambda(y)\\\
&=\int_{E_{2}}P_{X}^{y}(E_{1})\,\mathrm{d}P_{Y}(y)=\int_{E_{2}}P(X\in
E_{1}|Y=y)\,\mathrm{d}P_{Y}(y)=P((X\in E_{1})\cap(Y\in E_{2})). Hence, we have
the following: $$f_{X}^{y}(x|y)f_{Y}(y)=f_{X,Y}(x,y)$$ for any
$x,y\in\overline{\mathbb{R}}$.
\par\@@numbered@section{subsubsection}{toc}{Expectation of Continuous Random
Variables} \par Let $X$ be continuous and $f_{X}$ be the pdf of $X$. We
already know that $$\mathbb{E}(X)=\int_{\Omega}X(\omega)\,\mathrm{d}P.$$ By
the change of variable theorem, we have that
$$\mathbb{E}(X)=\int_{\overline{\mathbb{R}}}x\,\mathrm{d}P_{X}(x).$$
Therefore, we have that
$$\mathbb{E}(X)=\int_{\overline{\mathbb{R}}}xf_{X}(x)\,\mathrm{d}\lambda(x).$$
\par\par\@@numbered@section{subsubsection}{toc}{Conditional Expectation of
Continuous Random Variables} \par Let $X$ be continuous and $E\in\mathscr{F}$.
We already now that if $P^{E}$ is a conditional probability measure, then
$$\mathbb{E}(X|E)=\int_{\Omega}X(\omega)\,\mathrm{d}P^{E}(\omega).$$ Now, by
the change of variable theorem, we obtain
$$\mathbb{E}(X|E)=\int_{\overline{\mathbb{R}}}x\,\mathrm{d}P^{E}_{X}(x).$$
Assume that $P^{E}_{X}$ is absolutely continuous. Then, there exists a unique
measurable function $g^{E}:\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ for
which
$$P^{E}_{X}(E^{\prime})=\int_{E^{\prime}}g^{E}(x)\,\mathrm{d}\lambda(x)$$ for
any Borel set $E^{\prime}$ in $\overline{\mathbb{R}}$. We denote $g^{E}(x)$ by
$f_{X|E}(x|E)$ for any $x\in\overline{\mathbb{R}}$. It follows that
$$\mathbb{E}(X|E):=\int_{\overline{\mathbb{R}}}xf_{X|E}(x|E)\,\mathrm{d}\lambda(x).$$
\par\par\@@numbered@section{subsubsection}{toc}{Total Law of Probability for
Continuous Random Variables} \par Let $E\in\mathscr{F}$ and $X$ be continuous.
Assume that $f_{X}$ is the pdf of $X$. We already know the following version
of the total law of probability:
$$P(E)=\int_{\overline{\mathbb{R}}}P(E|X=x)\,\mathrm{d}P_{X}(x)$$ which
implies that
$$P(E)=\int_{\overline{\mathbb{R}}}P(E|X=x)f_{X}(x)\,\mathrm{d}\lambda(x).$$
\par\par\@@numbered@section{subsubsection}{toc}{Total Law of Probability for
Conditional Continuous Random Variables} \par Let $E,E_{0}\in\mathscr{F}$ and
$X$ be continuous. Assume that $P^{E}$ is a conditional probability measure,
and $P_{X}^{E}$ is absolutely continuous. Further, assume that $f_{X|E}(x|E)$
is the pdf of $P_{X}^{E}$. We already know that
$$P(E_{0}|E)=\int_{\overline{\mathbb{R}}}P(E_{0}|E,X=x)\,\mathrm{d}P_{X}^{E}(x).$$
It follows that
$$P(E_{0}|E)=\int_{\overline{\mathbb{R}}}P(E_{0}|E,X=x)f_{X|E}(x|E)\,\mathrm{d}\lambda(x).$$
\par\par\@@numbered@section{appendix}{toc}{Generalized Probability Density
Functions} \par In this appendix, we use the Dirac measure to unify both
discrete and continuous distributions (see
\cite[cite]{[\@@bibref{}{capinski2004measure}{}{}]} and
\cite[cite]{[\@@bibref{}{ccinlar2011probability}{}{}]}).
\par\par\@@numbered@section{subsubsection}{toc}{ Discrete Random Variable
Distributions via Dirac Measures} Let $X:\Omega\to\overline{\mathbb{R}}$ be a
discrete random variable and $E\subseteq\Omega$. We know that
$$P_{X}(E)=\sum_{x\in E}P_{X}(x).$$ Now, define the probability measure
$\mu:\mathcal{P}(\Omega)\to[0,1]$ as follows:
$$\mu:=\sum_{x\in\mathrm{Supp}(X)}P_{X}(x)\delta_{x}.$$ One could see that
$\mu=P_{X}$, and hence $$P_{X}(E)=\int_{E}\,\mathrm{d}\mu.$$
\par\par\@@numbered@section{subsubsection}{toc}{Definition of a Generalized
Probability Density Function} \par Let $(\Omega,\mathscr{F},P)$ be a
probability space, and $X:\Omega\to\overline{\mathbb{R}}$ be a random
variable. Assume that $$\mu=t\lambda+\sum_{n=1}^{\infty}t_{n}\delta_{x_{n}},$$
where $\lambda$ is the Lebesgue measure, and
$t_{n},x_{n}\in\overline{\mathbb{R}}$ are non-negative for any $n$. Then,
$\mu$ is a measure. If $P_{X}$ is absolutely continuous with respect to $\mu$,
then by the Radon-Nykodim theorem, there exists a unique non-negative
measurable function $f_{X}:\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ in
such a way that $$P_{X}(E)=\int_{E}f_{X}(x)\,\mathrm{d}\mu(x)$$ for any
$E\in\mathscr{B}(\overline{\mathbb{R}})$. The function $f_{X}$ is called the
generalized density function of $X$ with respect to $\mu$. Obviously, any pdf
is a generalized pdf. For an example, assume that
$\mathrm{Im}(f_{X})=\\{f_{X}(x):x\in\overline{\mathbb{R}}\\}\subseteq\mathbb{R}$,
and $(f_{X})_{|_{\mathbb{R}}}:\mathbb{R}\to\mathbb{R}$ is Riemann integrable,
and $\mu=\lambda+\delta_{a}$ for some $a\in\mathbb{R}$ with $a\leq t$, where
$t\in\mathbb{R}$. Then, we have that
\par\@@amsalign\int_{[-\infty,t]}f_{X}(x)\,\mathrm{d}\mu(x)&=\int_{[-\infty,t]}f_{X}(x)\,\mathrm{d}\lambda(x)+\int_{[-\infty,t]}f_{X}(x)\,\mathrm{d}\delta_{a}(x)\\\
&=\int_{(-\infty,t]}f_{X}(x)\,\mathrm{d}\lambda(x)+f_{X}(a)\\\
&=\int_{-\infty}^{t}f_{X}(x)\,\mathrm{d}x+f_{X}(a).
\par\par\@@numbered@section{subsubsection}{toc}{Dirac Delta Function} \par The
Dirac delta function is a function
$\delta:\overline{\mathbb{R}}\to\overline{\mathbb{R}}$ which takes the value
$0$ everywhere except for $x=0$ which takes $\infty$. It is assumed that
$\delta$ has the following properties:
\begin{itemize}\par\itemize@item@$\int_{-\infty}^{\infty}\delta(x-a)\,\mathrm{d}x=1$
for any $a\in\mathbb{R}$, and
\par\itemize@item@$\int_{-\infty}^{\infty}g(x)\delta(x-a)\,\mathrm{d}x=g(a)$
for any $a\in\mathbb{R}$ and any continuous function
$g:\mathbb{R}\to{\mathbb{R}}$ for which the smallest closed set of all points
$x\in\mathbb{R}$ with $g(x)\neq 0$ is bounded. \par\end{itemize} The first
property implies that $\delta(x-a)$ is a pdf. \par A main challenge here is
that $\delta$ is not a function on $\mathbb{R}$ to $\mathbb{R}$, and hence the
Riemann integral could not be defined for $\delta$. Hence, one could interpret
the above Riemann integrals as Lebesgue integrals. In this case, we note that
in the Lebesgue integral removing a set of zero measure (such as a point) does
not affect the value of an integral. Hence, we must have
$\int_{\overline{\mathbb{R}}}\delta(x)\,\mathrm{d}\lambda(x)=0$, a
contradiction! In fact, a true interpretation of the Dirac delta function is
to consider it as a certain linear functional. Here, we do not discuss the
details of a true mathematical definition of $\delta$ (for details, see
\cite[cite]{[\@@bibref{}{lax2014functional}{}{}]}). However, we can interpret
the integrals appeared above as symbolic integrals and define them as follows
(for a more detailed and comprehensive reference, one could see
\cite[cite]{[\@@bibref{}{loffler2019brownian}{}{}, Section 5]}):
$$\int_{-\infty}^{t}\delta(x-a)\,\mathrm{d}x:=\int_{[-\infty,t]}\,\mathrm{d}\delta_{a}(x),\quad\int_{-\infty}^{t}g(x)\delta(x-a)\,\mathrm{d}x:=\int_{[-\infty,t]}g(x)\,\mathrm{d}\delta_{a}(x),$$
where $\delta_{a}$ is the Dirac measure with respect to $a$. The two main
properties of the Dirac delta function are satisfies, since
$$\int_{\overline{\mathbb{R}}}\,\mathrm{d}\delta_{a}(x)=1,\qquad\int_{\overline{\mathbb{R}}}g(x)\,\mathrm{d}\delta_{a}(x)=g(a).$$
In general, we define
$$\int_{-\infty}^{t}g(x)\left(\sum_{n=1}^{\infty}t_{n}\delta(x-a_{n})\right)\,\mathrm{d}x:=\sum_{n=1}^{\infty}t_{n}\int_{[-\infty,t]}g(x)\,\mathrm{d}\delta_{a_{n}}(x),$$
where $t_{n},a_{n}\in\mathbb{R}$ for any $n$. Hence, we have that
\@@amsalign\int_{[-\infty,t]}g(x)\,\mathrm{d}\left(t_{0}\lambda+\sum_{n=1}^{\infty}t_{n}\delta_{a_{n}}\right)(x)&=\int_{[-\infty,t]}t_{0}g(x)\,\mathrm{d}\lambda(x)+\sum_{n=1}^{\infty}t_{n}\int_{[-\infty,t]}g(x)\,\mathrm{d}\delta_{a_{n}}(x)\\\
&=\int_{-\infty}^{t}t_{0}g(x)\,\mathrm{d}x+\int_{-\infty}^{t}g(x)\left(\sum_{n=1}^{\infty}t_{n}\delta(x-a_{n})\right)\,\mathrm{d}x\\\
&=\int_{-\infty}^{t}g(x)\left(t_{0}+\sum_{n=1}^{\infty}t_{n}\delta(x-a_{n})\right)\,\mathrm{d}x.
Indeed, we have that
$$\int_{-\infty}^{t}\delta(x-a)\,\mathrm{d}x=\int_{[-\infty,t]}\,\mathrm{d}\delta_{a}(x)=U_{a}(t),$$
where $U_{a}$ is the Heaviside step function defined as follows:
\par$$U_{a}(t)=\left\\{\begin{array}[]{ll}0&,t<a\\\ 1&,t\geq
a\end{array}\right..$$ It follows that $U_{a}(t)$ is the distribution function
associated to $\delta(x-a)$, and we have the following symbolic derivative:
$$\frac{\mathrm{d}U_{a}}{\mathrm{d}x}(t)=\delta(t-a).$$ Note that $U_{a}$ is
not really differentiable at $t=a$, that is why we used the word "symbolic"
above. \@add@PDF@RDFa@triples\LTX@newpage\par\end{document}}$