Timeline for Given a set of finite perimeter $\Omega$ s.t. $\partial ^* \Omega =\partial \Omega$, it's not true that $P(\Omega)= \mathcal{H}^{n-1} (\Omega)$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3 at 12:52 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Added link to article reviews
|
| Aug 31, 2023 at 9:28 | comment | added | tommy1996q | Or at least, that's how it works for me | |
| Aug 31, 2023 at 9:27 | comment | added | tommy1996q | Thanks! I read through your answer and it is crystal clear, exactly what I was thinking, better explained. It's certainly possible that the reduced boundary has finite Hausdorff measure but the topological boundary doesn't, but not if they are equal! I think the issue is simply that when you read something in an article that doesn't seem right, you always assume you're not understanding something, especially if it's written by famous people. You just forget everyone can make a mistake | |
| Aug 31, 2023 at 9:23 | vote | accept | tommy1996q | ||
| Aug 30, 2023 at 22:55 | comment | added | Piotr Hajlasz | @tommy1996q The comment by Anzelotti and Giaquinta is simply false. When we write a paper we and read our manuscript over and over again, we often find stupid mistakes before writing the final version. That requires many iterations of the process. In older days, without LaTeX and possibility of infinite iterations and corrections, it was easy to write something stupid and not correct it. | |
| Aug 30, 2023 at 22:24 | answer | added | Piotr Hajlasz | timeline score: 2 | |
| Aug 25, 2023 at 19:23 | comment | added | tommy1996q | @user378654 exactly, I get what they are doing but the removed segments are not the reduced boundary, as you say. I mean, I didn't go through the example in detail (that's why I didn't report it), what puzzles me is that they claim that the implication $P(\Omega)< +\infty$ + $\partial \Omega = \partial ^* \Omega \implies \mathcal{H}^{n-1} (\Omega)= P(\Omega)$ is false. | |
| Aug 25, 2023 at 16:37 | comment | added | user378654 | finite, and certainly now there are some points where the topological boundary is also reduced boundary. But you can't have it both ways at once: if you consider just one of the closures of $\cup_j B_i^j$, it's clear its boundary has $\mathcal{H}^1$ measure at least $1/2$. If the perimeter is controlled by the sum of perimeters of the balls, as they I guess claim, then there are points which are not in the reduced boundary. If not, then the perimeters are not summable, invalidating the example. I guess a reasonable exercise is to compute exactly what the perimeters of these sets are. | |
| Aug 25, 2023 at 16:29 | comment | added | user378654 | Reading through their example, I don't really follow what they gain by removing balls. If their $\Omega$ was instead taken as $A \setminus \cup S_i$, they will have (correctly) shown that $P(E; \Omega)$ finite does not imply $P(E; \mathbb{R}^n)$ finite, the caveat being that $P(\Omega) < \infty$ but $\mathcal{H}^1(\Omega) = \infty$ (which is not a contradiction, all the removed line segments are not reduced boundary). Instead of line segments they remove closures of unions of small balls centered on a dense set on each line segment: here it is less clear if the perimeter of $\Omega$ is ... | |
| Aug 25, 2023 at 14:03 | history | asked | tommy1996q | CC BY-SA 4.0 |