Timeline for Acting with all rational rotations on a subset of the circle having positive measure do you fill almost the whole circle?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 27, 2020 at 1:59 | comment | added | Claudio Rea | So, according with Josiah Park a gave a problem which is already at page 69 of a book. I Really apologize! I am going to give an harder one | |
| Apr 27, 2020 at 1:34 | comment | added | Claudio Rea | 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 $\Gamma E$. Use continuity of the $L_1$ norm $||\cdot||$ with respect to rotations and have $m(\Gamma E)=||f||=||f\circ R_n||\to ||f\circ R||=m[R(\Gamma E)]$. The sequence on the left is constant! Thus $\Gamma E$ has stable measure under the action of $R$ which is ergodic. Hence $m(\Gamma E)=2\pi$ because $m(\Gamma E)>0$ | |
| Apr 25, 2020 at 20:57 | comment | added | jcdornano | Let us continue this discussion in chat. | |
| Apr 25, 2020 at 20:56 | comment | added | jcdornano | if $m(E)=0$ $m(\Gamma E)=0$ by sigma additivity | |
| Apr 25, 2020 at 20:47 | comment | added | jcdornano | sorry i deleted the bad comment. i will ask a new question and might continue the discution in the chat , if you don't mind.... | |
| Apr 25, 2020 at 20:06 | comment | added | Josiah Park | No claims here about the nonmeasurable case. Maybe ask your question separately? | |
| Apr 25, 2020 at 19:55 | comment | added | Josiah Park | 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. | |
| Apr 25, 2020 at 19:49 | comment | added | jcdornano | 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? | |
| Apr 25, 2020 at 19:00 | comment | added | Josiah Park | @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 $m(\Gamma E)=2\pi$. | |
| Apr 25, 2020 at 18:56 | comment | added | jcdornano | "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) | |
| Apr 25, 2020 at 14:56 | history | answered | Josiah Park | CC BY-SA 4.0 |