# System with invariant measure, but no ergodic measure.

## 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.]

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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
What's wrong with using the Krein-Milman theorem in the general situation? – Jesse Peterson Oct 3 '11 at 15:49
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

First, I'd like to point out that asymptotic density is an ergodic and $T$-invariant probability measure on the set of integers $\mathbb Z$ with $T(x) = x+1$.
@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
Am I right in saying that no-one has actually answered either Q1 or Q2 yet? I'm particularly interested in the answer to Q1. (In fact, even ignoring a topology, I haven't managed to find anywhere the answer to the following basic question: Let $(X,\Sigma,\mu)$ be a probability space that is not a Lebesgue space, and let $T:X \to X$ be a $\mu$-preserving measurable map; does there necessarily exist a probability measure $\mu'$ on $(X,\Sigma)$ which is $T$-ergodic?) – Julian Newman Mar 28 at 0:49
@JulianNewman: It does not seem to me that Daniel defined his measure over the whole sigma algebra in his first attempt. In his second attempt, there is nothing that ensures the measure $\mu'$ is in fact ergodic. – André Caldas Jul 16 at 12:15
@DanielMansfield: Asymptotic density is not $\sigma$-additive: every singleton has density 0, and yet the union of all singletons (i.e. the whole space) has density 1. You are right that asymptotic density is an invariant finitely additive measure of the map $n \mapsto n+1$, but it is easy to show that this map has no invariant countably additive probability measures. (Indeed, this is an immediate consequence of the Poincaré recurrence theorem.) – Julian Newman Sep 22 at 18:49