The existence of invariant probability Radon measure I do not know any background of this topic. Feel free to edit my post.
Let X be a compact metric space. We call $\mu \in C(X,\mathbb{R})^\ast$ a probability radon measure if it is a positive linear functional and satisfies $\mu (\mathbb{1}) = 1$,where $\mathbb{1}$ denotes the constant function which assumes 1 everywhere. Suppose there exists at least one such measure. Let $\phi : X \to X$ be a homeomorphism. Show that there exists a probability radon measure $\mu$ s.t. $\phi^\ast\mu = \mu$ where $\phi^\ast$ is defined as $(\phi^\ast\mu)(f)=\mu(f\circ \phi)$
I tried to solve this exercise by using Schauder fixed point theorem which states that a continuous function from a compact convex subset of a Banach space to itself has a fixed point. Let $PR$ be the set of probability radon measure. $PR$ is obviously convex and $\phi^\ast$ is continuous (because it is a bounded linear function and it maps $PR$ to $PR$ ). What I tried to do next is of course to show that $PR$ is compact but in the strong topology I do not know if it is true. By using Ascoli theorem, I could show that $PR$ is compact in the compact convergence topology, but again there is no guarantee that the uniform topology and the compact convergence topology agree on $PR$ or this approach is effective.
By Ascoli theorem, I refer to the following one which is proved in Munkres.
Let Y be a metric space and X be a topological space and endow C(X,Y) with the compact convergence topology. Then a subset of C(X,Y) is precompact if it is pointwise precompact and equicontinuous.
In class, this problem was solved as follows. First apply Banach Alaoglu theorem and we have a weak* compact unit ball $B$ which includes $PR$. Let $F_\delta(f_1,f_2,\cdots f_n)=\left\{\mu\in PR: |(\phi^\ast\mu - \mu)(f_i)|<\delta     , i =1,2,\cdots n\right\}$ then this kind of set is non-empty and closed and running $\delta\in\mathbb{R}^{+}, f_i \in C(X,\mathbb{R}), n\in\mathbb{Z}$ and taking the intersection we can show the existence of a probability radon measure invariant under $\phi^\ast$
My question is whether my idea can be corrected or not. If impossible, I welcome other nice proofs.
 A: Let $Y$ denote the space of (signed) Radon measures over X, that is $C(X,\mathbb{R})^\ast$ equipped with the weak topology. This is a locally convex topological vector space (see the wikipedia article on weak topologies). Now the map $\Phi^\ast\colon Y \rightarrow Y$ is a continuous map, mapping the set $S := \{ \mu \in C(X,\mathbb{R})^\ast \,\colon\; \mu \ge 0, \|\mu\| \leq 1\}$ onto itself. Since $S$ is compact with respect to the weak topology, $\Phi^\ast\vert_S$ is a self-mapping of a convex compact subset of a locally convex topological vector space. Hence Tykhonov's Fixed Point Theorem (see here) implies that $\phi^\ast$ has a fixed point $\mu_0$. Set $\mu = \mu_0/\|\mu\|$, which proves the claim.
A: I always see this result (or the idea of the proof) attributed to Krylov and Bogoliubov.  Let $\nu$ be any Radon probability measure on $X$.  For each $n$, let $\varphi_{*}^{n}\nu$ be the measure given by $\varphi_{*}^{n}\nu(A) = \nu(\varphi^{n}(A))$, where $\varphi^{n}$ is the $n$th iterate of $\varphi$.  For each $N$, consider the Cesaro sum
$$\nu_{N} = N^{-1} \sum_{j = 0}^{N - 1} \varphi_{*}^{j} \nu.$$
Then the sequence $(\nu_{N})_{N \in \mathbb{N}}$ consists of Radon probability measures on $X$, and hence is pre-compact in the weak-$*$ topology on $C(X)^{*}$.  Since $X$ is compact, $C(X)$ is separable and, therefore, the closed unit ball in $C(X)^{*}$ is a compact metric space in the weak-$*$ topology.  In particular, the closed unit ball in $C(X)^{*}$ is sequentially compact in the weak-$*$ topology.  Thus, there is an increasing sequence $(N_{j})_{j \in \mathbb{N}}$ and a Radon probability measure $\mu$ such that $\mu = \lim_{j \to \infty} \nu_{N_{j}}$ in the weak-$*$ topology.  
I claim that $\varphi_{*}\mu = \mu$.  Indeed,
\begin{align*}
\varphi_{*}\mu - \mu &= \lim_{j \to \infty} N_{j}^{-1} \varphi_{*} \nu_{N_{j}} - \nu_{N_{j}} \\
&= \lim_{j \to \infty} N_{j}^{-1} \sum_{j = 0}^{N_{j} - 1} \left( \varphi_{*}^{j + 1}\nu - \varphi_{*}^{j}\nu \right) \\
&= N_{j}^{-1} ( \varphi_{*}^{N_{j}} \nu - \nu) \\
&= 0.
\end{align*}
This proves $\varphi_{*}\mu = \mu$.  
