Timeline for Doubling metrics, doubling measures, Lebesgue density
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Mar 6, 2018 at 17:42 | vote | accept | Aryeh Kontorovich | ||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 20, 2016 at 2:46 | answer | added | user95266 | timeline score: 3 | |
| Jul 19, 2016 at 21:51 | comment | added | Aryeh Kontorovich | I think I understand the source of my confusion in the 2nd paragraph -- I had confused $\mu\ll\nu$ with the stronger condition $||d\mu/d\nu||_\infty<\infty$. When I assume that both Radon-Nikodym derivatives are essentially bounded, the doubling property of $\mu$ follows, yes? But of course, that's not the main question I'm after... | |
| Jul 19, 2016 at 21:33 | comment | added | Nate Eldredge | Right. It's true that $d\nu/d\mu = 1/F'$ blows up near 0 but that's not a problem, it's still in $L^1(\mu)$. | |
| Jul 19, 2016 at 21:29 | history | edited | Aryeh Kontorovich | CC BY-SA 3.0 |
Removed some false claims in light of the comments.
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| Jul 19, 2016 at 21:26 | comment | added | Aryeh Kontorovich | Oh I see. I had mistakenly assumed that $F'(x)$ blows up at $0$, but now I see this isn't the case. I'll edit the question to remove that paragraph -- but the main question remains for now. | |
| Jul 19, 2016 at 21:16 | comment | added | Nate Eldredge | I believe my counterexample still applies. Note that the density $d\mu/d\nu$ is $F'(x)$ which is a.e. strictly positive. | |
| Jul 19, 2016 at 21:15 | comment | added | Aryeh Kontorovich | Ack, I see my mistake. Do you agree that the 2nd paragraph holds if I additionally assume $\nu\ll\mu$? | |
| Jul 19, 2016 at 21:08 | comment | added | Nate Eldredge | I don't think I agree with the second paragraph. Let $X = [0,1]$ with its usual metric, let $\nu$ be Lebesgue measure (which is doubling) and let $\mu$ be the probability measure with cdf $F(x) = e^{1-1/x}$. Then $\mu \ll \nu$ but $\mu$ is not doubling. | |
| Jul 19, 2016 at 18:46 | history | asked | Aryeh Kontorovich | CC BY-SA 3.0 |