[proofplan] We use the ergodic decomposition of $\mu$ to assign to $\mu$-almost every point $x \in X$ an ergodic invariant component measure $\mu_x$. Pushing $\mu$ forward under the component map gives a Borel probability measure on $\mathcal{M}_T$, and the ergodicity assertion in the decomposition makes this measure concentrated on $\mathcal{E}_T$. The disintegration identity in the Ergodic Decomposition Theorem is exactly the barycenter identity after applying the integral formula for pushforward measures. The uniqueness clause in the same theorem gives uniqueness of the representing measure. [/proofplan] [step:Apply the ergodic decomposition theorem to obtain the component map] Let $\mathcal{B}(X)$ denote the Borel $\sigma$-algebra of $X$, let $\mathcal{M}(X)$ denote the space of Borel probability measures on $X$ with the weak-$*$ topology, and let $\mathcal{B}(\mathcal{M}_T)$ denote the Borel $\sigma$-algebra of $\mathcal{M}_T$. Let $C(X;\mathbb{R})$ denote the real vector space of continuous maps from $X$ to $\mathbb{R}$, and let $\mathcal{E}_T \subseteq \mathcal{M}_T$ denote the set of ergodic $T$-invariant Borel probability measures. Since $X$ is compact metrizable, $T: X \to X$ is continuous, and $\mu \in \mathcal{M}_T$, the hypotheses of the [Ergodic Decomposition Theorem](/theorems/3453) are satisfied. Hence there exists a $\mathcal{B}(X)$-$\mathcal{B}(\mathcal{M}_T)$ measurable map \begin{align*} \Psi: X &\to \mathcal{M}_T, \\ x &\mapsto \mu_x, \end{align*} such that $\mu_x \in \mathcal{E}_T$ for $\mu$-almost every $x \in X$ and, for every $f \in C(X;\mathbb{R})$, \begin{align*} \int_X f(y) \, d\mu(y) = \int_X \left(\int_X f(y) \, d\mu_x(y)\right) \, d\mu(x). \end{align*} [/step] [step:Push $\mu$ forward to a probability measure concentrated on $\mathcal{E}_T$] Define \begin{align*} \tau_\mu: \mathcal{B}(\mathcal{M}_T) &\to [0,1], \\ A &\mapsto \mu(\Psi^{-1}(A)). \end{align*} The measurability of $\Psi$ makes this set function well-defined, and it is the pushforward probability measure $\Psi_\#\mu$ because \begin{align*} \tau_\mu(\mathcal{M}_T)=\mu(X)=1. \end{align*} Since $\mathcal{E}_T$ is Borel in $\mathcal{M}_T$ and $\Psi(x) \in \mathcal{E}_T$ for $\mu$-almost every $x \in X$, \begin{align*} \tau_\mu(\mathcal{E}_T) = \mu(\Psi^{-1}(\mathcal{E}_T)) = 1. \end{align*} Thus $\tau_\mu$ is a Borel probability measure on $\mathcal{M}_T$ supported on $\mathcal{E}_T$. [/step] [step:Convert the disintegration formula into the barycenter identity] Fix $f \in C(X;\mathbb{R})$, and define \begin{align*} I_f: \mathcal{M}_T &\to \mathbb{R}, \\ \nu &\mapsto \int_X f(y) \, d\nu(y). \end{align*} The weak-$*$ topology makes $I_f$ continuous, hence Borel measurable. Since $X$ is compact and $f$ is continuous, \begin{align*} M_f := \sup_{y \in X} |f(y)| \end{align*} is finite, and $|I_f(\nu)| \leq M_f$ for every $\nu \in \mathcal{M}_T$. Therefore $I_f$ is $\tau_\mu$-integrable. By the defining integral property of $\tau_\mu=\Psi_\#\mu$, \begin{align*} \int_{\mathcal{M}_T} I_f(\nu) \, d\tau_\mu(\nu) &= \int_X I_f(\Psi(x)) \, d\mu(x) \\ &= \int_X \left(\int_X f(y) \, d\mu_x(y)\right) \, d\mu(x) \\ &= \int_X f(y) \, d\mu(y), \end{align*} where the last equality is the disintegration formula from the [Ergodic Decomposition Theorem](/theorems/3453). Since $\tau_\mu(\mathcal{E}_T)=1$, \begin{align*} \int_X f(y) \, d\mu(y) = \int_{\mathcal{E}_T} \left(\int_X f(y) \, d\nu(y)\right) \, d\tau_\mu(\nu). \end{align*} This is exactly the weak-$*$ barycenter identity $\mu=\int_{\mathcal{E}_T}\nu \, d\tau_\mu(\nu)$. [/step] [step:Use the uniqueness clause of ergodic decomposition to identify any representing measure] Let $\rho: \mathcal{B}(\mathcal{M}_T) \to [0,1]$ be a Borel probability measure such that $\rho(\mathcal{E}_T)=1$ and, for every $f \in C(X;\mathbb{R})$, \begin{align*} \int_X f(y) \, d\mu(y) = \int_{\mathcal{E}_T} \left(\int_X f(y) \, d\nu(y)\right) \, d\rho(\nu). \end{align*} Then $\rho$ satisfies the representing-measure hypotheses in the uniqueness clause of the [Ergodic Decomposition Theorem](/theorems/3453): it is a Borel probability measure on $\mathcal{M}_T$, it is concentrated on $\mathcal{E}_T$, and its barycenter is $\mu$. That clause implies \begin{align*} \rho=\Psi_\#\mu=\tau_\mu. \end{align*} Hence the representing Borel probability measure $\tau_\mu$ is unique. [/step]