Timeline for Measurability of subspace of set of all functions
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2013 at 10:05 | history | edited | Davide Giraudo | CC BY-SA 3.0 |
added 530 characters in body
|
| Jun 4, 2013 at 12:34 | comment | added | Matthias Ludewig | Yes, sorry, I think I overread the $\mathbb{N}$ in the exponent. | |
| Jun 4, 2013 at 12:33 | vote | accept | Matthias Ludewig | ||
| Jun 4, 2013 at 8:21 | comment | added | Davide Giraudo |
I probably missed something, but here $B$ is a Borel subset of the space of sequences of elements of $\BBb R^n$. And if $(x_j)_j$, $(y_j)_j$ are two sequences, and $B_1$, $B_2$ two such Borel sets, define $z=(x_1,y_1,x_2,\dots,)$, and we can write $S((x_j)_j,B_1)\cup S((y_j)_j,B_2)=\{f,(f(z_{2n+1})\in B_1\mbox{ or }f(z_{2n})\in B_2\}$. For countable unions, we can proceed similarly taking a bijection from $\Bbb N^2$ to $\Bbb N$.
|
|
| Jun 4, 2013 at 8:18 | history | undeleted | Davide Giraudo | ||
| Jun 4, 2013 at 8:12 | history | deleted | Davide Giraudo | ||
| Jun 4, 2013 at 6:23 | comment | added | Matthias Ludewig | Also, the union of two such sets is generally not of this form anymore. | |
| Jun 4, 2013 at 6:21 | comment | added | Matthias Ludewig | This seems false. You at least need to allow different sets $B$ for each $x_j$, otherwise the Sigma-Algebra is too small. | |
| Jun 3, 2013 at 23:13 | history | answered | Davide Giraudo | CC BY-SA 3.0 |