Timeline for Class of integrable 0/1-functions "with no null sets."
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 9, 2013 at 7:01 | comment | added | Guido Kanschat | I do not think you can go beyond step functions | |
| Jan 31, 2013 at 0:04 | comment | added | Ben | Nik: Right, the set of all Riemann integrable functions is too large for my purposes. | |
| Jan 30, 2013 at 13:32 | comment | added | Pietro Majer | I don't know a name for this property, but note that any Lebesgue measurable set $F\subset\mathbb{R}^n$ has a subset $E$ (the set of its points of density $1$) with the even stronger property that |B(x,r)∩E|=|B(x,r)|(1+o(1)), and |F\E|=0. (see en.wikipedia.org/wiki/Lebesgue_density_theorem ). | |
| Jan 30, 2013 at 13:30 | comment | added | Nik Weaver | You might want to reconsider whether you're sure that ${\rm area}(B(x,r) \cap E) > 0$ for all $x \in E$ and $r > 0$. That's not even true for Riemann integrable functions into $\{0,1\}$. | |
| Jan 30, 2013 at 13:28 | comment | added | Jack Huizenga | On the Wikipedia page for the Lebesgue density theorem (which is very closely related to what you ask about) it is noted that if $E$ and its complement both have positive measure then there are always points where the "approximate density" is strictly between $0$ and $1$. It is then always possible change $E$ by either including or removing such points. | |
| Jan 30, 2013 at 13:02 | history | asked | Ben | CC BY-SA 3.0 |