[proofplan] We apply the Birkhoff-Khinchin ergodic theorem to the integrable stationary process $(X_t)_{t\in\mathbb Z}$, which identifies the almost sure limit of the time averages as the conditional expectation of $X_0$ with respect to the invariant $\sigma$-algebra. Ergodicity forces every invariant event to have probability $0$ or $1$, so every invariant random variable is almost surely constant. Finally, the defining property of conditional expectation identifies that constant with $\mathbb E[X_0]$. [/proofplan]

[step:Apply the ergodic theorem to identify the almost sure limit]Let $(\Omega,\mathcal F,\mathbb P)$ be the underlying probability space. Define the path map \begin{align*} X:\Omega&\to\mathbb R^{\mathbb Z}\\ \omega&\mapsto (X_t(\omega))_{t\in\mathbb Z}, \end{align*} and let $\mu:=\mathbb P\circ X^{-1}$ be the law of the two-sided process on the product measurable space $(\mathbb R^{\mathbb Z},\mathcal B(\mathbb R)^{\otimes\mathbb Z})$. Let $\theta:\mathbb R^{\mathbb Z}\to\mathbb R^{\mathbb Z}$ be the left shift map, defined by $(\theta x)_t=x_{t+1}$ for $x=(x_t)_{t\in\mathbb Z}$ and $t\in\mathbb Z$. Define the invariant $\sigma$-algebra \begin{align*} \mathcal I:=\{X^{-1}(B):B\in\mathcal B(\mathbb R)^{\otimes\mathbb Z}\text{ and }\theta^{-1}B=B\text{ modulo }\mu\}\subseteq\mathcal F. \end{align*} Since $\mathbb E[|X_0|]<\infty$, the [Birkhoff-Khinchin Ergodic Theorem for Stationary Processes](/theorems/???) applies to $(X_t)_{t\in\mathbb Z}$ and gives an integrable $\mathcal I$-measurable random variable $Y:\Omega\to\mathbb R$ such that \begin{align*} Y=\mathbb E[X_0\mid\mathcal I] \end{align*} and \begin{align*} \frac{1}{n}\sum_{t=1}^{n}X_t \to Y \end{align*} $\mathbb P$-a.s. as $n\to\infty$.[/step]

[guided]Let $(\Omega,\mathcal F,\mathbb P)$ be the underlying probability space. To make the invariant $\sigma$-algebra precise, define the path map \begin{align*} X:\Omega&\to\mathbb R^{\mathbb Z}\\ \omega&\mapsto (X_t(\omega))_{t\in\mathbb Z}, \end{align*} and let $\mu:=\mathbb P\circ X^{-1}$ be the law of the two-sided process on $(\mathbb R^{\mathbb Z},\mathcal B(\mathbb R)^{\otimes\mathbb Z})$. Let $\theta:\mathbb R^{\mathbb Z}\to\mathbb R^{\mathbb Z}$ be the left shift map, defined by $(\theta x)_t=x_{t+1}$ for $x=(x_t)_{t\in\mathbb Z}$ and $t\in\mathbb Z$. We define \begin{align*} \mathcal I:=\{X^{-1}(B):B\in\mathcal B(\mathbb R)^{\otimes\mathbb Z}\text{ and }\theta^{-1}B=B\text{ modulo }\mu\}\subseteq\mathcal F. \end{align*} Thus $\mathcal I$ consists of events determined by the two-sided trajectory whose occurrence is unchanged, up to null sets, by shifting the trajectory. The hypothesis $\mathbb E[|X_0|]<\infty$ is exactly the integrability hypothesis needed to apply the [Birkhoff-Khinchin Ergodic Theorem for Stationary Processes](/theorems/???). That theorem states that the time averages of an integrable stationary process converge almost surely to the conditional expectation of one coordinate with respect to the invariant $\sigma$-algebra. Therefore there exists an integrable $\mathcal I$-measurable random variable $Y:\Omega\to\mathbb R$ satisfying \begin{align*} Y=\mathbb E[X_0\mid\mathcal I] \end{align*} and \begin{align*} \frac{1}{n}\sum_{t=1}^{n}X_t \to Y \end{align*} $\mathbb P$-a.s. as $n\to\infty$. This is the only place where the convergence theorem is used. The remaining work is to show that ergodicity turns the conditional expectation $Y$ into the constant $\mathbb E[X_0]$.[/guided]

[step:Use ergodicity to show the invariant conditional expectation is constant]Because the process is ergodic, every event $A\in\mathcal I$ satisfies $\mathbb P(A)\in\{0,1\}$. Since $Y$ is $\mathcal I$-measurable, for every $a\in\mathbb R$ the event \begin{align*} A_a:=\{\omega\in\Omega:Y(\omega)\le a\} \end{align*} belongs to $\mathcal I$, and hence $\mathbb P(A_a)\in\{0,1\}$. Let $\mathbb Q\subset\mathbb R$ denote the set of rational numbers. Define \begin{align*} c:=\inf\{q\in\mathbb Q:\mathbb P(Y\le q)=1\}. \end{align*} The defining set is nonempty: since $Y$ is finite $\mathbb P$-a.s., the events $\{Y\le m\}$ for $m\in\mathbb N$ increase to an event of probability $1$, so continuity from below gives \begin{align*} \lim_{m\to\infty}\mathbb P(Y\le m)=1. \end{align*} Each event $\{Y\le m\}$ lies in $\mathcal I$, hence each probability $\mathbb P(Y\le m)$ belongs to $\{0,1\}$. Therefore $\mathbb P(Y\le m_0)=1$ for some $m_0\in\mathbb N$. The defining set is not all of $\mathbb Q$: since $Y$ is finite $\mathbb P$-a.s., the events $\{Y\le -m\}$ for $m\in\mathbb N$ decrease to an event of probability $0$, so continuity from above gives \begin{align*} \lim_{m\to\infty}\mathbb P(Y\le -m)=0. \end{align*} Again $\mathbb P(Y\le -m)\in\{0,1\}$ for every $m\in\mathbb N$, so $\mathbb P(Y\le -m_1)=0$ for some $m_1\in\mathbb N$. Thus $c\in\mathbb R$ is well-defined. If $q\in\mathbb Q$ and $q>c$, then by the definition of the infimum there exists $r\in\mathbb Q$ with $c<r<q$ and $\mathbb P(Y\le r)=1$, so $\mathbb P(Y\le q)=1$. If $q\in\mathbb Q$ and $q<c$, then $\mathbb P(Y\le q)\ne1$, and therefore $\mathbb P(Y\le q)=0$. Thus \begin{align*} \mathbb P(Y\le q)=0 \quad \text{for all } q\in\mathbb Q,\ q<c, \end{align*} and \begin{align*} \mathbb P(Y\le q)=1 \quad \text{for all } q\in\mathbb Q,\ q>c. \end{align*} Taking a countable union over rational $q<c$ and a countable intersection over rational $q>c$ gives \begin{align*} \mathbb P(Y<c)=0, \end{align*} and \begin{align*} \mathbb P(Y\le c)=1. \end{align*} Therefore $Y=c$ $\mathbb P$-a.s.[/step]

[guided]The key point is that an $\mathcal I$-measurable random variable can only depend on invariant information. Under ergodicity, invariant information is probabilistically trivial. For each $a\in\mathbb R$, define the event \begin{align*} A_a:=\{\omega\in\Omega:Y(\omega)\le a\}. \end{align*} Because $Y$ is $\mathcal I$-measurable, $A_a\in\mathcal I$. Since the process is ergodic, every event in $\mathcal I$ has probability either $0$ or $1$, so \begin{align*} \mathbb P(A_a)\in\{0,1\}. \end{align*} We now prove that this forces $Y$ to be almost surely constant. Let $\mathbb Q\subset\mathbb R$ denote the set of rational numbers, and define \begin{align*} c:=\inf\{q\in\mathbb Q:\mathbb P(Y\le q)=1\}. \end{align*} The rational numbers are used so that the later unions and intersections are countable. We must justify that this infimum is finite. Because $Y:\Omega\to\mathbb R$ is finite $\mathbb P$-a.s., the events $\{Y\le m\}$ for $m\in\mathbb N$ increase to an event of probability $1$. Continuity from below for the probability measure $\mathbb P$ gives \begin{align*} \lim_{m\to\infty}\mathbb P(Y\le m)=1. \end{align*} For every $m\in\mathbb N$, the threshold event $\{Y\le m\}$ lies in $\mathcal I$, so ergodicity gives $\mathbb P(Y\le m)\in\{0,1\}$. A sequence taking only the values $0$ and $1$ and converging to $1$ must equal $1$ for at least one index; hence there exists $m_0\in\mathbb N$ such that $\mathbb P(Y\le m_0)=1$. Thus the defining set is nonempty. We also need to know that the defining set is not all of $\mathbb Q$. Since $Y$ is real-valued $\mathbb P$-a.s., the events $\{Y\le -m\}$ for $m\in\mathbb N$ decrease to an event of probability $0$. Continuity from above for $\mathbb P$ gives \begin{align*} \lim_{m\to\infty}\mathbb P(Y\le -m)=0. \end{align*} Each probability $\mathbb P(Y\le -m)$ is again either $0$ or $1$, so there exists $m_1\in\mathbb N$ such that $\mathbb P(Y\le -m_1)=0$. Therefore $-m_1$ does not belong to the defining set, and $c\in\mathbb R$. If $q\in\mathbb Q$ and $q>c$, then by the definition of infimum there exists $r\in\mathbb Q$ such that $c<r<q$ and $\mathbb P(Y\le r)=1$. Since $\{Y\le r\}\subseteq\{Y\le q\}$, we get \begin{align*} \mathbb P(Y\le q)=1. \end{align*} If $q\in\mathbb Q$ and $q<c$, then $q$ cannot belong to the defining set for $c$, so $\mathbb P(Y\le q)\ne1$. But this probability is either $0$ or $1$, hence \begin{align*} \mathbb P(Y\le q)=0. \end{align*} Now countability of $\mathbb Q$ lets us pass from rational thresholds to the random variable itself: \begin{align*} \{Y<c\}=\bigcup_{\substack{q\in\mathbb Q\\ q<c}}\{Y\le q\}, \end{align*} so $\mathbb P(Y<c)=0$. Also, \begin{align*} \{Y>c\}=\bigcup_{\substack{q\in\mathbb Q\\ q>c}}\{Y>q\}, \end{align*} and each event $\{Y>q\}$ has probability $0$, because $\mathbb P(Y\le q)=1$ for every rational $q>c$. Therefore $\mathbb P(Y>c)=0$. Combining the two estimates gives $Y=c$ $\mathbb P$-a.s.[/guided]

[step:Identify the constant as the expectation of $X_0$] Since $Y=\mathbb E[X_0\mid\mathcal I]$, the defining property of conditional expectation applied to the event $\Omega\in\mathcal I$ gives \begin{align*} \mathbb E[Y]=\mathbb E[X_0]. \end{align*} Because $Y=c$ $\mathbb P$-a.s., we also have $\mathbb E[Y]=c$. Hence \begin{align*} c=\mathbb E[X_0]. \end{align*} Combining this identity with the almost sure convergence from the first step gives \begin{align*} \bar X_n=\frac{1}{n}\sum_{t=1}^{n}X_t \to \mathbb E[X_0] \end{align*} $\mathbb P$-a.s. as $n\to\infty$. [/step]