Timeline for Condition for $f^\prime$ to be absolute integrable
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 21 at 2:51 | vote | accept | Mingzhou Liu | ||
| Mar 23, 2023 at 8:04 | comment | added | Pietro Majer | @Mingzhou of course; if not compact add "locally" | |
| Mar 23, 2023 at 3:37 | comment | added | Mingzhou Liu | @fedja See the EDIT. | |
| Mar 23, 2023 at 3:35 | comment | added | Mingzhou Liu | @PietroMajer Hi, what do you mean by "exists a.e."? Besides, as far as I know, "absolutely continuity indicates Lebesgue integrability" only holds on compact intervals. | |
| Mar 23, 2023 at 3:31 | history | edited | Mingzhou Liu | CC BY-SA 4.0 |
added 112 characters in body
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| Mar 22, 2023 at 18:57 | comment | added | Pietro Majer | If f is absolutely continuous then f' [exists a.e.] and is Lebesgue integrable | |
| Mar 22, 2023 at 18:41 | comment | added | Pietro Majer | Actually $|f'|$ is measurable and non-negative so the integral does exist. Maybe you mean "is finite" ? | |
| Mar 22, 2023 at 18:13 | answer | added | Christophe Leuridan | timeline score: 3 | |
| Mar 22, 2023 at 12:10 | comment | added | Mingzhou Liu | @CarloBeenakker Indeed. Then, I can not offer an example where the integral does not exist. Could you prove that the integral always exists? | |
| Mar 22, 2023 at 11:37 | comment | added | Carlo Beenakker | @MingzhouLiu --- the function in the earlier post you mention is not continuous at the points $n+2^{-n}$, which is what you require in your question... | |
| Mar 22, 2023 at 11:30 | comment | added | fedja | Provide a non-trivial condition Define "non-trivial". | |
| Mar 22, 2023 at 11:08 | comment | added | Mingzhou Liu | See this post. | |
| Mar 22, 2023 at 9:23 | comment | added | Carlo Beenakker | can you give an example where this integral does not exist? | |
| S Mar 22, 2023 at 8:01 | review | First questions | |||
| Mar 22, 2023 at 8:40 | |||||
| S Mar 22, 2023 at 8:01 | history | asked | Mingzhou Liu | CC BY-SA 4.0 |