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?

  • $\begingroup$ Isn't an invertible minimal system typically of type ii? Even always? $\endgroup$
    – Ville Salo
    Jun 10, 2021 at 9:56
  • $\begingroup$ If the answer is yes, I don't understand your choice of terminology but the answer to your question is "no". $\endgroup$
    – Ville Salo
    Jun 10, 2021 at 9:58

1 Answer 1


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.

  • $\begingroup$ Ah, very nice - I did miss that any invertible minimal system is already “uniformly transitive” or uniformly minimal might be the better terminology like you say. $\endgroup$
    – Nate River
    Jun 10, 2021 at 12:12

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