Timeline for When is the infimum of an arbitrary family of measurable functions also measurable?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 31, 2018 at 5:35 | comment | added | Piotr Hajlasz | @FrançoisG.Dorais I added an answer (with a complete proof) based on the notion of the lattice infimum. In guess it is related to the argument of Bill. | |
| Dec 31, 2018 at 3:05 | answer | added | Piotr Hajlasz | timeline score: 3 | |
| Jul 27, 2012 at 23:11 | vote | accept | Vidit Nanda | ||
| Jul 15, 2012 at 19:08 | answer | added | Gerald Edgar | timeline score: 15 | |
| Jul 15, 2012 at 17:59 | comment | added | François G. Dorais | Bill, I think that's worth posting as an answer... | |
| Jul 15, 2012 at 16:32 | comment | added | Bill Johnson | ...take a decreasing sequence $(f_n)$ from the net s.t. $d(f_n,0)$ converges the infimum over $f$ in the net of $d(f,0)$ and consider the pointwise a.e. limit of $(f_n)$. | |
| Jul 15, 2012 at 16:30 | comment | added | Bill Johnson | I am thinking of the $\sigma$-finite case, Vel; probably it is not true without that assumption. In various books you can find that $L_\infty(\mu)$ (or, in fact, any dual $C(K)$) space) is an order complete Banach lattice (the Albiac-Kalton book comes to mind). Some of these arguments can be modified to work for $L_0$ with the topology of convergence in measure when $\mu$ is finite, which easily gives also the $\sigma$-finite case. Take your favorite metric for generating convergence in measure. If you have (for simplicity) a downward directed net of non negative measurable functions... | |
| Jul 14, 2012 at 23:26 | comment | added | Vidit Nanda | Anton, of course that makes sense. Thank you. Yemon, I have made some changes that I hope will clarify matters somewhat. Bill: this sounds perfect for my needs. Could you please provide a reference? | |
| Jul 14, 2012 at 23:23 | history | edited | Vidit Nanda | CC BY-SA 3.0 |
Clarifications regarging minimal conditions
|
| Jul 14, 2012 at 23:18 | answer | added | Andreas Blass | timeline score: 13 | |
| Jul 14, 2012 at 23:17 | comment | added | Bill Johnson | The measurable functions modulo functions that are 0 a.e. form a complete lattice. Maybe that is what you are really looking for. | |
| Jul 14, 2012 at 23:08 | comment | added | Yemon Choi | I guess you mean "minimal conditions" rather than "the minimal conditions"? It would IMHO really help if you gave some idea of what you are thinking of (e.g. monotonicity) - this is not a question of "not being research material" but just a matter of having a more well-defined target. | |
| Jul 14, 2012 at 23:08 | answer | added | Otis Chodosh | timeline score: 5 | |
| Jul 14, 2012 at 23:07 | comment | added | Anton Petrunin | Q1: yes you are right --- the answer is NO. Take a nonmeasurable negative function $q(x)$ and set $f_x(y)=0$ if $x\ne y$ and $f_x(y)=q(x)$ if $x=y$. | |
| Jul 14, 2012 at 22:58 | history | asked | Vidit Nanda | CC BY-SA 3.0 |