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\begin{theorem*}
    Let $P$ be irreducible and aperiodic with
    invariant distribution $\pi$. Let $X \sim
    \Markov(\lambda, P)$. Then $\forall y$,
    \[ \PP(X_n = y) \to \pi(y) \]
    as $n \to \infty$. In particular, for all $x$
    and $y$,
    \[ P^n(x, y) \to \pi(y) \]
    as $n \to \infty$. (Taking $\lambda = \delta_x$)
\end{theorem*}

\begin{proof}
    Coupling of Markov chains. Let $(Y_n)_{n \ge
    0} \sim \Markov(\pi, P)$ independent of $X$.
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/73398f2e64ce11ed.png}
    \end{center}
    Consider $((X_n, Y_n))_{n \ge 0} \sim
    \Markov(\lambda \times \pi, \tilde{P})$ where
    \[ \tilde{P}((x, y), (x', y')) = P(x, x')P(y,
    y') \]
    We claim that $\tilde{P}$ is irreducible. So
    we want to show that there exist $l, m$ such
    that
    \[ P^l(x, x') > 0 \quad \text{ and } \quad
    P^m(y, y') > 0 \]
    Then note
    \[ P^n(x, x') \ge P^l(x, x') P^{n - l}(x', x')
    > 0 \]
    for all $N$ sufficiently large by aperiodicity
    of $P$. Similarly
    \[ P^n(y, y') \ge P^m(y, y') P^{n - m}(y', y')
    > 0 \]
    for sufficiently large $N$. So
    \[ \tilde{P}^n((x, y), (x', y')) = P^n(x, x')
    P^n(y, y') > 0 \]
    for all $n$ large enough. So $\tilde{P}$ is
    irreducible. \\
    $\tilde{P}$ has invariant distribution
    $\tilde{\pi}(x, y) = \pi(x)\pi(y)$. So
    $\tilde{P}$ is positive recurrent. Fix $\alpha
    \in I$. Define $T = \inf\{n \ge 1 : (X_n, Y_n)
    = (a, a)\}$. $T$ is a stopping time for $(X,
    Y)$. $\tilde{P}$ is positive recurrent so
    $\PP(T < \infty) = 1$. Define
    \[ Z_n = \begin{cases}
        X_n & n < T \\
        Y_n & n \ge T
    \end{cases} \]
    Now claim that $Z \sim \Markov(\lambda, P)$.
    Note that $Z_0 \sim \lambda$ because
    \[ \PP(Z_0 = x) = \PP(X_0 = x) = \lambda(x) .\]
    Let $A = \{Z_{n - 1} = z_{n - 1}, \dots, Z_0 =
    z_0\}$. Need to show
    \[ \PP(Z_{n + 1} = y \mid Z_n = x, A) = P(x,
    y) \]
    So we calculate:
    \begin{align*}
        \PP(Z_{n + 1} = y \mid Z_n = x, A)
        &= \PP(Z_{n + 1} = y, T > n \mid Z_n = x,
        A) + \PP(Z_{n + 1} = y, T \le n \mid Z_n =
        x, A) \\
        \PP(Z_{n + 1} = y, T > n \mid Z_n = x, A)
        &= \PP(X_{n + 1} = y \mid T > n, Z_n = x,
        A)\PP(T > n \mid Z_n = x, A)
    \end{align*}
    Now note that $\{T > n\} = \{T \le n\}^c$, but
    $\{T \le n\}^c$ only depends on $(X_0, Y_0),
    (X_n, Y_n)$ so by Strong Markov Property,
    \[ = P(x, y) \PP(T > n \mid Z_n = x, A) \]
    Similarly
    \[ \PP(Z_{n + 1} = y, T \le n \mid Z_n = x, A)
    = P(x, y) \PP(T \le n \mid Z_n = x, A) \]
    So
    \[ \PP(Z_{n + 1} = y \mid Z_n = x, A) = P(x,
    y) \]
    So $Z \sim \Markov(\lambda, P)$. \\
    Now we want to show that $|\PP(X_n = y) =
    \pi(y)| \to 0$ as $n \to \infty$. But since $Y
    \sim \Markov(\pi, P)$ so $\PP(Y_n = y) =
    \pi(y)$ for all $n$. So
    \begin{align*}
        |\PP(X_n = y) - \PP(Y_n = y)|
        &= |\PP(Z_n = y) - \PP(Y_n = y)| \\
        &= |\PP(Z_n = y, n < T) + \PP(Z_n = y, n
        \ge T) - \PP(Y_n = y)| \\
        &= |\PP(X_n = y, n < T) + \PP(Y_n = y, n
        \ge T) - \PP(Y_n = y)| \\
        &= |\PP(X_n = y, n < T) - \PP(Y_n = y, n <
        T)| \\
        &\le \PP(T > n)
    \end{align*}
    Take $n \to \infty$ we get $\PP(T > n) \to 0$
    because $\PP(T < \infty) = 1$. So $\PP(X_n =
    y) \to \PP(Y_n = y) = \pi(y)$.
\end{proof}

\begin{theorem*}
    Let $P$ be irreducible and aperiodic. Suppose
    $P$ is null recurrent. Then for all $x, y$,
    \[ P^n(x, y) \to 0 \]
    as $n \to \infty$.
\end{theorem*}

\begin{proof}
    Consider $\tilde{P}((x, y), (x', y')) = P(x,
    x') P(y, y')$. As before, $\tilde{P}$ is
    irreducible.
    \begin{itemize}
        \item If $\tilde{P}$ transient, then
            \[ \sum_n \tilde{P}^n((x, x), (y, y))
            = \sum_n (P^n(x, y))^2 < \infty \]
            so $P^n(x, y) \to 0$ as $n \to
            \infty$.
        \item If $\tilde{P}$ is recurrent then
            \[ \nu_y(z) = \EE_y \left[ \sum_{i =
            0}^{\tau_y - 1} \mathbbm{1}(x_i =
            z)\right] \]
            $\nu_y$ is invariant, i.e. $\nu_y P =
            \nu_y$. $P$ is null-recurrent so
            $\EE_y[\tau_y] = \infty$, so $\nu_y(I)
            = \infty$. Fix $M \in \NN$. Since
            $\nu_y(I) = \infty$, we can find a
            finite set $A$ such that $\nu_y(A) >
            M$. Define
            \[ \mu(x) = \frac{\nu_y(x)}{\nu_y(A)}
            \mathbbm{1}(x \in A) \]
            probability measure.
            \begin{align*}
                \mu P^n(z)
                &= \sum_x \mu(x) P^n(x, z) \\
                &\le \sum_x
                \frac{\nu_y(x)}{\nu_y(A)} P^n(x,
                z) \\
                &= \frac{\nu_y(x)}{\nu_y(A)}
            \end{align*}
            ($\nu_y P = \nu_y$) So
            \[ \boxed{\mu P^n(z) \le
            \frac{\nu_y(z)}{\nu_y(z)}} \]
            Consider $(X, Y) \sim \Markov(\mu
            \times \delta_x, \tilde{P})$. Define
            $T = \inf\{n \ge 0: (X_n, Y_n) = (x,
            x)\}$. $\PP(T < \infty) = 1$, since
            $\tilde{P}$ is recurrent.
            \[ Z_n = \begin{cases}
                X_n & n < T \\
                Y_n & n \ge T
            \end{cases} \]
            Then $Z_n \sim \Markov(\mu, P)$
            \begin{align*}
                \PP(Z_n = y)
                &= \mu P^n(y) \\
                &\le \frac{\nu_y(y)}{\nu_y(A)} \\
                &= \frac{1}{\nu_y(A)} \\
                &< \frac{1}{M}
            \end{align*}
            Need to show: $P^n(x, y) \to \infty$
            as $n \to \infty$.
            \begin{align*}
                P^n(x, y)
                &= \PP(Y_n = y) \\
                &= \PP(Y_n = y, n < T) + \PP(Y_n =
                y, n \ge T) \\
                &\le \PP(T > n) + \PP(Z_n = y)
            \end{align*}
            Let $n \to \infty$ then $\PP(T > n)
            \to 0$ ($\PP(T < \infty) = 1$). So
            \[ \lim_{n \to \infty} P^n(x, y) <
            \frac{1}{M} \]
            Taking $M \to \infty$ finishes that
            proof. \qedhere
    \end{itemize}
\end{proof}
