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Feb 24, 2013 at 16:52 comment added Ian Morris @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.
Feb 24, 2013 at 14:38 comment added Julian Newman @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 $X$?)
Feb 24, 2013 at 12:08 comment added Ian Morris @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".
Feb 24, 2013 at 3:26 comment added Anthony Quas @Julian: This is hopeless. On any reasonable space, there will be transformations that are minimal, but not strictly ergodic.
Feb 24, 2013 at 2:44 comment added Julian Newman 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 $X=\mathbb{R}^n$?)
Feb 24, 2013 at 2:25 vote accept Julian Newman
Feb 24, 2013 at 1:35 history answered Ian Morris CC BY-SA 3.0