Timeline for Elementary proof that an open subset of $\Bbb{R}^n$ does not have measure zero?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2021 at 11:16 | comment | added | Ben McKay | That is the argument in Guillemin and Pollack, Differential Topology. | |
| Apr 3, 2021 at 6:54 | comment | added | Stefan Friedl | Thanks! That is also the proof in Guillemin-Pollack that Ben McKay referred to. In Guillemin-Pollack the proof gets attributed to John von Neumann. | |
| Apr 3, 2021 at 6:52 | vote | accept | Stefan Friedl | ||
| Apr 2, 2021 at 20:36 | comment | added | Iosif Pinelis | Nice argument! As I see it, you reduce the subadditivity of the Lebesgue measure to that of the approximating rescaled counting measure on a fine enough lattice. | |
| Apr 2, 2021 at 20:27 | comment | added | Reid Barton | A minor variant: by a Lebesgue number argument we can choose $N$ large enough that each of the little cubes is contained in some member of the cover, and then counting the little cubes directly gives us the inequality. | |
| Apr 2, 2021 at 20:18 | history | answered | Will Sawin | CC BY-SA 4.0 |