Timeline for Generalized density functions on the natural numbers
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 6, 2015 at 5:13 | history | edited | tmh | CC BY-SA 3.0 |
added 128 characters in body
|
| Oct 6, 2015 at 5:12 | comment | added | tmh | Right, sorry, I was misremembering the construction. | |
| Oct 6, 2015 at 0:28 | comment | added | Anthony Quas | I don't think this is the case. I believe the shift-invariant means are produced using the Hahn-Banach theorem: you start off with defining a linear functional on the set of things that do have a limit; and then use Hahn-Banach (plus lots of axiom of choice) to extend it to the rest. To see the means are not subsequential, once you know the subsequence, it's pretty easy to find a set so that you don't get convergence along the subsequence. | |
| Oct 5, 2015 at 23:23 | comment | added | tmh | At least in the case that $\theta$ is a subsequential limit of the functions $\frac{1}{n}\sum_{x\in[0,n]}1(x\in A)$, it is a limit of measurable functions and hence measurable (the same subsequence is used for every $A$). I think that every shift-invariant mean on $\mathbb{N}$ will arise as a similar sort of limit, giving measurability, but I'm not an expert on this. | |
| Oct 5, 2015 at 15:01 | comment | added | Anthony Quas | Don't you need measurability of $\theta$ for $\theta(A^+)$ and $\theta(A^-)$ to be constant? | |
| Oct 5, 2015 at 0:38 | history | answered | tmh | CC BY-SA 3.0 |