Timeline for Approximation of the radon-derivative
Current License: CC BY-SA 3.0
11 events
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| Nov 18, 2018 at 11:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 19, 2018 at 11:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 19, 2018 at 10:40 | answer | added | pietro siorpaes | timeline score: 1 | |
| Jan 12, 2012 at 9:24 | history | edited | Pietro Majer | CC BY-SA 3.0 |
TeX etc
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| Jan 11, 2012 at 16:11 | comment | added | Yulia Kuznetsova | @fedja: thank you, I forgot much of it since using. @Klaus: I remember reading a good review close to this topic: A. Bruckner, “Differentiation of integrals,” Amer. Math. Monthly 78 (9, part II) (1971). | |
| Jan 11, 2012 at 15:44 | comment | added | Klaus | @yulia: many thank's, embarassingly I did not know the notion of Lebesgue points. As X is also metric and one measure is the Hausdorff measure there might be some theorem that almost every point is lebesgue. @fedja: Many thank's for correcting me, my question is missleading. Actually, I am looking for a criterion for the shape of $U_i$. | |
| Jan 11, 2012 at 15:22 | comment | added | fedja | @Yulia: By Luzin's theorem, the restriction of $f$ to some set $E$ of almost full measure is continuous or, if you prefer, $f$ equals to some other continuous function $g$ outside a set of small measure. This is very different from the continuity of $f$ itself. The theorem on Lebesgue points also requires very particular shapes of $U_i$ to employ covering lemmas. @Klaus The desired property is currently stated so sloppily that it doesn't hold even at points of continuity: take $X=\mathbb R$, $x=0$, and $U_i=(-1/i,1/i)\cup (i,+\infty)$. | |
| Jan 11, 2012 at 14:03 | comment | added | Yulia Kuznetsova | Try to search for material on Lebesgue points. If $\nu$ is the Lebesgue measure in $\mathbb R^n$, then almost every point is a Lebesgue point and so has your property. | |
| Jan 11, 2012 at 13:32 | comment | added | Yulia Kuznetsova | By Luzin's theorem, $f$ is continuous except for a set of arbitrarily small measure; if $f$ is continuous at $x$ then your limit equals $f(x)$. This has to be made more accurate though. | |
| Jan 11, 2012 at 13:19 | history | edited | Klaus |
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| Jan 11, 2012 at 10:50 | history | asked | Klaus | CC BY-SA 3.0 |