Timeline for Is the space of Radon measures a Polish space or at least separable?
Current License: CC BY-SA 4.0
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| Sep 10, 2019 at 10:36 | comment | added | Mark | Reminder of Lusin's theorem was nice. Especially that the classical statement has property that looks similar to tightness which I am trying to prove. By the end of the week I will know more about my problem and I'll maybe sent you additional question. Thanks again. | |
| Sep 10, 2019 at 9:50 | comment | added | Michael Greinecker | Yes, the Borel $\sigma$-algebra in Euclidean spaces (or any separable metric space) is countably generated. In a bounded domain, one can do some nice approximations that show you can basically use all bounded measurable functions in the definition f your norm and that gives you the variation norm; basically one can use Lusin's theorem to show it is enough to use continuous functions and then Stone-Weierstrass to go to smooth functions with compact support. This might fail for general domains. I simply don't know, those parts are outside my knowledge. | |
| Sep 10, 2019 at 9:43 | comment | added | Mark | Thanks for the follow up. I am not sure if my measurable space is countably generated but I guess I could add it as a request. It is interesting that you mentioned variational norm. Why do I need a variation norm? This is connected to the fact that in my SPDE problem I have a solution that is in BV space - and that space is not separable so I can't use it for the Prokhorov's theorem. But I know that BV functions are functions whose distributional derivative is finite Radon measure. I just don't know how to use that. | |
| Sep 10, 2019 at 9:25 | comment | added | Michael Greinecker | I'm just not sure that this will give you the variation norm in general. If it does, then it still holds that a subspace is separable if and only if it is a subspace of $L_1(\tau)$ for an appropriate measure $\tau$. All that is needed that the underlying measurable space is countably generated. | |
| Sep 10, 2019 at 8:21 | comment | added | Mark | Thank you for your answer. Nice simple example in the first paragraph that explains the problem of separability. Could you elaborate more your second paragraph? $S$ is subspace with some countable dense subset of measures... Knowing that sounds useful. Also I should be using a bounded domain, but just for curiosity, what would be additional complications in the unbounded $\mathbb{R}^d$ case? | |
| Sep 9, 2019 at 17:36 | history | edited | Michael Greinecker | CC BY-SA 4.0 |
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| Sep 9, 2019 at 17:10 | history | answered | Michael Greinecker | CC BY-SA 4.0 |