Timeline for Who first found this characterization of Lebesgue integration?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 7, 2011 at 17:21 | vote | accept | Tom Leinster | ||
| Jan 3, 2011 at 19:27 | comment | added | Qiaochu Yuan | @Mariano: L^1([0, 1]) can be defined as the abstract Banach space completion of, say, the step functions on [0, 1] with the L^1 norm (which has a purely finitary description). This definition technically doesn't require any knowledge of integration. | |
| Jan 3, 2011 at 19:21 | answer | added | Todd Trimble | timeline score: 4 | |
| Jan 3, 2011 at 16:58 | comment | added | Gerald Edgar | If instead of $[0,1]$ we think of binary expansions and use $\{0,1\}^\mathbb{N}$, then we want the unique Borel measure invariant under the shift map with measures $(1/2,1/2)$ for the sets specified by the first coordinates. Or something like that. Maybe related to Borel around 1910 who proposed using Lebesgue measure to model independence in probability? | |
| Jan 3, 2011 at 16:23 | comment | added | Pete L. Clark | @Tom: my mistake, there it is, and was, when I looked at your page. Actually I think your webpage is very nice, and I look forward to doing some browsing. | |
| Jan 3, 2011 at 16:17 | comment | added | Tom Leinster | Thanks, Omar. I know about Freyd's paper, but I hadn't thought to look at it for references. (And I don't think I've posted about it!) | |
| Jan 3, 2011 at 16:14 | comment | added | Omar Antolín-Camarena | I don't know about that result, but the the same result for continuous functions is due to Freyd (tac.mta.ca/tac/volumes/20/10/20-10abs.html). If I learned about that paper from a blog post of yours on the n-Category Café, I apologize. | |
| Jan 3, 2011 at 16:07 | comment | added | Tom Leinster | Pete: yes, that's what I thought you might have meant. It's here: maths.gla.ac.uk/~tl/glasgowpssl Guess I should tidy up my web page a bit... | |
| Jan 3, 2011 at 16:00 | comment | added | Pete L. Clark | @Tom: no, I really thought you had some note or something where this result was proved. I just searched for it and didn't find it. Am I mistaken? | |
| Jan 3, 2011 at 15:47 | comment | added | Tom Leinster | Mariano: re the integral having been used to define $L^1$, that's a good point; thanks. I'd definitely be interested to know any reference to the characterization of Lebesgue measure that you mention. On the other hand, putting historical questions aside, there's at least one way to define $L^1$ without knowing what integration is (which I suspect is what Pete's referring to). | |
| Jan 3, 2011 at 15:32 | comment | added | Mariano Suárez-Álvarez | Well, the integral is already been used to define what $L^1$ is. A simpler way to state this would be to say that the Lebesgue measure is the unique one such that the measure of a set $A\subseteq[0,1]$ is the same as the average of those of $A/2$ and $(A+1)/2$. | |
| Jan 3, 2011 at 15:31 | comment | added | Pete L. Clark | The only writeup on this topic I've seen is yours. Probably you know about that one already... | |
| Jan 3, 2011 at 15:23 | history | asked | Tom Leinster | CC BY-SA 2.5 |