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.

  • $\begingroup$ Does $T$ preserve the measure? $\endgroup$ – Joel Moreira Feb 24 '13 at 1:07
  • $\begingroup$ Good point! Let's assume it does. (I'll now edit the question accordingly.) $\endgroup$ – Julian Newman Feb 24 '13 at 1:36

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.

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    $\begingroup$ Thank you. This is most helpful. Do you know any conditions on $X$ under which, if $\mu$ is a strictly positive probability measure on $X$, then every minimal $\mu$-preserving continuous transformation is ergodic? (E.g. is this true for Euclidean space $X=\mathbb{R}^n$?) $\endgroup$ – Julian Newman Feb 24 '13 at 2:44
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    $\begingroup$ @Julian: This is hopeless. On any reasonable space, there will be transformations that are minimal, but not strictly ergodic. $\endgroup$ – Anthony Quas Feb 24 '13 at 3:26
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    $\begingroup$ @Julian: this is equivalent to asking for a condition on $X$ such that every minimal transformation on $X$ is uniquely ergodic, i.e. has only one invariant measure. (If a transformation has two distinct invariant measures then a strict linear combination of the two is never ergodic.) Such conditions do exist: finite spaces $X$ have this property, as does the circle (I think) but as Anthony says this is a severly restrictive requirement. The broader stroke of your question seems to be whether ergodicity can be easily characterised using only topological concepts. The answer to this is "No". $\endgroup$ – Ian Morris Feb 24 '13 at 12:08
  • $\begingroup$ @Ian and Anthony: Just to be clear, I did not say that I require every invariant probability measure of a minimal transformation to be ergodic - I just required that every strictly positive invariant probability measure of a minimal transformation had to be ergodic. (By strictly positive, I mean that its support is the whole of $X$). Is this still equivalent to requiring that every minimal transformation is uniquely ergodic? (And in the case $X=\mathbb{R}^n$, if the requirement still is not satisfied, what about if we weaken the requirement by restricting to, say, diffeomorphisms on $X$?) $\endgroup$ – Julian Newman Feb 24 '13 at 14:38
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    $\begingroup$ @Julian: Every invariant probability measure of a minimal transformation is fully supported, because otherwise its support would be a nonempty closed invariant proper subset, contradicting minimality. So the two statements are equivalent. $\endgroup$ – Ian Morris Feb 24 '13 at 16:52

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