Timeline for Does a positive-measure subset of the unit interval almost surely intersect a random translation of some countable subgroup of $\mathbb{R}$?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 15, 2015 at 19:02 | comment | added | Nate Eldredge | @ChristianRemling: In other words, having $U_n \downarrow A$ does not imply $U_n - G \downarrow A - G$. | |
| Sep 15, 2015 at 19:01 | answer | added | Nate Eldredge | timeline score: 1 | |
| Sep 15, 2015 at 18:34 | comment | added | Nate Eldredge | @ChristianRemling: I can't seem to make this argument work. Let's say $G = \mathbb{Q}$. For a measurable set $B$, let $f_B(t) = \sup_{q \in \mathbb{Q}} 1_B(t+q)$. Then $f_A$ is the indicator of the displayed set. And if $U$ is any open set then $f_U = 1$ everywhere. So it can't in general be true that if $1_{U_n} \downarrow 1_A$ then $f_{U_n} \to f_A$ almost everywhere, because it will fail when $A$ has measure zero. There is a sup and inf that we can't interchange. | |
| Jun 28, 2015 at 11:09 | comment | added | Julian Newman | @ChristianRemling: Thank you for your answer. This is also a very nice answer. | |
| Jun 28, 2015 at 10:42 | comment | added | Julian Newman | @PabloShmerkin: Thank you very much for this, this is a nice simple solution. | |
| Jun 28, 2015 at 5:15 | comment | added | Christian Remling | By monotone convergence, we can compute the measure of the displayed set as a limit where we replace $A$ by $U_n$, if $\chi_{U_n}$ decreases to $\chi_A$ a.e. Now use outer regularity to approximate $A$ by open sets, and take any dense $G$. | |
| Jun 28, 2015 at 2:45 | comment | added | Pablo Shmerkin | Take $G=\mathbb{Q}$. Then $A-G$ has full measure (say, by the Lebesgue density theorem), so $(t+G)\cap A\neq\varnothing$ for almost all $t$. | |
| Jun 28, 2015 at 1:52 | history | asked | Julian Newman | CC BY-SA 3.0 |