Timeline for System with invariant measure, but no ergodic measure.
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 22, 2015 at 18:49 | comment | added | Julian Newman | @DanielMansfield: Asymptotic density is not $\sigma$-additive: every singleton has density 0, and yet the union of all singletons (i.e. the whole space) has density 1. You are right that asymptotic density is an invariant finitely additive measure of the map $n \mapsto n+1$, but it is easy to show that this map has no invariant countably additive probability measures. (Indeed, this is an immediate consequence of the Poincaré recurrence theorem.) | |
| Jul 19, 2015 at 3:13 | history | edited | Daniel Mansfield | CC BY-SA 3.0 |
removal of errors
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| Jul 16, 2015 at 12:15 | comment | added | André Caldas | @JulianNewman: It does not seem to me that Daniel defined his measure over the whole sigma algebra in his first attempt. In his second attempt, there is nothing that ensures the measure $\mu'$ is in fact ergodic. | |
| Jul 12, 2015 at 22:51 | comment | added | Julian Newman | @DanielMansfield: Sorry, I did not see your replies until just now. Thank you for this. Your answer is quite perplexing to me - are you sure it is correct? (And are you perhaps working with a non-standard definition of "conservative"?) | |
| Mar 28, 2015 at 0:49 | comment | added | Julian Newman | Am I right in saying that no-one has actually answered either Q1 or Q2 yet? I'm particularly interested in the answer to Q1. (In fact, even ignoring a topology, I haven't managed to find anywhere the answer to the following basic question: Let $(X,\Sigma,\mu)$ be a probability space that is not a Lebesgue space, and let $T:X \to X$ be a $\mu$-preserving measurable map; does there necessarily exist a probability measure $\mu'$ on $(X,\Sigma)$ which is $T$-ergodic?) | |
| Oct 4, 2011 at 1:54 | comment | added | André Caldas | @Daniel: I will correct the post to emphasize that the measure is over the Borel sets and the transformation is continuous. If you are free to choose the $\sigma$-algebra, then you can just take $\{\emptyset, X\}$. | |
| Oct 4, 2011 at 1:19 | history | edited | Daniel Mansfield | CC BY-SA 3.0 |
fixed up non-periodic case
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| Oct 4, 2011 at 0:53 | history | answered | Daniel Mansfield | CC BY-SA 3.0 |