Timeline for Are arbitrary collections of ergodic measures "strongly mututally singular"?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2022 at 15:29 | comment | added | Ronnie Pavlov | Is this an answer or an intuition, i.e. do you have a counterexample for a non-Lebesgue space? I've been unable to construct one. | |
| Nov 30, 2022 at 9:06 | comment | added | coudy | It should not hold without a separability assumption. The ergodic decomposition needs it. Ergodic decomposition holds on Lebesgue spaces, which are spaces isomorphic to an interval together with a countable set of atoms. The proof makes use of the fact that the sigma-algebra is countably generated mod 0 in a crucial way. | |
| Nov 6, 2022 at 0:26 | answer | added | Ronnie Pavlov | timeline score: 1 | |
| Nov 5, 2022 at 17:24 | comment | added | Anthony Quas | Oops! I guess that's what happens if I think I remember "facts" rather than proofs. | |
| Nov 5, 2022 at 14:11 | comment | added | Ronnie Pavlov | Followup since I ran out of room: I guess (from Googling) that $[0, \omega_1]$ with the order topology is compact, Hausdorff, but not 2nd countable. I think $[0,1]^I$ for uncountable $I$ and the product topology is another one, but didn't check details. | |
| Nov 5, 2022 at 14:05 | comment | added | Ronnie Pavlov | I won't be surprised if it is something basic like this; I'm bad at remembering these basic topology things since I always work in nice spaces. 1. Of course I think you're right about Lebesgue; I didn't consider the fact that this implicitly restricts the $\sigma$-algebra as well! (It has a countable generating set) 2. That being said, I'm still curious if this fails in the non-Lebesgue case. I guess you need compact, Hausdorff and second countable for $C(X)$ separable, right? So maybe the question is: if $X$ is not necessarily 2nd ctable, does strong mutual singularity still hold? | |
| Nov 5, 2022 at 7:45 | comment | added | Anthony Quas | If $C(X)$ is separable, say then you can take $(g_k)$ to be a dense subset and set $f(\mu)=\{x\colon \frac 1N\sum_{n=0}^{N-1}g_k(T^nx)\to\int\g\,d\mu\}$. But this is essentially repeating what Ian Morris said. What do you mean by a topological dynamical system? Are you just assuming that $X$ is compact and $T$ is continuous? or is there a Hausdorff assumption also? (in that case $C(X)$ is separable). BTW: isn't there a standard assumption that $(X,\mu)$ is a Lebesgue space for ergodic decomposition? | |
| Nov 5, 2022 at 2:24 | history | asked | Ronnie Pavlov | CC BY-SA 4.0 |