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.
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.
