why a minimal dynamical system is a ergodic measure-preserving system?

Question

A dynamical system(DS) is a map $(X,T)$ where $X$ is a compact metric space and $T：X-->X$ is a continuous transformation.

A minimal DS means for any point $x$ belongs to $X$, $x$ is a (topological) transitive point, that is to say, the orbit of $x$(donote $orb(x)$) is dense in $X$.

A measure-preserving system $ (X,B(X),m,T)$ is a measure-preserving system,that is to say, m is probability measure, and if $ B\in B(X) $ , then $ m(T^{-1}(B))=m(B)$.

A ergodic measure-preserving system $ (X,B(X),m,T)$ is a measure-preserving system, where $m$ is a ergodic measure,that is to say if $ T^{-1}(B)=B $, then $ m(B)=0 $ or $1$.

My question is in a minimal dynamical system, does there exist a ergodic measure m?

Answer 1

The proof consists of two steps.

1. Let $M$ be the convex set (equipped with the weak *-topology) of Borel probability measures on $X$. The transformation acts on $M$ and the main thing is to show that the fixed-point set $F\subset M$ is non-empty. This is the content of so called Krylov-Bogolyubov theorem and can be seen as a special case of Markov-Kakutani theorem, http://en.wikipedia.org/wiki/Markov-Kakutani_fixed-point_theorem.

See also https://mathoverflow.net/questions/66669/proof-of-krylov-bogoliubov-theorem and https://mathoverflow.net/questions/71683/alternative-proofs-of-the-krylov-bogolioubov-theorem for a collection of alternative arguments.

2. The set $F$ is closed and convex and it is easy to see that ergodicity of the measure is equivalent to the assumption that the measure is an extreme point of $F$. Since each nonempty closed convex set has at least one extreme point (say, by Krein-Millman theorem), existence of an invariant ergodic measure follows.

See the discussion https://mathoverflow.net/questions/15654/extreme-point-compact-convex-set for simple proofs of existence of extreme points.

Note that minimality of the dynamical system is not needed.