Timeline for Is the set of functions between two arbitrarily close L^1 functions closed?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 20, 2022 at 20:27 | comment | added | Willie Wong | uh... my previous comment is a bit facetious. (It is not that useful since the pseudometric itself now depends on the set F, so is not external to the set F as what I am guessing you would prefer.) | |
| Feb 19, 2022 at 16:05 | comment | added | coudy | This is a good idea. I will try to sort it out. Still I am surprised that it has not a well known answer. | |
| Feb 18, 2022 at 17:29 | history | edited | coudy | CC BY-SA 4.0 |
add details concerning measurability
|
| Feb 18, 2022 at 16:42 | comment | added | Willie Wong | What if you define $d(f,f') = \inf \int h-g$ where the inf is taken over $\{(g,h)\in F^2: g \leq f-f' \leq h\}$? This probably defines a pseudometric. | |
| Feb 18, 2022 at 16:35 | comment | added | coudy | Good point. The space $F$ should be a subspace of Borel integrable functions without identification, in order to define $\hat{F}$ with inequality holding everywhere. | |
| Feb 18, 2022 at 16:23 | comment | added | Willie Wong | Are a.e. functions identified in $L^1(X)$? And if so, is the inequality $g \leq f \leq h$ understood a.e.? | |
| Feb 18, 2022 at 14:52 | history | asked | coudy | CC BY-SA 4.0 |