Timeline for A set whose Hausdorff dimension gradually changes?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2020 at 10:45 | comment | added | Yaakov Baruch | @Wojowu you are right. So it must be that the procedure I had in mind doesn't converge (a.e. pointwise) to a well defined set. | |
| Jan 20, 2020 at 8:46 | comment | added | Wojowu | @YaakovBaruch Are you sure? Doesn't that contradict Lebesgue Density Theorem? | |
| Jan 20, 2020 at 8:31 | comment | added | Yaakov Baruch | Cool. I think a similar idea (reversing subintervals of $[k/2^n,(k+1)/2^n]$ as needed, for $n\rightarrow \infty, 0\le k<n$) can also be used to produce a set $A$ with density $x$ at $x$, i.e. $\mu(A\cap[0,x])=x^2/2$. | |
| Jan 19, 2020 at 20:05 | comment | added | Pietro Majer | In fact for all $0<x\le1$ $A\cap[0,x]$ is an $x$-dimensional set of $x$-dimensional null measure, for the same reason. | |
| Jan 18, 2020 at 23:11 | comment | added | Wojowu | @Anixx Each $A_n$ has Hausdorff dimension below $1$, so their (1-dimensional) Lebesgue measure is zero. So $A$ has Lebesgue measure zero. You can certainly integrate functions over null sets, but the results are not too interesting... | |
| Jan 18, 2020 at 23:06 | comment | added | Anixx | Can we Lebesgue-integrate over such set? | |
| Jan 18, 2020 at 22:27 | vote | accept | Anixx | ||
| Jan 18, 2020 at 22:08 | history | answered | Christian Remling | CC BY-SA 4.0 |