Timeline for Weak*-convergence of signed measures
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2018 at 9:08 | vote | accept | user3522356 | ||
| Oct 22, 2018 at 21:39 | comment | added | Nate Eldredge | In fact, it seems to me that it might be possible to adapt Darst's "gliding hump" argument to prove this statement. But it may take some thought. | |
| Oct 22, 2018 at 4:09 | comment | added | Nate Eldredge | This question has connections to an old question of mine. In particular, if the condition here had "Borel sets" instead of "open sets" then the claim would be true by the result of Darst cited in my answer (there might be an earlier result of Nikodym that suffices in this case). And if the answer to this question is yes, then it will give another counterexample to my Q2. | |
| Oct 21, 2018 at 19:24 | vote | accept | user3522356 | ||
| Nov 1, 2018 at 9:08 | |||||
| Oct 21, 2018 at 19:10 | answer | added | Dirk Werner | timeline score: 2 | |
| Oct 21, 2018 at 18:33 | comment | added | Nate Eldredge | To illustrate why this is tricky, the following non-example is interesting. Let $X = \mathbb{N} \cup \{\infty\}$, and let $\nu_n = n(\delta_{n+1} - \delta_n)$ with $\nu = 0$. This sequence doesn't converge weakly. It doesn't satisfy the hypothesis either - take $O$ to be the set of odd integers - but it is not so clear how to "produce" that set. | |
| Oct 21, 2018 at 18:14 | comment | added | Christian Remling | This is essentially the same as asking if this condition implies that $\nu_n$ is bounded, that is, $\sup |\nu_n|(X)<\infty$, since the claim is almost trivial in that case ("see" my failed attempt at an answer), and obviously this boundedness is necessary for weak $*$ convergence. | |
| Oct 21, 2018 at 16:35 | comment | added | André Porto | So, my answer lacks some arguments, so I deleted it. It is true for positive measures, though. But I don't know for signed measures | |
| Oct 21, 2018 at 15:34 | comment | added | user3522356 | @AndréPorto, mustn't we have $\sup_O |\nu_n(O) - \nu(O)|\to 0$ for strong convergence? | |
| Oct 21, 2018 at 15:33 | comment | added | André Porto | By the outter regularity, this condition implies that $\nu_n$ is strongly convergent to $\nu$, doesn't it? | |
| Oct 21, 2018 at 15:28 | history | edited | user3522356 | CC BY-SA 4.0 |
added 2 characters in body
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| Oct 21, 2018 at 15:10 | review | Close votes | |||
| Oct 27, 2018 at 3:05 | |||||
| Oct 21, 2018 at 12:20 | history | asked | user3522356 | CC BY-SA 4.0 |