\section{Transition Functions}
\label{section:markov-transition}

\begin{definition}[Transition Function]
\label{definition:transition-function}
    Let $(X, \cm)$ be a measurable space. A \textbf{transition function} $\bracs{P_t| t \ge 0}$ on $X$ is a collection of maps
    \[
    P_t: X \times \cm \to [0, 1]
    \]

    such that:
    \begin{enumerate}
        \item For each $t \ge 0$ and $x \in X$, $P_t(x, \cdot)$ is a probability measure on $X$.
        \item For each $t \ge 0$ and $A \in \cm$, $x \mapsto P_t(x, A)$ is $(\cm, \cb_\real)$-measurable.
        \item For any $s, t \ge 0$, $x \in X$, and $A \in \cm$,
        \[
        P_{t + s}(x, A) = \int P_t(y, A)P_s(x, dy)
        \]

    \end{enumerate}
\end{definition}

\begin{definition}[Semigroup of Transition Function]
\label{definition:transition-function-semigroup}
    Let $(X, \cm)$ be a measurable space, $\bracs{P_t|t \ge 0}$ be a transition function on $X$. For each $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable, let
    \[
    (\bp_t f)(x) = \int f(y)P_t(x, dy)
    \]

    then $\bracs{\bp_t|t \ge 0}$ is a semigroup such that:
    \begin{enumerate}
        \item For each $t \ge 0$, $\bp_t \one = \one$.
        \item For each $t \ge 0$, $\bp_t$ is a positive linear functional.
    \end{enumerate}

    known as the \textbf{semigroup} of $\bracs{P_t|t \ge 0}$.
\end{definition}
\begin{proof}
    Let $s, t \ge 0$, then for any $A \in \cm$,
    \[
    (\bp_{s+t}\one_A)(x) = \int \one_A(y)P_{s+t}(x, dy) = \int P_t(y, A)P_s(x, dy) = \int \int \one_A(z)P_t(y, dz)P_s(x, dy)
    \]

    Let $f: X \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable. By \autoref{proposition:measurable-simple-separable-norm}, there exists $\seq{f_n} \subset \Sigma(X; \complex)$ such that $\abs{f_n} \le \abs{f}$ for all $n \in\nat$, and $f_n \to f$ pointwise. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
    \begin{align*}
    (\bp_{s+t}f)(x) &= \limv{n}\int f_n(y)P_{s+t}(x, dy) = \limv{n}\iint f_n(z)P_t(y, dz)P_s(x, dy) \\
    &= \int \int f_n(z)P_t(y, dz)P_s(x, dy) = (\bp_s\bp_t) f(x)
    \end{align*}
\end{proof}

\begin{definition}[Markov Process]
\label{definition:markov-process}
    Let $(\Omega, \cf, \bp)$ be a probability space, $(Y, \cm)$ be a measurable space, $\bracs{X_t|t \ge 0}$ be a stochastic process, and $\bracs{\cf_t|t \ge 0}$ be its natural filtration.

    The process $\bracs{X_t|t \ge 0}$ is \textbf{Markov} if there exists a transition function $\bracs{P_t|t \ge 0}$ such that for every $f: Y \to \complex$ bounded and $(\cm, \cb_\complex)$-measurable,
    \[
    \ev\braks{f(X_{t+s})|\cf_s} = (\bp_tf)(X_s)
    \]

\end{definition}


\begin{theorem}[{{\cite[Theorem 3.14]{Baudoin}}}]
\label{theorem:markov-existence}
    Let $(Y, \cm)$ be a measurable space, $\nu: \cm \to [0, 1]$ be a probability measure on $Y$, and $\bracs{P_t|t \ge 0}$ be a transition function on $Y$, then there exists a probability space $(\Omega, \cf, \bp)$ and a stochastic process $\bracs{X_t|t \ge 0}$ such that:
    \begin{enumerate}
        \item The distribution of $X_0$ is $\nu$.
        \item $\bracs{X_t|t \ge 0}$ is a Markov process associated with $\bracs{P_t|t \ge 0}$.
    \end{enumerate}
\end{theorem}
\begin{proof}
    For each $N \in \natp$, $A \in \cm$ and $B \in \bigotimes_{n = 1}^N\cm$, and $0 = t_0 < t_1 < \cdots < t_N$, let
    \[
    \mu_{t_0, t_1, \cdots, t_N} = \int_A\int_B P_{t_1}(z, dx_1)P_{t_2 - t_1}(x_1, dx_2) \cdots P_{t_N - t_{N-1}}(x_{N_1}, dx_N) \nu(dz)
    \]

    then $\mu_{t_0, t_1, \cdots, t_N}$ is a probability measure on $\cm \times $
\end{proof}