Timeline for If a measure $\mu$ and Lebesgue measure $\lambda$ are singular, is the derivative of $\mu$ with respect to $\lambda$ $\infty$, $\mu$-a.e.?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2018 at 18:36 | comment | added | ohliv | Yes, you are right. As the answer show, if $\mu\bot\lambda$ and $B=\{x\in A:\ \lim \lambda(B(x,r))/\mu(B(x,r))>c\}$, then $c\mu(B)\le \lambda(B)=0$. Thanks! | |
| Mar 15, 2018 at 18:32 | vote | accept | ohliv | ||
| Mar 15, 2018 at 15:06 | answer | added | user83457 | timeline score: 6 | |
| Mar 15, 2018 at 12:09 | answer | added | Iosif Pinelis | timeline score: 7 | |
| S Mar 15, 2018 at 11:05 | history | suggested | Hannes | CC BY-SA 3.0 |
Improved formatting and title
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| Mar 15, 2018 at 10:07 | review | Suggested edits | |||
| S Mar 15, 2018 at 11:05 | |||||
| Mar 15, 2018 at 6:30 | comment | added | Anthony Quas | Doesn’t mutual singularity guarantee you that $\lambda(B(x,r))/\mu(B(x,r))\to 0$, $\mu$-a.e., which is the same thing? | |
| Mar 15, 2018 at 4:02 | history | edited | ohliv | CC BY-SA 3.0 |
added 266 characters in body
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| Mar 15, 2018 at 3:58 | review | First posts | |||
| Mar 15, 2018 at 4:07 | |||||
| Mar 15, 2018 at 3:56 | history | asked | ohliv | CC BY-SA 3.0 |