Timeline for Measure Theories with a different convention to $\infty\cdot 0 =0$ [closed]
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2020 at 3:56 | review | Reopen votes | |||
| May 14, 2020 at 10:10 | |||||
| May 14, 2020 at 2:59 | history | closed |
R W user44191 Alex M. David C LeechLattice |
Not suitable for this site | |
| May 13, 2020 at 23:05 | comment | added | Alan | @NikWeaver there are many questions here that are not research level though they don't get closed! | |
| May 13, 2020 at 21:10 | comment | added | Alan | @NikWeaver sorry, this is my last post or thread in this cursed QA website. I will still ask and perhaps also answer questions in MSE, but for me mathoverflow is a forgotten memory. | |
| May 13, 2020 at 20:54 | comment | added | Alan | So @LSpice am I right in saying we can take $\infty \cdot 0$ to be any number and thus we might get uncountable number of measure theories? | |
| May 13, 2020 at 14:59 | comment | added | Nik Weaver | One of my answers was downvotes to oblivion, far worse than -2. Don't place more importance on this than it deserves. | |
| May 13, 2020 at 14:56 | comment | added | Nik Weaver | @Alan I checked your activity history and it seems to me that you contribute a lot of value to mathoverflow, and the community here has shown its appreciation by giving you many upvotes. The current question isn't appropriate but I wish you wouldn't take that judgement too personally. | |
| May 13, 2020 at 14:52 | comment | added | Alan | @NikWeaver ok, I won't ask anymore questions on overflow; I know when I am not welcomed. | |
| May 13, 2020 at 13:16 | comment | added | Nik Weaver | @Alan I didn't vote to close but yes, it is not research level and should have been asked at math.stackexchange. | |
| May 13, 2020 at 6:13 | comment | added | user44191 | I think the precise convention needs to be explained a bit more; is it saying that the integral of the zero-function over a set of measure infinity is 0? Or that the integral of the infinite function over a set of measure 0 is 0? Or something else entirely? | |
| May 13, 2020 at 3:51 | comment | added | Alan | @LSpice so it seems that my friend who proved his argument was partially correct. Thanks for pointing it to me LSpice! | |
| May 13, 2020 at 2:46 | comment | added | LSpice | $\infty\cdot0 = \infty$ I can imagine, but $\infty\cdot0 = -\infty$? Then $-\infty = \infty\cdot(0\cdot0) = (\infty\cdot0)\cdot0 = (-\infty)\cdot0 = -(\infty\cdot0) = \infty$, if multiplication is still associative (and, if not, then why not, say, $\infty\cdot0 = 1$?). | |
| May 13, 2020 at 2:38 | answer | added | Nik Weaver | timeline score: 8 | |
| May 13, 2020 at 1:39 | comment | added | R W | Precisely where in measure theory you find this convention? | |
| May 13, 2020 at 1:26 | review | Close votes | |||
| May 14, 2020 at 2:59 | |||||
| May 13, 2020 at 0:03 | comment | added | Sam Hopkins | Finitely additive functions are certainly studied: see e.g. en.wikipedia.org/wiki/… | |
| May 13, 2020 at 0:03 | comment | added | Alan | @SamHopkins either way I'd like to see those theories. Some say all of maths when done correctly is trivial. cheers! | |
| May 13, 2020 at 0:00 | comment | added | Sam Hopkins | I feel like any measure theory where you get rid of countable additively and insist all infinite sets have infinite measure is gonna be a very trivial theory... | |
| May 12, 2020 at 23:55 | comment | added | user76284 | @SamHopkins $\mathbb{R}$ is not a countable union of singletons, though. | |
| May 12, 2020 at 23:51 | comment | added | Sam Hopkins | Arguably $\infty \cdot 0 = \infty$ is consistent with usual measure theory: the (usual Lebesgue) measure of a singleton subset of $\mathbb{R}$ is 0 but the measure of all of $\mathbb{R}$ is infinity. | |
| May 12, 2020 at 23:48 | history | asked | Alan | CC BY-SA 4.0 |