Timeline for Can a nowhere differentiable function preserve measurability?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 28, 2016 at 9:58 | comment | added | Anton Petrunin | @Eric Oh, right, try to apply Lebegues density theorem --- hope it works. | |
| May 27, 2016 at 23:41 | comment | added | user92157 | @Anton Yeah I see how to do it if the set $A$ is dense somewhere but sets of positive measure can be nowhere dense and that's what I'm not sure about. | |
| May 27, 2016 at 11:17 | comment | added | Anton Petrunin | @Eric I think so. You have to start with an interval in which most of the points have undefined derivative. | |
| May 27, 2016 at 3:34 | comment | added | user92157 | Could we modify this to show that if the derivative of $f$ fails to exist on some set of positive measure, then it fails to have the Luzin N property? | |
| May 26, 2016 at 13:48 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
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| May 26, 2016 at 13:22 | vote | accept | CommunityBot | ||
| May 26, 2016 at 12:55 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
added 192 characters in body
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| May 26, 2016 at 12:50 | comment | added | Anton Petrunin | @FedorPetrov I made an update. | |
| May 26, 2016 at 12:45 | history | edited | Anton Petrunin | CC BY-SA 3.0 |
added 254 characters in body
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| May 26, 2016 at 12:29 | comment | added | Fedor Petrov | Hm, how do you apply Vitali theorem? Intervals $I$ with image $f(I)$ much greater than $I$ should be everywhere dense, but they should not cover the whole segment or even most of it. | |
| May 26, 2016 at 8:36 | history | answered | Anton Petrunin | CC BY-SA 3.0 |