Timeline for Does the Doob-Dynkin lemma hold for any measurable space that separates points?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Mar 9, 2017 at 23:56 | vote | accept | Jack M | ||
| Mar 7, 2017 at 19:41 | history | edited | Jack M | CC BY-SA 3.0 |
deleted 81 characters in body
|
| Mar 7, 2017 at 17:44 | answer | added | Derived Cats | timeline score: 12 | |
| Mar 7, 2017 at 15:46 | answer | added | user95282 | timeline score: 10 | |
| Mar 7, 2017 at 2:10 | comment | added | user95282 | @JackM I don't know any reference for this result. | |
| Mar 6, 2017 at 21:25 | history | edited | Jack M | CC BY-SA 3.0 |
deleted 98 characters in body
|
| Mar 6, 2017 at 21:23 | comment | added | Jack M | @NateEldredge I'd love a counter-example to prove that surjectivity can't be removed from the premises. I'll update the question. | |
| Mar 6, 2017 at 19:20 | comment | added | Nate Eldredge | I don't see anything wrong with your proof. But as to why you don't see references for this, your added restriction that $f$ is surjective is a substantial weakening of the lemma. You can't remove this restriction by replacing $Y$ with the image $f(X)$ because $f(X)$ may not be measurable. | |
| Mar 6, 2017 at 17:40 | history | edited | Jack M | CC BY-SA 3.0 |
added 909 characters in body
|
| Mar 6, 2017 at 17:10 | comment | added | Jack M | @user95282 Do you have a reference for the fact that the conjecture is true? It's just surprising to me, since proofs of the Doob-Dynkin lemma that I've seen involve the monotone convergence theorem, which seems like such overkill if all you need to do is say "$\mathbb R^n$ separates points". | |
| Mar 6, 2017 at 17:08 | history | edited | Jack M | CC BY-SA 3.0 |
edited title
|
| Mar 6, 2017 at 16:03 | comment | added | user95282 | I think you mean a measurable (not measure) space $Z$. And as Nate Eldredge points out, "separable" means something else; "separated" would be OK for what you mean. You are correct that the answer is yes, so you may just as well post your own answer. | |
| Mar 6, 2017 at 15:16 | comment | added | Nate Eldredge | Interesting question. A terminology note: the property you call "separable" is more commonly called "separates points". "Separable measure space" means something else. | |
| Mar 6, 2017 at 13:37 | comment | added | Jochen Wengenroth | Why don't you sketch your simple proof here? | |
| Mar 6, 2017 at 11:36 | history | asked | Jack M | CC BY-SA 3.0 |