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Oct 9 at 13:25 vote accept rfloc
Oct 9 at 5:51 comment added Fedor Petrov If $A$ is not null, it contains a bounded subset $A_1$ which is not null. If you consider only translates of $A_1$ on vectors of length at most 1,you can not have more then finitely if them which are disjoint, by pigeonhole principle
Oct 9 at 3:40 comment added rfloc The set of all $v\in\mathbb{R}^n$ with $\sum v=0$ has measure zero. Are you certain that what you wrote is in fact true?
Oct 9 at 1:16 comment added rfloc Could please explain why $A$ is necessarily null?
Oct 9 at 1:07 history answered Wojowu CC BY-SA 4.0