I am trying to use the coarea formula to get estimates on the measure of an epsilon-neighbourhood of a set. Specificly, given a compact 'nice' set $A\subseteq \mathbb{R^d}$, possibly with more than one connected components which are not convex, I'm hoping to get an upper estimate on $$\lambda(A^\epsilon)-\lambda(A),$$ where $\lambda$ is the Lebesgue measure on $\mathbb{R}^d$ and $A^\epsilon= \cup_{a\in A} B_\epsilon(a)$. The notion of niceness is something that should be not just convex. I'm pretty sure it should be possible to write $$ \lambda(A^\epsilon)-\lambda(A)= \int_0^\epsilon \nu(A^{(t)})dt $$ by a Fubini like argument, where $A^{(t)}=\{x: d(x,A)=t \}$ and $\nu$ is some measure. I found this thread and related threads, which seem to rely on the Minkowski-Steiner formula. Is there perhaps some results of this nature using the co-formula area for non convex sets?
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2$\begingroup$ You will definitely need some sufficient notion of niceness, with the obvious standard counterexample being the open, dense sets of arbitrary small measure you can construct as the union of a sequence of smaller and smaller balls around any countable dense set. (Btw, this set should really have a name, since it comes up so often in similar questions...) $\endgroup$– mlkCommented Aug 5, 2022 at 17:13
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$\begingroup$ $A^\varepsilon$ and $A^{(t)}$ cannot distinguish $A$ from its closure, so you may have to assume that $A$ is closed. $\endgroup$– user95282Commented Aug 6, 2022 at 22:09
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$\begingroup$ I revised the question and added the condition of being compact, since I was hoping for something more general, but I do not yet how to exclude the dense sets. $\endgroup$– Keen-ameteurCommented Aug 7, 2022 at 8:15
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$\begingroup$ @mlk When you say this set should have a name, do you mean $A^\epsilon$ or my $A^{(t)}$? $\endgroup$– Keen-ameteurCommented Aug 7, 2022 at 8:16
1 Answer
Your formula is true (up to appropriate constants) if you take $\nu$ to be the Minkowski content, assuming that the volume of the tubular neighborhood is locally Lipschitz as a function of the tube radius. This is because the Minkowski content is defined as the derivative of the volume of the tubular neighborhood.
One condition ensuring that the tube volume is locally Lipschitz is that the distance function to $A$ has Clarke gradient bounded below in the $\epsilon$-neighborhood of $A$. It also ensures that Minkowski contents coincide with the Hausdorff measures of the boundaries.
Finally, although that's not the question, it is possible to bound the volume of the tubular neighborhood without any regularity conditions, using the $\epsilon$-covering numbers of $A$.
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$\begingroup$ First, thank you for your response. Do you know of a reference for the first property you stated? Do you know if there exist informative estimates for the Minkowski content aside from the Minkowski-Steiner formula? I think when $A\subseteq \mathbb{R}$ is compact, the Minkowski content of $A^\epsilon$ is the number of connected components of $A^\epsilon$. I was wonedring whether there are similar estitmates in $\mathbb{R}^d$. $\endgroup$ Commented Aug 8, 2022 at 12:14
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$\begingroup$ which first property? As for the Minkowski content, it is basically the area of the boundary $\endgroup$– alesiaCommented Aug 8, 2022 at 15:23
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$\begingroup$ I was asking how to apply the coarea formula when the tubular neighborhood is locally Lipschitz. I am not sure how to derive it. $\endgroup$ Commented Aug 9, 2022 at 7:54
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$\begingroup$ ah, it isn't even the coarea formula if you use minkowski contents, it's just the fact that a (locally) lipschitz function (of a real variable = tube radius) is the integral of its derivative $\endgroup$– alesiaCommented Aug 9, 2022 at 16:44
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$\begingroup$ Or actually you could use coarea formula as well, but the issue again is that you need your isosurfaces to have finite area $\endgroup$– alesiaCommented Aug 9, 2022 at 16:51