Characterising ergodicity of continuous maps - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T03:32:32Z http://mathoverflow.net/feeds/question/122764 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-maps Characterising ergodicity of continuous maps Julian Newman 2013-02-24T00:34:30Z 2013-02-25T13:25:32Z <p>Hello all.</p> <p>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 <em>not</em> ergodic.</p> <p>Does there necessarily exist a Borel set $A \subset X$ such that</p> <ul> <li><p>$\mu(A) \in (0,1)$;</p></li> <li><p>$\mu(A \ \triangle \ T^{-1}(A)) = 0$;</p></li> <li><p>$A$ has non-empty interior?</p></li> </ul> <p>What about if we replace the third point with the stronger requirement that $A$ is open?</p> <p>Many thanks, Julian.</p> http://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-maps/122769#122769 Answer by Ian Morris for Characterising ergodicity of continuous maps Ian Morris 2013-02-24T01:35:38Z 2013-02-24T01:35:38Z <p>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 <em>Ergodic Theory on Compact Spaces</em> by Denker, Grillenberger and Sigmund, or in John Oxtoby's classic 1952 article <em>Ergodic sets</em>.)</p> <p>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.</p>