Timeline for Borel sigma algebra on measures generated by distance inducing weak convergence and the one generated by weak topology
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2021 at 15:05 | vote | accept | Bremen000 | ||
| Feb 3, 2021 at 12:49 | comment | added | Bremen000 | Actually I can define $\{f_k\}_{k \ge 1} = \{ 1 \} \cup \left (\cup_n \cup_m \cup_j f_{m,n,j} \right )$ where $f_{m,n,j}$ is a smooth function identically one on $B_{1/m}(x_n)$ and identically $0$ outside $B_{1/m +1/j}(x_n)$, where $\{x_n\}_{n \ge 1} = \mathbb{Q}^d$. In this way all the functions are already nonnegative and uniformly bounded. I didn't specify it because I thought it was not important... | |
| Feb 3, 2021 at 12:45 | comment | added | Dieter Kadelka | Since $(f_k)$ seems to be rather arbitrary, is it possible to replace this sequence by $(f_1^+,f_1^-,f_2^+,f_2^-,...)$. Maybe the arguments become a little bit easier. | |
| Feb 3, 2021 at 12:34 | history | edited | Bremen000 | CC BY-SA 4.0 |
added 39 characters in body
|
| Feb 3, 2021 at 12:28 | comment | added | Bremen000 | Yes, I can take it with norm bounded by 1. | |
| Feb 3, 2021 at 12:25 | comment | added | Taras Banakh | By the way, if the sequence $\{f_k\}_{k\in\mathbb N}$ is not carefully chosen, then the series in the definition of the metric $d_p$ can be divergent. Maybe one should impose a condition that $\sup_{k\in\mathbb N}\|f_k\|<\infty$? | |
| Feb 3, 2021 at 10:07 | vote | accept | Bremen000 | ||
| Feb 3, 2021 at 12:57 | |||||
| Feb 3, 2021 at 7:17 | answer | added | Taras Banakh | timeline score: 3 | |
| Feb 2, 2021 at 21:07 | comment | added | Bremen000 | I see... maybe there is some other argument..... | |
| Feb 2, 2021 at 20:51 | comment | added | Michael Greinecker | That only gives you the compactly supported measures. | |
| Feb 2, 2021 at 20:41 | comment | added | Bremen000 | I see...thank you! What if I take $B_n := B \cap \{ |\mu| \le n \} \cap \{ \text{supp}(\mu) \subset \{ |x| \le n\} \}$? Seems too simple... | |
| Feb 2, 2021 at 20:38 | comment | added | Michael Greinecker | It turns out, that space is not metrizable either. According to Proposition 3.1.8. of Bogachev's new book on weak convergence, the space of signed Radon measures of variation at most 1 on a metric space is not metrizable if the metric space is not compact. | |
| Feb 2, 2021 at 20:30 | comment | added | Bremen000 | This is exactly the point where I am not sure.. | |
| Feb 2, 2021 at 20:29 | comment | added | Michael Greinecker | I'm not completely sure. If you would replace $\mathbb{R}^d$ by a compact metric space, the metrizability of $\tau|B_n$ follows from functional analytic machinery. Here, I don't know. | |
| Feb 2, 2021 at 20:16 | comment | added | Bremen000 | Yes, in case I restrict $\tau$ to positive measures this is true. I thought to the following argument but I am not sure: if $B \in \sigma(\tau_d)$ I can consider the sets $B_n := B \cap \{ |\mu| \le n \}$. I think that $\tau |_{B_n} = \tau_d |_{B_n}$ (I am not completely sure) so that $B_n \in \sigma(\tau)$. However $B= \cup_n B_n$, so that $B \in \sigma(\tau)$ and then $\sigma(\tau_d) \subset \sigma(\tau)$. Let me know if it convinces you... | |
| Feb 2, 2021 at 20:10 | comment | added | Michael Greinecker | You are right, I misread what you wrote and thought you were writing about positive measures. | |
| Feb 2, 2021 at 20:09 | comment | added | Bremen000 | Are you sure? I think $\tau$ is not metrisable... | |
| Feb 2, 2021 at 20:01 | comment | added | Michael Greinecker | Yes. Actually, $\tau=\tau_d$. | |
| Feb 2, 2021 at 18:06 | history | asked | Bremen000 | CC BY-SA 4.0 |