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

Question

  1. Examples of continuous transformations $T: X \to X$ such that the family of invariant probability measures $M(T)$ is NOT empty but there is no ergodic measure ($E(T) = \emptyset$).
    Notice that the measures considered are defined over the Borel sets of $X$.

  2. Example of a dynamical system where the following inequality is strict: $$\sup_{m \in E(T)} h_m(T) < \sup_{\mu \in M(T)} h_\mu(T)$$.

Background

Consider $T(x) = x + 1$ over the set of integers $\mathbb{Z}$. In this case, $E(T) = M(T) = \emptyset$. The first question asks for a $\emptyset = E(T) \subsetneq M(T)$ example.

In the locally-compact metrizable case, the set of positive invariant measures $\mu$ with $0 \leq \mu(X) \leq 1$ is compact (weak* topology) with extremals with total measures equal to $0$ or $1$. That is, according to Krein-Milman Theorem, if $M(T) \neq \emptyset$, then $E(T) \neq \emptyset$. So, an answer to Question 1 is not supposed to be locally-compact metrizable.

[Edit: The question only makes sense if the $\sigma$-algebra is fixed. So the post was edited, making $X$ a topological space, $T$ continuous and the $\sigma$-algebra is the family of Borel sets.]

share|improve this question
1  
This post is related to mathoverflow.net/questions/76908/… –  André Caldas Oct 3 '11 at 11:44
    
Are you interested in finite measures or infinite measures? What is the notion of entropy you are referring to in the infinite case? Anyways, usually at any question in ergodic theory (especially in entropy theory), one usually deals with "standard" probability spaces (and maybe even Lebesgue spaces). –  Asaf Oct 3 '11 at 15:13
    
@Asaf, I am interested on probability measures. The entropy is the Kolmogorov-Sinai entropy. –  André Caldas Oct 3 '11 at 15:46
1  
What's wrong with using the Krein-Milman theorem in the general situation? –  Jesse Peterson Oct 3 '11 at 15:49
3  
I don't understand the problem here. Metrizability of $X$ doesn't enter in any of your arguments as long as you have local compactness. By its definition the set of positive measures is a weak$^∗$-closed cone in $M(X)$, and thus it cuts out a compact set out of the unit ball, so as soon as you have invariant measures you have invariant ergodic measures by Krein-Milman. –  Theo Buehler Oct 4 '11 at 2:26

1 Answer 1

To answer question 1, I think there will always be a measure for which $T$ is ergodic.

Rotation of the unit circle $T(z) = az$ is measure preserving, (with Haar measure) and ergodic only when $a$ is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8).

However, we can still define a measure for which $T$ is ergodic. Say $\mu(\{a^k\}) = 1/n$ where $a$ an $n$-th root of unity, $k=0,\ldots, n-1$. Then the $T$-invariant sets either have measure 0 or 1.

When there are periodic points we can use the above argument to create an ergodic measure.

The argument is not very different in the absence of periodic points. For any measure preserving system $(X,\mathcal B, \nu)$ take any element $a \in X$ and define $orb_T(a) = \{T^na: n \in \mathbb Z\}$. Let $\mu$ be a probability measure on $\mathcal B / orb_T(a)$ with $\mu(0 + orb_T(a)) = 1$ and zero otherwise. Then $T$ is ergodic on $(X, \mathcal B/ orb_T(a), \mu)$.

share|improve this answer
    
So in the example of $Tx = x+1$ on $\mathbb R$ with measurable sets $\mathit B$. You can actually define a invariant measure and an ergodic measure. Say we let measure $\mu$ be 1 on the integers; 0 elsewhere. Then $(X, \mathit B / \mathbb N, \mu, T)$ is ergodic. The catch here is that subsets of the integers are not measurable. –  Daniel Mansfield Oct 4 '11 at 1:34
    
@Daniel: I will correct the post to emphasize that the measure is over the Borel sets and the transformation is continuous. If you are free to choose the $\sigma$-algebra, then you can just take $\{\emptyset, X\}$. –  André Caldas Oct 4 '11 at 1:54

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.