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
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2021 at 15:28 | comment | added | Bremen000 | Do you have any reference for the non sigma compactness of $\mathcal{M}$? It's a shame, the argument with the $\mathcal{M}_{f,n}$ seemed to be correct... | |
| Feb 3, 2021 at 15:26 | comment | added | Taras Banakh | @Bremen000 The problem with such sets like $\mathcal M_n$ is that the space $\mathcal M$ does not seem to be $\sigma$-compact, so the problem cannot be reduced to the compact case. | |
| Feb 3, 2021 at 15:05 | vote | accept | Bremen000 | ||
| Feb 3, 2021 at 14:57 | comment | added | Bremen000 | Yes, thank you. I found Theorem 6.8.6 in "Measure Theory - Volume 2" by Bogachev, this should work. Finally I think that my argument with $\mathcal{M}_n$ above is not correct. A possible correct one could be the following: for every $f: \mathbb{N} \to \mathbb{N}$ and every $n \in \mathbb{N}$ I can consider the set $\mathcal{M}_{f,n} := \cap_k\{ |\mu| \le n, \mu ( \{x > f(k)\}) < 1/k ) \}$. Then $\overline{\mathcal{M}_n}$ is compact in $\mathcal{M}$ hence metrisable. It should also be that $\mathcal{M} = \cup_{f,n}\mathcal{M}_{f,n}$ so the argument with $\tau | \mathcal{M}_{f,n}$ should work. | |
| Feb 3, 2021 at 14:37 | comment | added | Taras Banakh | @Bremen000 The space $\mathbb R^\omega$ is necessary if we would like to apply the classical form of Lusin-Souslin Theorem, which is formulated in Kechris for Polish spaces. But of course, there are generalizations of Lusin-Souslin Theorem applicable to injective maps with values in functionally Hausdorff Souslin spaces. But in the latter case one should find a proper reference. Instead of finding such a reference I just made this short reduction via $\mathbb R^\omega$. | |
| Feb 3, 2021 at 14:33 | comment | added | Bremen000 | Thank you very much for the edit. Just a last question: why do you have to introduce the space $\mathbb{R}^{\omega}$? Is it not enough to work with the Souslin space $\mathcal{M}$? | |
| Feb 3, 2021 at 14:31 | comment | added | Taras Banakh | @Bremen000 I have adjusted the argument. Now it should work for your metric (with uniformly bounded sequence $(f_k)_{k\in\omega}$). | |
| Feb 3, 2021 at 14:27 | history | edited | Taras Banakh | CC BY-SA 4.0 |
adjusted the argument
|
| Feb 3, 2021 at 14:15 | comment | added | Bremen000 | Yes, see the comments to the question, the $f_k$ are uniformly bounded. | |
| Feb 3, 2021 at 12:45 | comment | added | Bremen000 | Moreover, may the argument of the answer with $U \subset \mathcal{M}$ open in $\tau_d$ work? I should be able to prove in the same way that $U \in \sigma(\tau)$... | |
| Feb 3, 2021 at 12:42 | comment | added | Bremen000 | No, $d$ does not generate the coarsest topology for which the maps $\mu \mapsto \int f_k d \mu$ are continuous. This last topology indeed coincides with $\tau$, I think. Then it is not metrisable. Unfortunately I am working on some other people work and I cannot change the metric $d$. Maybe a possible choice for the $\mathcal{M}_{n}$ is the following: $\mathcal{M}_{n} = \text{cl} \left (\cap_m \cup_k \{ |\mu| \le n , \mu(\{x > m\}) \le 1/k \} \right )$. These should be compact subsets which union is $\mathcal{M}$ and $\tau | \mathcal{M}_n$ should me metrisable. What do you think? | |
| Feb 3, 2021 at 12:28 | comment | added | Taras Banakh | A substantial amount of energy should be spent for struggle with this unnaturally chosen metric $d_p$. As I understand that had to be a metric generating the weakest topology in which all integrals $\int f_kd\mu$ are continuous? So, maybe just redefine this metric and then everything will be a bit easier? | |
| Feb 3, 2021 at 12:19 | comment | added | Taras Banakh | Then the argument I have written in my answer should work, maybe modulo decomposition of $\mathcal M$ into the union $\bigcup_{n\in\omega}\mathcal M_n$ in order to guarantee that the metric $\tau_d$ restricted to $\mathcal M_n$ generates a weaker topology than $\tau$. | |
| Feb 3, 2021 at 12:17 | comment | added | Taras Banakh | It seems that you are right: the weak topology on $\mathcal M_n$ is non-metrizable (because of non-compactness of $\mathbb R^d$). | |
| Feb 3, 2021 at 12:17 | comment | added | Bremen000 | It is not compact in the $\tau$ topology, since it is not tight. Notice that $\mathcal{M}$ is not the dual of $C_b(\mathcal{R}^d)$. Michael Greinecker pointed this out too in the above comments. | |
| Feb 3, 2021 at 12:06 | comment | added | Bremen000 | Even if it seems to suggest that on $\mathcal{M}_n$ the two topologies coincide, which is impossibile, since $\tau$ is not metrisable even when restricted to $\mathcal{M}_n$. | |
| Feb 3, 2021 at 12:04 | comment | added | Bremen000 | Oh nice! I will think about it! | |
| Feb 3, 2021 at 11:59 | comment | added | Taras Banakh | But in the Borel sense this does not change anything. Just decompose $\mathcal M$ into the countable union $\bigcup_{n\in\omega}\mathcal M_n$ where $\mathcal M_n=\{\mu\in\mathcal M:\|\mu\|\le n\}$. On the sets $\mathcal M_n$ the metric $d_p$ does generate the weakest topology generated by the integrals $\int f_nd \mu$. So, on these sets $\tau_d\subseteq\tau$ and the converse follows from the metrizability of $\tau_d$. So, the topologies $\tau$ and $\tau_d$ generate the same Borel $\sigma$-algebras on the closed sets $\mathcal M_n$ and because of that the Borel $\sigma$-algebras are the same. | |
| Feb 3, 2021 at 11:57 | comment | added | Bremen000 | Maybe the same argument with $U$ open in $\tau_d$ may work... | |
| Feb 3, 2021 at 11:55 | comment | added | Bremen000 | Oh thanks god: I started questioning everything! | |
| Feb 3, 2021 at 11:54 | comment | added | Taras Banakh | Ups! I have just understood that your metric $d_p$ does not generate the weakest topology in which all integrals $\int f_kd\mu$ are continuous. So, this changes the situation. Let me think a bit more. | |
| Feb 3, 2021 at 11:53 | comment | added | Bremen000 | I’m sorry, I am confused: if $E$ is $\tau$-closed, then $E$ is sequentially $\tau$-closed. Then it is sequentially $\tau_d$-closed i.e. it is $\tau_d$-closed hence $\tau\subset \tau_d$. Where is my mistake? | |
| Feb 3, 2021 at 11:47 | comment | added | Taras Banakh | The obvious direction is $\sigma(\tau_d)\subseteq \sigma(\tau)$ because the metrizable topology is contained in the weak topology. What I have written is the non-obvious direction $\sigma(\tau)\subseteq \sigma(\tau_d)$. | |
| Feb 3, 2021 at 11:13 | comment | added | Bremen000 | The argument above allows me to conclude that $\sigma(\tau) \subset \sigma(\tau_d)$. Is it not already obvious from $\tau \subset \tau_d$? I need the opposite inclusion... | |
| Feb 3, 2021 at 10:42 | history | edited | Taras Banakh | CC BY-SA 4.0 |
added 3 characters in body
|
| Feb 3, 2021 at 10:07 | vote | accept | Bremen000 | ||
| Feb 3, 2021 at 12:57 | |||||
| Feb 3, 2021 at 7:17 | history | answered | Taras Banakh | CC BY-SA 4.0 |