Timeline for Couplings as generalized functions
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Nov 30, 2021 at 9:53 | vote | accept | Danny Stoll | ||
| Nov 7, 2021 at 0:30 | vote | accept | Danny Stoll | ||
| S Nov 30, 2021 at 9:53 | |||||
| Nov 4, 2021 at 7:56 | comment | added | Alex M. | @RW: Of course I meant the opposite, that was the whole point of that sentence! Thank you for noticing it, it was late and I was tired when I wrote that. | |
| Nov 4, 2021 at 3:19 | comment | added | R W | @AlexM Are you sure you don't confuse pull-back and push-forward? Measures are covariant! | |
| Nov 4, 2021 at 3:15 | answer | added | R W | timeline score: 1 | |
| Nov 3, 2021 at 21:08 | review | Close votes | |||
| Nov 4, 2021 at 7:56 | |||||
| Nov 3, 2021 at 20:44 | comment | added | Danny Stoll | @AlexM. $\pi$ restricted to the graph of $f$ is the inverse of the measurable map $(\mathrm{id} \times f):X \rightarrow X\times Y$. If it’s more clear, just go by the second version I wrote, which will clearly always be well-defined for $f$ measurable. | |
| Nov 3, 2021 at 20:40 | comment | added | Alex M. | You've got a serious problem here: who guarantees that $\pi (E \cap \Gamma_f)$ is measurable? Remember that measurable maps don't always take measurable subsets into measurable subsets; there are cases when this is true, but they are very restrictive (they require $\pi$ to be injective, among other conditions). So your formula does not define a measure $\gamma$. And remember, the natural operation on measures is the pull-back, not the push-forward. So I am very pessimistic about your project; I believe that it is hopeless, save for some trivial cases. | |
| Nov 3, 2021 at 20:11 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Nov 3, 2021 at 18:20 | comment | added | Danny Stoll | We would take $\gamma(E) = \mu_X(\pi_1(E \cap \Gamma_f))$, where $\Gamma_f \subset X\times Y$ is the graph of $f$. In particular, $\gamma(A \times B) = \mu_X(A \cap f^{-1}(B))$. | |
| Nov 3, 2021 at 18:04 | comment | added | Alex M. | How is a (measurable?) map $f:X \to Y$ a particular case of your proposed definition? | |
| S Nov 3, 2021 at 17:54 | review | First questions | |||
| Nov 3, 2021 at 18:04 | |||||
| S Nov 3, 2021 at 17:54 | history | asked | Danny Stoll | CC BY-SA 4.0 |