# Acting with a finite number of rotations on a set of positive measure can you fill almost the whole circle?

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.

• Do you have an exmple of a compact set with positive mesure but empty interior? Apr 27, 2020 at 11:00
• You can take a Cantor set of positive measure Apr 27, 2020 at 11:03
• Of course I wanted to delete my comment but to late, you answered already. Also, your answer to the main question is very nice! Apr 27, 2020 at 11:20
• 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 $2\pi-\varepsilon$? Apr 27, 2020 at 14:29
• @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/…). Apr 27, 2020 at 15:01