Timeline for Why is there a $\mathcal{H}^d$-null set in the definition of d-rectifiable set?
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8 events
| when toggle format | what | by | license | comment | |
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| Aug 31, 2023 at 9:38 | vote | accept | tommy1996q | ||
| Aug 30, 2023 at 23:02 | answer | added | Piotr Hajlasz | timeline score: 4 | |
| Apr 20, 2023 at 7:48 | comment | added | mlk | I think there are simple practical reasons. Rectifiable sets are often used to integrate wrt. the corresponding Hausdorff-measure, so having an extra null-set does not matter. Also a common way to prove rectifiability is by exhaustion, i.e. cover 1% (in measure) of the set by a Lipschitz map and then iterate. This procedure naturally leaves over a null-set. | |
| Apr 20, 2023 at 3:14 | comment | added | Aidan Backus | I think here is a counterexample to sets of locally finite perimeter having rectifiable reduced boundary for your definition. Consider the even extension of $y = x^{1/2}$ to $[-1, 1]$. Its graph $\gamma$ has finite perimeter (you could even give the problem of computing its length to a slightly competent calc student). But the germ of $\gamma$ at $(0, 0)$ probably cannot be covered by Lipschitz curves. | |
| Apr 19, 2023 at 22:31 | comment | added | tommy1996q | @DaveLRenfro yes, some d-measure zero sets cannot be covered by a countable union of Lipshitz graphs (I think a product of suitable Cantor sets does the trick), however the issue I am having is another: since every set can be decomposed in a rectifiable part and in an unrectifiable one, why did they choose to keep this null set in the rectifiable part instead of the unrectifiable one? My guess is that you want to have some sort of compactness which you wouldn't have if you dropped this condition. For example, sets of finite perimeter wouldn't have a rectifiable reduced boundary (I think) | |
| Apr 19, 2023 at 18:51 | comment | added | Dave L Renfro | that are also first category. However, there exist $d$-measure zero sets that are not first category -- even more, ${\mathbb R}^d$ can be expressed as the union of a $d$-measure zero set and a first category set. Incidentally, being cone free is a stronger smallness notion than nowhere dense -- a cone free set is lower porous (see end of this MSE answer), so their countable unions are even $\sigma$-lower porous (hence have Hausdorff dimension less than $d).$ | |
| Apr 19, 2023 at 18:45 | comment | added | Dave L Renfro | This is something I should know better, but I think the reason is that for measure-theoretic stuff like this you're going to get a "left over" $d$-measure zero set, and some $d$-measure zero sets cannot be covered by a countable union of $d$-measure zero images of Lipschitz maps. Regarding this last part, I believe the latter are countable unions of what I called "cone free sets" near the end of this sci.math post, which are first Baire category sets, and so we're looking at $d$-measure zero sets (continued) | |
| Apr 19, 2023 at 14:32 | history | asked | tommy1996q | CC BY-SA 4.0 |