# Ergodic decomposition and integral representation of functions that depends on a measure

Let $X$ be a compact metric space, $T:X \to X$ continuous, $M_T(X)$ the set of borel measure that are $T$-invariant and $E_T(X)\subseteq M_T(X)$ the set of ergodic measures. The ergodic decomposition theorem tell us that for every $\mu \in M_T(X)$ exists a probability measure $\tau : \mathcal{B}(M_T(X))\to [0,1]$ such that for every continuos function $f \in C(X)$ $$\int_X f d\mu = \int_{E_T(X)} \left( \int_X f d\nu \right) d\tau(\nu)$$ The latter can be extend to every measurable function $f$.

Let now $F: M_T(X)\to \mathbb{R}$ be a $\mathcal{B}(M_T(X))$-measurable function.

It is true that $F(\mu)= \int_{E_T(X)} F(\nu)d\tau(\nu)$ ?

An interesting case is when $F(\nu)=h_\nu(T)$ but I would like to know the general case as well.

Thanks!

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The answer to your main question is negative, take $F:=\chi_{ E_T(X)}$.
This question is closely connected with the metrisable case of Choquet's theorem, which states: if $K$ is a metrisable compact convex subset of a locally convex topological space $V$, and $E$ is the set of extreme points of $K$, then for every $v \in K$ there exists a Borel probability measure $\mu$ on $K$ such that $\mu(E)=1$ and such that for every continuous affine function $f \colon K \to \mathbb{R}$ we have $f(v)=\int f\,d\mu$. To obtain an ergodic decomposition theorem we define $V:=C(X)^*$ with the weak-* topology and $K:=M_T(X)$, noting that the set of extreme points on $K$ is precisely $E_T(X)$ (see e.g. Theorem 6.1 in Walters' An Introduction to Ergodic Theory).
Yes, $X$ need to be compact, I'll edit. Thanks for your answer! –  TV2323 May 22 '14 at 21:36
But Choquet's theorem needs $f$ to be continuous and affine and the entropy map is not necessarily continuos. Maybe every affine function can be approximated by affine continuous function but I really don't know convex analysis. –  TV2323 May 23 '14 at 19:13