Timeline for Weyl's Equidistribution Theorem and Measure Theory
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 13, 2019 at 6:04 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
edited Mathjax (the question has been bumped anyway)
|
| Feb 5, 2012 at 13:51 | vote | accept | George Lazou | ||
| Feb 5, 2012 at 13:46 | vote | accept | George Lazou | ||
| Feb 5, 2012 at 13:51 | |||||
| Jan 28, 2012 at 16:36 | answer | added | Asaf | timeline score: 3 | |
| Jan 26, 2012 at 23:37 | comment | added | Goldstern | Related question: mathoverflow.net/questions/75777 | |
| Jan 26, 2012 at 23:33 | answer | added | Goldstern | timeline score: 5 | |
| Jan 26, 2012 at 22:37 | comment | added | Vaughn Climenhaga | @George: In that case it does. Lebesgue measure is invariant and ergodic for any irrational rotation, and in particular, Birkhoff's ergodic theorem implies that the limit frequency of visits to any measurable set is equal to the Lebesgue measure of that set for Lebesgue-a.e. starting point $x$. Thus if $A$ is measurable and $E$ has positive Lebesgue measure and the limiting frequency of visits exists and is equal for every $x\in E$, then that limit must equal $\mu(A)$. | |
| Jan 26, 2012 at 21:44 | comment | added | George Lazou | Sorry if I am flogging a dead horse, but what happens if we require the limit to be equal for all irrational $x$? i.e. if $lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N, \tilde{(kx)} \in A$} = $lim_{n \rightarrow \infty}$$\frac{1}{N}$$card${$k : 1 \leq k \leq N, \tilde{(ky)} \in A$} for all $x, y$ irrational, then does the limit equal $\mu(A)$? | |
| Jan 26, 2012 at 19:47 | comment | added | Noam D. Elkies | Sorry, this is way too much to hope for: ${\bf Z} x \bmod 1 = \lbrace k x \bmod 1 | x \in {\bf Z} \rbrace$ is countable, and thus of measure zero. Hence we could add it to or remove it from any measurable subset $A$, leaving $\mu(A)$ fixed but making the limit $1$ or $0$. Or, by having $A$ meet ${\bf Z} x \bmod 1$ in a set of multiples whose density in $[1,N]$ comes arbitrarily close to $0$ and $1$ as $N \rightarrow \infty$, we could prevent the proportion of multiples in $A$ from having any nontrivial upper lower bounds while keeping $A$ measurable. | |
| Jan 26, 2012 at 19:32 | history | asked | George Lazou | CC BY-SA 3.0 |