Let $E\subset S^1$ have positive Lebesgue measure. Do there exist finitely many rotations $r_1, r_2, \dots ,r_n$ such that $r_1E\cup r_2E\cup \dots\cup r_nE$ has measure $2\pi$? Or is there a counterexample?
1 Answer
The answer is no. Take a compact set $E$ with positive measure buth empty interior and assume that $K=r_1 E\cup r_2 E \cdots \cup r_n E$ has measure $2\pi$. Then $K$ would be dense in $S^1$, hence equal to $S^1$, since it is closed. But this is impossible, since $K$ has empty interior, too.

$\begingroup$ Do you have an exmple of a compact set with positive mesure but empty interior? $\endgroup$ Apr 27, 2020 at 11:00

4$\begingroup$ You can take a Cantor set of positive measure $\endgroup$ Apr 27, 2020 at 11:03

$\begingroup$ Of course I wanted to delete my comment but to late, you answered already. Also, your answer to the main question is very nice! $\endgroup$ Apr 27, 2020 at 11:20

$\begingroup$ Nice answer! What can be said if we relax the condition to: for every $\varepsilon>0$, there exist $n$ rotations such that $r_1E\cup\ldots r_nE$ has measure greater than $2\pi\varepsilon$? $\endgroup$ Apr 27, 2020 at 14:29

4$\begingroup$ @Capublanca Thanks. In that case it is true, by a limit argument, since using a countable dense set of rotations one can cover almost all the circle (see mathoverflow.net/questions/358506/…). $\endgroup$ Apr 27, 2020 at 15:01