Timeline for Is there a Borel measurable $f:\mathbb{R}^d \to \mathbb{R}^d$ such that $f(x) \in \partial \varphi (x)$ for all $x$?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2023 at 7:10 | vote | accept | Akira | ||
| Sep 5, 2023 at 2:24 | answer | added | Piotr Hajlasz | timeline score: 6 | |
| Sep 3, 2023 at 22:55 | comment | added | Akira | @MichaelGreinecker Thank you so much for your help! I will check it out. | |
| Sep 3, 2023 at 22:53 | comment | added | Michael Greinecker | A correspondence $\phi$ is upper hemicontinuous if for every open set $O$ in the codomain, the set $\{x\mid \phi(x)\subseteq O\}$ is open. Eqivalently, for ever closed set $C$ in the codomain, the set $\{x\mid \phi(x)\cap C\neq\emptyset\}$ is closed. If the latter set is only required to be Borel, you have the definition of a measurable correspondence. In a metric space, each open set is a countable union of closed sets, so a measurable correspondence is weakly measurable too. A good source for this stuff is the book by Aliprantis and Border. | |
| Sep 3, 2023 at 22:41 | comment | added | Akira | @MichaelGreinecker I could not see how upper hemicontinuity implies weakly measurability. Could you elaborate more? | |
| Sep 3, 2023 at 22:27 | comment | added | Michael Greinecker | This should follow from this answer together with the selection theorem of Kuratowski and Ryll-Nardzewski. | |
| Sep 3, 2023 at 21:47 | history | edited | Akira | CC BY-SA 4.0 |
added 73 characters in body
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| Sep 3, 2023 at 21:41 | history | asked | Akira | CC BY-SA 4.0 |