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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 $2\pi$?

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  • $\begingroup$ If there is a counterexample, E can be taken to be a union of orbits of Gamma without being of full measure. I think one can show a la Vitali that such an E is non-measurable, but I am unsure. If not, consider Bernstein sets. Gerhard "Taking Measure Of This Question" Paseman, 2020.04.25. $\endgroup$
    – Gerhard Paseman
    Commented Apr 25, 2020 at 14:45
  • $\begingroup$ By the aswer of Nik Weaver we have the same result for any infinite $\Gamma$. $\endgroup$
    – jcdornano
    Commented Apr 26, 2020 at 2:13

3 Answers 3

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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 $m(\Gamma E) > (1-\epsilon)2\pi$.

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  • $\begingroup$ Thanks for answering. Could you please explain better the last step? In particular which lower bound you have for $m(I)$? $\endgroup$
    – Claudio Rea
    Commented Apr 27, 2020 at 1:45
  • $\begingroup$ $I$ is an interval of the circle and $m(I)$ is its arc length ... not sure what you're asking. $\endgroup$
    – Nik Weaver
    Commented Apr 27, 2020 at 2:29
  • $\begingroup$ In the first $4,5$ rows of your answer you tell us what are density points as we learnt in the kindergarten. The last $0,5+0,5$ rows are the conclusion which should be proved. Where is the proof? The only nontrivial step is that $m(E \cap I) > (1-\epsilon)m(I)$ implies $m(\Gamma E) > (1-\epsilon)2\pi$. But this is absent. Could you write, please, an extended proof of that? $\endgroup$
    – Claudio Rea
    Commented Apr 27, 2020 at 15:58
  • $\begingroup$ Okay, if you learned about density points in kindergarten then you really shouldn't have trouble with this last very easy step. Hint: if $m(E\cap I) > (1-\epsilon)m(I)$ then $m(\Gamma E\cap I') > (1-\epsilon)m(I)$ for any arc $I'$ of the same length as $I$. $\endgroup$
    – Nik Weaver
    Commented Apr 27, 2020 at 18:49
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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, $m(\Gamma E)=2\pi$.

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  • $\begingroup$ "The action of G is ergodic " Does this mean that for any $E\subset S$ either $m(\Gamma E)=2\pi$, either $m(\Gamma E)=0$ ?(sorry about my inculture) $\endgroup$
    – jcdornano
    Commented Apr 25, 2020 at 18:56
  • $\begingroup$ @jcdornano It means that any positive non full-measure set must map to a larger measure set under the action of $G$, and this in turn implies $m(\Gamma E)=2\pi$. $\endgroup$
    – Josiah Park
    Commented Apr 25, 2020 at 19:00
  • $\begingroup$ sorry for my bad english, but what i mean by "does it mean ..." is if one can quickly deduce from ergodicity that we cannot have $0<m(\Gamma E)< 2\pi$ for some $E\subset S$. If not, is it true? $\endgroup$
    – jcdornano
    Commented Apr 25, 2020 at 19:49
  • $\begingroup$ An action is ergodic if the only sets $E$ which are mapped to themselves (up to zero measure sets) are either of full or zero measure. Since rational rotations include the identity, $E\subset \Gamma E$, and so for $E$ of neither full nor zero measure, $m(\Gamma E)>m(E)$. But this shows in turn that $m(\Gamma E)=2\pi$ by repetition. $\endgroup$
    – Josiah Park
    Commented Apr 25, 2020 at 19:55
  • $\begingroup$ No claims here about the nonmeasurable case. Maybe ask your question separately? $\endgroup$
    – Josiah Park
    Commented Apr 25, 2020 at 20:06
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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 $m(\Gamma E)>0$

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