Timeline for If a sequence of measures is weakly convergent outside each compact ball, the sequence itself is weakly convergent
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Dec 29, 2020 at 13:24 | vote | accept | 0xbadf00d | ||
| Dec 29, 2020 at 13:18 | comment | added | 0xbadf00d | BTW, I don't see that you've used separability at some point. We only need to assume that $(\lambda_n)_{n\in\mathbb N}$ is bounded in total variation norm and that each $\lambda_n$ is tight (equivalently, Radon) in order to apply Prohorov's theorem. | |
| Dec 29, 2020 at 13:16 | comment | added | 0xbadf00d | Thank you for your answer! Just a minor correction which doesn't break your argument: You can "only" conclude that $S$ is contained in $B_{1/m}^c\color{red}{\cup\{0\}}$ and hence in $K_i'\color{red}{\cup\{0\}}$. Of course, it doesn't matter, since $\{0\}$ is compact as well. | |
| Dec 29, 2020 at 1:39 | history | answered | Nate Eldredge | CC BY-SA 4.0 |