Hello all.


Suppose $X$ is a Polish space, $\mu$ is a Borel probability measure on $X$, and $T:X \to X$ is a continuous $\mu$-preserving map which is *not* ergodic.


Does there necessarily exist a Borel set $A \subset X$ such that


* $\mu(A) \in (0,1)$;
* $\mu(A \ \triangle \ T^{-1}(A)) = 0$;
* $A$ has non-empty interior?


What about if we replace the third point with the stronger requirement that $A$ is open?


Many thanks,
Julian.



Characterising ergodicity of continuous maps
Let $T \colon X \to X$ be a minimal transformation of a compact metric space which is not uniquely ergodic, let $\mu$ be a non-ergodic $T$-invariant measure on $X$, and let $A$ be a set with nonempty interior such that $\mu(A \triangle T^{-1}A)=0$. I claim that necessarily $\mu(A)=1$, contradicting the above conjecture. (Some constructions of transformations with the above combination of properties may be found for example in the textbook *Ergodic Theory on Compact Spaces* by Denker, Grillenberger and Sigmund, or in John Oxtoby's classic 1952 article *Ergodic sets*.)


Let $U \subseteq A$ be open and nonempty. Since $T$ is minimal we have $\bigcup\_{n=0}^\infty T^{-n}U=X$, and indeed even $\bigcup\_{n=0}^NT^{-n}U=X$ for some integer $N$ since $X$ is compact. In particular $\bigcup\_{n=0}^N T^{-n}A=X$. Let us write
$$\bigcup\_{n=0}^N T^{-n}A = A \cup \bigcup\_{n=1}^N \left(\left( T^{-n}A\right)\setminus \bigcup\_{k=0}^{n-1} T^{-k}A\right)=A \cup \bigcup\_{n=1}^N B\_n,$$
say, which is a disjoint union. We would like to show that this union has measure identical to that of $A$. For each $n$ we have
$$\mu(B\_n)=\mu\left(T^{-n}A\setminus \bigcup\_{k=0}^{n-1} T^{-k}A\right)\leq \mu\left(T^{-n}A \setminus T^{-(n-1)}A\right)=\mu\left(T^{-1}A \setminus A\right)=0$$ by invariance and the hypothesis $\mu(A \triangle T^{-1}A)=0$. It follows that
$$\mu(A)=\mu\left(\bigcup\_{n=0}^N T^{-n}A \right)=\mu(X)=1$$
so the desired situation can not occur.


