Let $X$ be a compact metric space, and $\mu$ a probability measure on $X$ with $\text{supp} \ \mu = X$. Suppose $T: X \to X$ is continuous, measure preserving and *uniformly transitive* in the sense that it satisfies the following two conditions:


i) The orbit of every point is dense in $X$.


ii) For every $\varepsilon > 0$, there exists some $\delta > 0$ such that for every $x \in X$, and $r > 0$ with $\mu(B\_r (x)) > \varepsilon$, we have $T(B\_s (T^{-1} (x)) \subset B\_r (x)$ for all $ s > 0$ such that $\mu(B\_s (T^{-1} (x)) < \delta$.



> 
> Question: Does it follow that $T$ is ergodic?
> 
> 
> 



Is a “uniformly minimal” dynamical system ergodic?
The answer is no, I think any of the usual examples works. Some argument below.



> 
> Lemma. Suppose $X$ is a compact metric space, $\mu$ a nonatomic probability measure on $X$, and $T : X \to X$ is a minimal measure-preserving homeomorphism. Then $(X, \mu, T)$ is "uniformly transitive" (did you mean to write "uniformly minimal" as in the title?)
> 
> 
> 


Proof. i) The orbit of every point is dense, that's the definition of minimality.


ii) Let $S\_r$ be the set of measures of all balls $B\_r(x)$ as $x$ ranges over $X$. Clearly $S\_r$ does not contain $0$ for any $r > 0$: apply compactness to the cover by $B\_{r/2}(x)$'s and you get that for any $x$, every point in $X$ enters $B\_r(x)$ after some uniform time $N$ meaning the measure of $B\_r(x)$ is at least $1/N$. In fact this means even the closure of $S\_r$ does not contain $0$. On the other hand, $\sup S\_r$ tends to zero as $r \rightarrow 0$ by nonatomicity.


The fact $0$ is not in the closure means $\mu(B\_r(x)) > \epsilon$ puts some lower bound on $r$.
On the other hand by uniform continuity of $T^{-1}$, $\mu(B\_s(T^{-1}(x)) < \delta$ puts a uniform upper bound on $s$.
By uniform continuity of $T$, there is then a uniform upper bound on the diameter of $T(B\_s(T^{−1}(x))$. This set
contains $x$, so we have $T(B\_s(T^{−1}(x)) \subset B\_r(x)$ for small enough $s$, meaning any small enough $\delta$ works. Square.


Then we just recall the classical result.



> 
> Theorem. There is a compact metric space $X$, a nonatomic probability measure $\mu$ on $X$, and a minimal measure-preserving homeomorphism $T : X \to X$, such that $(X, \mu, T)$ is not ergodic.
> 
> 
> 


Proof. It's enough to find a system $(X, T)$ with two different invariant measures. Take words $w\_0^0 = 0$ and $w\_1^0 = 1$, and construct $w^{n+1}\_0$ and $w^{n+1}\_1$ inductively from $w^n\_0$ and $w^n\_1$ by concatenating them around. Do this inductively so that the frequency of $a$-symbols in $w^{n+1}\_a$ is strictly more than $2/3$, and take at least $3$ copies of both $w^n\_0$ and $w^n\_1$ when constructing $w^n\_a$ for $a \in \{0,1\}$. Let $X \subset \{0,1\}^{\mathbb{Z}}$ be the subshift (topologically closed set invariant under the left shift action $T(x)\_i = x\_{i+1}$) consisting of all bi-infinite words whose finite subwords all appear in the words $w^n\_a$. By a compactness argument $X$ is nonempty, and $X$ is minimal by the following standard argument: if a word $w$ appears in some point of $X$, it appears in some $w^n\_a$, thus it appears in $w^{n+1}\_0$ and $w^{n+1}\_1$, and every point in $X$ is a concatenation of these, so $w$ appears with bounded gaps in every point.


Now, for a fixed $n$ construct two measures $\mu^n\_0, \mu^n\_1$ on $X$ as follows: for $\mu^n\_a$ take any configuration $x$ in the cylinder $[w^n\_a]$ (any word containing $w^n\_a$ at the origin), and if $w^n\_a$ has length $k$ define $\mu^n\_a = \sum\_{i = 0}^{k-1} T^i(\delta\_x)/k$ where $\delta\_x$ is the atomic measure on $\{x\}$. Clearly $\mu^n\_a([a]) > 2/3$ and $\mu^n\_a([1-a]) < 1/3$. Any weak-\* limit point of $\mu^n\_a$ is $T$-invariant by a standard computation, and by the definition of the topology we get a measure $\mu\_a$ with $\mu\_a([a]) \geq 2/3$ and $\mu\_a([1-a]) \leq 1/3$. So we have two different invariant measures on $X$. Square.


