Timeline for Disintegration of von Neumann algebra
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 12, 2012 at 19:39 | vote | accept | m07kl | ||
| Oct 18, 2011 at 16:57 | comment | added | m07kl | ok, I see the trouble. | |
| Oct 18, 2011 at 15:35 | comment | added | Jesse Peterson | @m07kl: So what is wrong with choosing $q_y' = p_y$? | |
| Oct 18, 2011 at 15:18 | comment | added | m07kl | For each $y\in Y$ we can find a $q_y$ which is equivalent to $p_y$ (we don't know whether $q_y$ equals $p_y$ or not, let's say they are different), but $y\mapsto q_y$ is not necessarily measurable in the sense that $Q=\int_Y^{\oplus}q_ydy$ makes sense. So I want to find $q'_y$ which is equivalent to $q_y$ in $B_y$ and $y\mapsto q'_y$ is measurable. I need a kind of measurable selection theorem. Thanks for your attention :) | |
| Oct 18, 2011 at 12:46 | comment | added | Jesse Peterson | That's certainly one interpretation. I wonder if the OP would mind clarifying what is meant by "good", "choice" (since the $q_y$ are already defined), and "up to equivalence"? | |
| Oct 18, 2011 at 7:47 | comment | added | Dmitri Pavlov | @Jesse: I presume that when the OP says “good choice of $q_y$ up to equivalence” he means the choice of a representative of the equivalence class of a section with respect to sets of measure 0. | |
| Oct 17, 2011 at 21:40 | comment | added | Jesse Peterson | I don't think the question makes sense as is. Why not choose $q_y = p_y$? | |
| Oct 17, 2011 at 19:58 | answer | added | Dmitri Pavlov | timeline score: 3 | |
| Oct 17, 2011 at 17:50 | history | asked | m07kl | CC BY-SA 3.0 |