Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question

1 Answer 1

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).

The entropy map is affine (Theorem 8.1 in Walters) but in general it is not continuous, so I am not sure whether Choquet's theorem is sufficient to obtain the desired result in the case of entropy.

share|improve this answer
    
Yes, $X$ need to be compact, I'll edit. Thanks for your answer! –  TV2323 May 22 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 at 19:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.