Timeline for measure zero in R but not in R^2
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 19, 2014 at 19:07 | comment | added | alich | yes, what about a not measurable set? | |
| Mar 14, 2014 at 19:34 | comment | added | Noah Schweber | As Joseph's answer shows, the Fubini argument only shows that there is no such measurable set. | |
| Mar 14, 2014 at 14:00 | vote | accept | alich | ||
| Mar 14, 2014 at 12:29 | answer | added | Joseph Van Name | timeline score: 9 | |
| Mar 14, 2014 at 11:59 | comment | added | Johannes Hahn | @alich: Because such a set $A\subseteq\mathbb{R}^2$ would satisfy $0\neq \lambda^2(A) = \int_{\mathbb{R}^2} \chi_A(x,y) d\lambda^2(x,y) = \int_\mathbb{R} \int_\mathbb{R} \chi_A(x,y) d\lambda^1(x) d\lambda^1(y) = \int_\mathbb{R} 0 d\lambda^1(y) = 0$. (where $\lambda^{1,2}$ denotes the 1-dimensional and 2-dimensional lebesgue meausure respectively) | |
| Mar 14, 2014 at 11:43 | review | First posts | |||
| Mar 14, 2014 at 11:43 | |||||
| Mar 14, 2014 at 11:35 | comment | added | alich | Why does It contradict Fubini theorem? | |
| Mar 14, 2014 at 11:31 | comment | added | Alex Degtyarev | Wouldn't this contradict Fubini's theorem? Of course, assuming that the set is measurable in the first place. | |
| Mar 14, 2014 at 11:29 | comment | added | alich | because of some reasons in integrability of multivariable functions | |
| Mar 14, 2014 at 11:27 | comment | added | Asaf Karagila♦ | Why?${}{}{}{}{}$ | |
| Mar 14, 2014 at 11:23 | history | asked | alich | CC BY-SA 3.0 |