[proofplan] We first reduce equidistribution to checking the convergence identity on a countable uniformly dense family in $C(X)$. The [Birkhoff ergodic theorem](/theorems/518) gives a full-measure set on which the desired identity holds for every function in that countable family. Finally, uniform approximation transfers the identity from the dense family to an arbitrary continuous [test function](/page/Test%20Function). [/proofplan]

[step:Choose a countable uniformly dense family of continuous test functions] Let $d: X \times X \to [0,\infty)$ denote the metric on $X$. For $a \in X$ and $r > 0$, let $B(a,r) := \{x \in X : d(x,a) < r\}$ denote the open ball in the [metric space](/page/Metric%20Space) $(X,d)$. For a subset $E \subset X$, define the distance-to-set function $\operatorname{dist}(\cdot,E): X \to [0,\infty]$ by \begin{align*} \operatorname{dist}(x,E) := \inf_{y \in E} d(x,y). \end{align*} Let $C(X)$ denote the real [vector space](/page/Vector%20Space) of continuous functions $f: X \to \mathbb{R}$ equipped with the uniform norm \begin{align*} \|f\|_{\infty} := \sup_{x \in X} |f(x)|. \end{align*} Since $(X,d)$ is compact and metric, $C(X)$ is separable in the uniform norm. Fix a countable [dense subset](/page/Dense%20Subset) \begin{align*} \mathcal{D} := \{f_j : X \to \mathbb{R} \mid j \in \mathbb{N}\} \subset C(X). \end{align*} To justify the separability assertion, for each $m \in \mathbb{N}$ choose a finite set $F_m = \{a_{m,1},\dots,a_{m,k_m}\} \subset X$ such that the open balls $B(a_{m,i},1/(4m))$ cover $X$. For each $m$ and $i$, define the [continuous function](/page/Continuous%20Function) $\psi_{m,i}: X \to [0,\infty)$ by \begin{align*} \psi_{m,i}(x) := \operatorname{dist}\bigl(x, X \setminus B(a_{m,i},1/(2m))\bigr). \end{align*} The function $\Psi_m: X \to (0,\infty)$ given by \begin{align*} \Psi_m(x) := \sum_{i=1}^{k_m} \psi_{m,i}(x) \end{align*} is strictly positive because the smaller balls $B(a_{m,i},1/(4m))$ cover $X$. Hence the functions $\varphi_{m,i}: X \to [0,1]$ defined by \begin{align*} \varphi_{m,i}(x) := \frac{\psi_{m,i}(x)}{\Psi_m(x)} \end{align*} are continuous and satisfy \begin{align*} \sum_{i=1}^{k_m} \varphi_{m,i}(x) = 1 \end{align*} for every $x \in X$. The collection of all rational linear combinations \begin{align*} \sum_{i=1}^{k_m} q_i \varphi_{m,i} \end{align*} with $m \in \mathbb{N}$ and $q_i \in \mathbb{Q}$ is countable. Let $f: X \to \mathbb{R}$ be continuous and let $\varepsilon > 0$. Since $X$ is compact, $f$ is uniformly continuous. Choose $m \in \mathbb{N}$ so large that $d(x,y) < 1/m$ implies $|f(x)-f(y)| < \varepsilon/2$. Choose rational numbers $q_i \in \mathbb{Q}$ satisfying $|q_i - f(a_{m,i})| < \varepsilon/2$ for $1 \leq i \leq k_m$, and define $g: X \to \mathbb{R}$ by \begin{align*} g(x) := \sum_{i=1}^{k_m} q_i\varphi_{m,i}(x). \end{align*} If $\varphi_{m,i}(x) \neq 0$, then $x \in B(a_{m,i},1/(2m))$, so $d(x,a_{m,i}) < 1/m$ and therefore $|f(x)-f(a_{m,i})| < \varepsilon/2$. Using the partition-of-unity identity, \begin{align*} |f(x)-g(x)| \leq \sum_{i=1}^{k_m} \varphi_{m,i}(x)|f(x)-q_i| < \varepsilon \end{align*} for every $x \in X$. Thus the countable family of these rational linear combinations is dense in $C(X)$, and we may enumerate a dense subset as $\mathcal{D} = \{f_j\}_{j \in \mathbb{N}}$. [/step]

[step:Apply Birkhoff's theorem to every function in the dense family] For each $j \in \mathbb{N}$, the function $f_j: X \to \mathbb{R}$ is continuous on compact $X$, hence bounded and Borel measurable. Since $\mu$ is a probability measure, $f_j \in L^1(X,\mathcal{B}(X),\mu)$. By the Birkhoff Ergodic Theorem (citing a result not yet in the wiki: Birkhoff Ergodic Theorem), applied to the measure-preserving system $(X,\mathcal{B}(X),\mu,T)$ and the integrable function $f_j$, there exists a measurable set $A_j \in \mathcal{B}(X)$ with $\mu(A_j)=1$ such that for every $x \in A_j$, \begin{align*} \lim_{N \to \infty} \frac{1}{N}\sum_{n=0}^{N-1} f_j(T^n x) = \int_X f_j(y)\,d\mu(y). \end{align*} Here ergodicity is used in the Birkhoff theorem to identify the invariant [conditional expectation](/page/Conditional%20Expectation) of $f_j$ with the constant function equal to $\int_X f_j(y)\,d\mu(y)$. Define the measurable set $A \in \mathcal{B}(X)$ by \begin{align*} A := \bigcap_{j=1}^{\infty} A_j. \end{align*} Since $A$ is a countable intersection of full-measure sets, \begin{align*} \mu(A)=1. \end{align*} For every $x \in A$ and every $j \in \mathbb{N}$, the Birkhoff average identity above holds for $f_j$. [/step]

[step:Extend the convergence identity from the dense family to every continuous function]Fix $x \in A$ and let $f: X \to \mathbb{R}$ be continuous. Let $\varepsilon > 0$. Since $\mathcal{D}$ is dense in $C(X)$ with respect to $\|\cdot\|_{\infty}$, choose $j \in \mathbb{N}$ such that \begin{align*} \|f-f_j\|_{\infty} < \varepsilon. \end{align*} For each $N \in \mathbb{N}$, define the time averages $M_N(f,x) \in \mathbb{R}$ and $M_N(f_j,x) \in \mathbb{R}$ by \begin{align*} M_N(f,x) := \frac{1}{N}\sum_{n=0}^{N-1} f(T^n x) \end{align*} and \begin{align*} M_N(f_j,x) := \frac{1}{N}\sum_{n=0}^{N-1} f_j(T^n x). \end{align*} Then the triangle inequality gives \begin{align*} \left|M_N(f,x)-\int_X f(y)\,d\mu(y)\right| \leq \left|M_N(f,x)-M_N(f_j,x)\right| + \left|M_N(f_j,x)-\int_X f_j(y)\,d\mu(y)\right| + \left|\int_X f_j(y)\,d\mu(y)-\int_X f(y)\,d\mu(y)\right|. \end{align*} The first term is bounded by $\varepsilon$ because \begin{align*} \left|M_N(f,x)-M_N(f_j,x)\right| \leq \frac{1}{N}\sum_{n=0}^{N-1} |f(T^n x)-f_j(T^n x)| \leq \|f-f_j\|_{\infty} < \varepsilon. \end{align*} The third term is bounded by $\varepsilon$ because $\mu(X)=1$ and \begin{align*} \left|\int_X f_j(y)\,d\mu(y)-\int_X f(y)\,d\mu(y)\right| \leq \int_X |f_j(y)-f(y)|\,d\mu(y) \leq \|f-f_j\|_{\infty}\mu(X) < \varepsilon. \end{align*} Since $x \in A$, the middle term tends to $0$ as $N \to \infty$. Therefore \begin{align*} \limsup_{N \to \infty}\left|M_N(f,x)-\int_X f(y)\,d\mu(y)\right| \leq 2\varepsilon. \end{align*} Because $\varepsilon > 0$ was arbitrary, \begin{align*} \lim_{N \to \infty} M_N(f,x) = \int_X f(y)\,d\mu(y). \end{align*}[/step]

[guided]We now explain why convergence on the countable dense family is enough. Fix a point $x \in A$ and a continuous function $f: X \to \mathbb{R}$. The set $A$ was chosen so that the Birkhoff average identity is already known for every function $f_j$ in the dense family $\mathcal{D}$. The goal is to compare $f$ with one such $f_j$ uniformly on all of $X$, because uniform control automatically controls both the time averages along the orbit and the space averages against $\mu$. Let $\varepsilon > 0$. Since $\mathcal{D}$ is dense in $C(X)$ for the uniform norm, there exists $j \in \mathbb{N}$ such that \begin{align*} \|f-f_j\|_{\infty} < \varepsilon. \end{align*} For each $N \in \mathbb{N}$, define the time averages \begin{align*} M_N(f,x) := \frac{1}{N}\sum_{n=0}^{N-1} f(T^n x) \end{align*} and \begin{align*} M_N(f_j,x) := \frac{1}{N}\sum_{n=0}^{N-1} f_j(T^n x). \end{align*} We compare the desired quantity with the known one by adding and subtracting $M_N(f_j,x)$ and $\int_X f_j(y)\,d\mu(y)$. The triangle inequality gives \begin{align*} \left|M_N(f,x)-\int_X f(y)\,d\mu(y)\right| \leq \left|M_N(f,x)-M_N(f_j,x)\right| + \left|M_N(f_j,x)-\int_X f_j(y)\,d\mu(y)\right| + \left|\int_X f_j(y)\,d\mu(y)-\int_X f(y)\,d\mu(y)\right|. \end{align*} The first term is small because every orbit point $T^n x$ lies in $X$, and the uniform norm controls the difference at every point of $X$: \begin{align*} \left|M_N(f,x)-M_N(f_j,x)\right| \leq \frac{1}{N}\sum_{n=0}^{N-1}|f(T^n x)-f_j(T^n x)| \leq \|f-f_j\|_{\infty} < \varepsilon. \end{align*} The third term is small for the same uniform reason, now integrated against the probability measure $\mu$: \begin{align*} \left|\int_X f_j(y)\,d\mu(y)-\int_X f(y)\,d\mu(y)\right| \leq \int_X |f_j(y)-f(y)|\,d\mu(y) \leq \|f-f_j\|_{\infty}\mu(X) < \varepsilon. \end{align*} The middle term is the one for which we use the definition of $A$: since $x \in A \subset A_j$, the Birkhoff average identity for $f_j$ holds at $x$, so \begin{align*} \lim_{N \to \infty}\left|M_N(f_j,x)-\int_X f_j(y)\,d\mu(y)\right| = 0. \end{align*} Taking the limit superior in the triangle inequality therefore gives \begin{align*} \limsup_{N \to \infty}\left|M_N(f,x)-\int_X f(y)\,d\mu(y)\right| \leq 2\varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, the limit superior is $0$, and hence \begin{align*} \lim_{N \to \infty} M_N(f,x) = \int_X f(y)\,d\mu(y). \end{align*} This proves the equidistribution identity for the arbitrary continuous test function $f$.[/guided]

[step:Conclude equidistribution on a full-measure set] We have constructed a measurable set $A \subset X$ with $\mu(A)=1$ such that for every $x \in A$ and every continuous function $f: X \to \mathbb{R}$, \begin{align*} \lim_{N \to \infty} \frac{1}{N}\sum_{n=0}^{N-1} f(T^n x) = \int_X f(y)\,d\mu(y). \end{align*} This is precisely the definition that the orbit $(T^n x)_{n \geq 0}$ is equidistributed with respect to $\mu$. Hence the orbit of $\mu$-almost every point $x \in X$ is equidistributed with respect to $\mu$. [/step]