Timeline for Definition of area
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2013 at 0:11 | comment | added | Anton Petrunin | Say, this idea does not beat the definition with measuring grid. | |
| Jan 26, 2013 at 22:06 | comment | added | Anton Petrunin | methodologically, proving that the area does not depend on triangulation is wrong thing to do --- it moves you to discreet geometry, which is irrelevant to the question. The price for keeping things "elementary" is too high. | |
| Jan 26, 2013 at 21:46 | comment | added | Andy Putman | @Anton Petrunin : My answer not quite the same as the others. For instance, it works correctly in higher dimensions. | |
| Jan 26, 2013 at 21:37 | comment | added | Anton Petrunin | You keep adding the same answer :) | |
| Jan 26, 2013 at 21:27 | comment | added | Andy Putman | (there are also issues with non-locally-finite triangulations that arise if you try to deal with sets that are not open) | |
| Jan 26, 2013 at 20:46 | comment | added | Andy Putman | @Loïc Teyssier : Countable triangulations are harder for two reasons. One, you have to worry about limits converging and what not. Two, it is much more technical to prove that countable triangulations have common subdivisions than finite triangulations, and then it is harder to prove that subdividing doesn't change the area since you can't express a subdivision as a finite sequence of "vertex additions". BTW, I agree that if you want to construct Lebesgue measure then rectangles are best -- for that, I recommend the treatment in Frank Jones's book "Lebesgue integration in Euclidean Space". | |
| Jan 26, 2013 at 20:33 | comment | added | Loïc Teyssier | You can also allow for a countable triangulation, by taking the limit (you of course need to check that the limit is independant on the sequence of triangulation, but you already did the dirty work beforehand in proving the common subdivision lemma). Then you can measure any bounded element of the borelian tribe. It's easy to prove that it is the unique measure giving the usual formula for triangles, which satisfies the usual axiom of a measure. Personally I prefer using rectangles with sides parallel to a given system of cartesian coordinates. It's the same, but easier for students... | |
| Jan 26, 2013 at 20:10 | history | answered | Andy Putman | CC BY-SA 3.0 |