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Jul 2 at 21:37 comment added alesia also, if you don't want to assume regularity, you can bound the volume of tubular neighborhoods using covering numbers of the boundary (consider union of balls)
Jul 2 at 21:23 history edited alesia CC BY-SA 4.0
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Jul 2 at 21:21 comment added alesia yes the idea is essentially as @unwissen describes. The tube formula provides an exact formula refining the above area based first order expansion, which is useful if you want an upper bound and not a first order behavior
Jul 2 at 17:43 comment added unwissen Just to have my understanding verified, not because I think I can explain @alesia's nice idea better: the measure of all points which have distance less than $\epsilon$ to $\Omega_u$ (but are not included in $\Omega_u$) is $\lesssim \epsilon \vert \partial \Omega_u \vert \lesssim \epsilon \vert \partial \Omega \vert $. Therefore the measure of all points which are in $\Omega_v$ but not in $\Omega_u$ is $\lesssim d_H(\Omega_u, \Omega_v) \vert \partial \Omega \vert$. By symmetry, $\vert \Omega_u \Delta \Omega_v \vert \lesssim d_H(\Omega_u, \Omega_v) \vert \partial \Omega \vert$.
Jul 2 at 16:57 comment added tommy1996q Could you elaborate a little? In particular bounding the volume using the Hausdorff distance and the argument using the tube formula and the reach (never heard of neither of them to be honest)
Jul 2 at 16:17 history answered alesia CC BY-SA 4.0