Timeline for Continuity of the Lebesgue measure w.r.t the Hausdorff metric
Current License: CC BY-SA 4.0
14 events
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| Jun 25, 2020 at 21:33 | comment | added | Redeldio | @PierrePC I posted the question both here and in math.SE. You're right, the answers in math.SE are rather detailed but they were posted after my question on mathoverflow. Anyway, your comments were useful for me! Thank you! | |
| Jun 25, 2020 at 15:11 | comment | added | Pierre PC | @VilleSalo Yes, precisely the interval $[0,\lambda(K)]$. I must say though, the answers on math.SE are rather detailed already (I just checked it), so I'm not sure what the OP has in mind. | |
| Jun 25, 2020 at 14:56 | comment | added | Ville Salo | @PierrePC and is it actually an interval? | |
| Jun 25, 2020 at 14:54 | comment | added | Pierre PC | Conjecture (that I think is not difficult to prove with the above comments): for all neighbourhood $U$ for some fixed $K$, the set $\lambda_*(U)=\lbrace \lambda(K'),\ K'\in U\rbrace$ contains $[0,\lambda(K)]$; more precisely, the intersection of all $\lambda_*(U)$ where $U$ ranges over all neighbourhoods of $K$ is precisely $[0,\lambda(K)]$. I think I can write that down if it is of interest. | |
| Jun 25, 2020 at 14:49 | comment | added | Ville Salo | Proof by Socratic method | |
| Jun 25, 2020 at 14:46 | history | edited | Martin Sleziak |
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| Jun 25, 2020 at 14:46 | comment | added | Pierre PC | If $K_k$ converges to $K$, then for any $\varepsilon>0$, $K_k$ is eventually included in $K+B(0,\varepsilon)$. Since $\lambda(K+B(0,\varepsilon))\to0$ as $\varepsilon\to0$ (Lebesgue dominated convergence), the Lebesgue measure is semicontinuous (I want to say upper?). | |
| Jun 25, 2020 at 14:42 | comment | added | Ville Salo | $K$ has to be zero measure since finite sets are dense. Is there a zero measure compact that is a limit of $>\epsilon$ measure compacts? | |
| Jun 25, 2020 at 14:30 | history | edited | LSpice | CC BY-SA 4.0 |
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| Jun 25, 2020 at 14:22 | history | edited | Redeldio | CC BY-SA 4.0 |
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| Jun 25, 2020 at 14:21 | comment | added | Redeldio | @erz I don't know, maybe yes. But on the set $K$ too. I was asking if there are special cases in which the condition $(\star)$ holds. | |
| Jun 25, 2020 at 11:31 | comment | added | erz | what do you want to put conditions on? the sequence, right? | |
| Jun 25, 2020 at 10:57 | review | First posts | |||
| Jun 25, 2020 at 12:42 | |||||
| Jun 25, 2020 at 10:51 | history | asked | Redeldio | CC BY-SA 4.0 |