Acting with all rational rotations on a subset of the circle having positive measure do you fill almost the whole circle? Set $\Gamma$ for the group of the roots of the identity: $\Gamma=\{z\in \Bbb C | z^n=1$, for some $n\geq 0\}$ and for $E\subset S^1$ set 
$\Gamma E=\{z\zeta, z\in \Gamma, \zeta\in E \}$
A trivial but very frequently used fact is that if $E$ is open then $\Gamma E=S^1$.
A measure-theoretic version of that is the following
QUESTION. If $E$ has positive Lebesgue measure, does $\Gamma E$ have measure $2\pi$?
 A: Let $S$ be the circle with Haar-Lebesgue measure and let $G$ be the group of rotations of $S$ through angles that are rational multiples of $\pi$. The action of $G$ is ergodic (according to pg. 69 in The Legacy of John von Neumann
edited by James Glimm, John Impagliazzo, Isadore Singer).
So for any measurable subset $E\subset S$ of positive measure, $m(\Gamma E)=2\pi$.
A: An easy way to see this is by using the Lebesgue density theorem. Any set of positive measure has a density point $t$ (indeed, almost every element of the set is a density point). This means that for any $\epsilon > 0$ there is an interval $I$ containing $t$ such that $m(E \cap I) > (1-\epsilon)m(I)$. This pretty much immediately implies that $m(\Gamma E) > (1-\epsilon)2\pi$.
A: I just found this proof. Take a sequence $\Bbb Q\ni x_n\to x\notin\Bbb Q$ and set $R_n$ and $R$ for the rotations of angles $x_n$ and $x$. Set $f$ for the characteristic function of $\Gamma E$, hence $f\circ R_n=f$ . Use continuity of the $L_1$ norm $||\cdot||$ with respect to rotations, so for $n\to\infty$ we have  $m(\Gamma E)=||f||=||f\circ R_n||\to ||f\circ R||=m[R(\Gamma E)]$. Thus $\Gamma E$ has stable measure under the action of $R$ which is ergodic. Hence $m(\Gamma E)=2\pi$ because 
$m(\Gamma E)>0$ 
