[proofplan]
We prove the three assertions separately. Non-emptiness is exactly the content of the Krylov-Bogolyubov existence theorem for continuous self-maps of compact spaces. Convexity follows by checking the defining invariance identity against continuous test functions, and compactness follows because $\mathcal{M}(X)$ is weak* compact while $\mathcal{M}_T$ is the intersection of closed zero-sets of weak* continuous affine functionals.
[/proofplan]
[step:Use Krylov-Bogolyubov to obtain at least one invariant probability measure]
By the theorem hypotheses, $X$ is non-empty and compact and $T: X \to X$ is continuous. Therefore the [Krylov-Bogolyubov Theorem](/theorems/3423) applies to $T$ and gives a Borel probability measure $\mu \in \mathcal{M}(X)$ satisfying $T_\#\mu = \mu$. By the definition of $\mathcal{M}_T$, this means $\mu \in \mathcal{M}_T$. Hence $\mathcal{M}_T \neq \varnothing$.
[/step]
[step:Check convex combinations preserve invariance]
Let $\mu, \nu \in \mathcal{M}_T$ and let $t \in [0,1]$. Define the measure $\lambda \in \mathcal{M}(X)$ by
\begin{align*}
\lambda(B) := t\mu(B) + (1-t)\nu(B)
\end{align*}
for every Borel set $B \in \mathcal{B}(X)$. Since $\mu$ and $\nu$ are Borel probability measures and $t,1-t \geq 0$ with $t+(1-t)=1$, $\lambda$ is again a Borel probability measure on $X$.
For every $B \in \mathcal{B}(X)$, continuity of $T$ implies $T^{-1}(B) \in \mathcal{B}(X)$. Since $\mu,\nu \in \mathcal{M}_T$, we have $\mu(T^{-1}(B))=\mu(B)$ and $\nu(T^{-1}(B))=\nu(B)$. Therefore
\begin{align*}
\lambda(T^{-1}(B))
&= t\mu(T^{-1}(B)) + (1-t)\nu(T^{-1}(B)) \\
&= t\mu(B) + (1-t)\nu(B) \\
&= \lambda(B).
\end{align*}
Thus $\lambda \in \mathcal{M}_T$. Hence $\mathcal{M}_T$ is convex.
[/step]
[step:Express invariance as vanishing of weak* continuous functionals]
For each function $f \in C(X)$, define the affine functional
\begin{align*}
\Phi_f: \mathcal{M}(X) &\to \mathbb{R} \\
\eta &\mapsto \int_X f(T(x)) \, d\eta(x) - \int_X f(x) \, d\eta(x).
\end{align*}
Since $T: X \to X$ is continuous and $f: X \to \mathbb{R}$ is continuous, the composition $f \circ T: X \to \mathbb{R}$ is continuous. By definition of the weak* topology on $\mathcal{M}(X)$, the maps
\begin{align*}
\eta \mapsto \int_X f(T(x)) \, d\eta(x),
\qquad
\eta \mapsto \int_X f(x) \, d\eta(x)
\end{align*}
are continuous. Hence $\Phi_f$ is weak* continuous.
Moreover, by the characterization of pushforward measures through continuous test functions on compact metrizable spaces, a measure $\eta \in \mathcal{M}(X)$ satisfies $T_\#\eta=\eta$ if and only if
\begin{align*}
\int_X f(T(x)) \, d\eta(x) = \int_X f(x) \, d\eta(x)
\end{align*}
for every $f \in C(X)$. Therefore
\begin{align*}
\mathcal{M}_T
=
\bigcap_{f \in C(X)} \Phi_f^{-1}(\{0\}).
\end{align*}
Since $\{0\} \subset \mathbb{R}$ is closed and each $\Phi_f$ is weak* continuous, each set $\Phi_f^{-1}(\{0\})$ is weak* closed. Hence $\mathcal{M}_T$ is weak* closed in $\mathcal{M}(X)$.
[/step]
[step:Combine closedness with compactness of the ambient measure space]
By weak-$*$ compactness of the space of Borel probability measures on a compact metrizable space, $\mathcal{M}(X)$ is compact in the weak-$*$ topology. Since $\mathcal{M}_T$ is weak-$*$ closed in $\mathcal{M}(X)$, the topological fact that a closed subset of a compact space is compact implies that $\mathcal{M}_T$ is compact in the weak-$*$ topology. Combining non-emptiness, convexity, and compactness proves that $\mathcal{M}_T$ is a non-empty compact convex subset of $\mathcal{M}(X)$.
[/step]
