Characterising ergodicity of continuous maps - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T03:32:32Zhttp://mathoverflow.net/feeds/question/122764http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/122764/characterising-ergodicity-of-continuous-mapsCharacterising ergodicity of continuous mapsJulian Newman2013-02-24T00:34:30Z2013-02-25T13:25:32Z
<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#122769Answer by Ian Morris for Characterising ergodicity of continuous mapsIan Morris2013-02-24T01:35:38Z2013-02-24T01:35:38Z<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>