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Let $\Omega$ be a bounded domain in $\Bbb R^n$. Define $$ \Omega_r=\{x\in\Omega: \text{dist}(x,\partial\Omega)<r \}, $$ i.e. it the ring of thickness $r$ at the boundary of $\Omega$. Intuitively, if the domain $\Omega$ is nice enough, says of class $C^1$, then we can expect $$ \mathcal H^n(\Omega_r) \approx r \mathcal H^{n-1}(\partial\Omega) $$ to hold in some loose sense, e.g. perhaps $\mathcal H^n(\Omega_r) \le c r \mathcal H^n(\partial\Omega)$ for some $c\ge 1$ and all sufficiently small $r>0$.

Q: Does this kind estimate hold in a more general setting, says a Lipschitz domain or when $\Omega$ is a sublevel set of a function $f\in BV(U) $ (when reinterpret the Hausdorff measure appropriately)?

If such an estimate exists could anyone please tell me where can I read more about it?

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    $\begingroup$ If $\Omega$ is a bounded Lipschitz domain, we have $\lim_{r \to 0}(1/r)\int_{\Omega_r}f(y)\,dy=\int_{\partial \Omega}f(y)\,d\mathcal{H}^{n-1}(y)$ for any bounded continuous function $f \colon \overline{\Omega} \to \mathbb{R}$. See Lemma 7.1 in this article projecteuclid.org/download/pdfview_1/euclid.aop/1485421330. $\endgroup$
    – sharpe
    Commented Jan 4, 2021 at 11:21
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    $\begingroup$ @sharpe very interesting, thanks! This piqued my interest quite a bit. Do you know if a similar result holds when $f$ is a Sobolev function instead? $\endgroup$
    – BigbearZzz
    Commented Jan 4, 2021 at 18:56
  • $\begingroup$ I don't know well. At least if $p=2$, it follows that $C(\overline{D}) \cap W^{1,2}(D)$ is a dense subspace of $W^{1,2}(D)$. Wouldn't it work if we used this fact? $\endgroup$
    – sharpe
    Commented Jan 5, 2021 at 2:08
  • $\begingroup$ @sharpe unfortunately, for any continuous approximation of $f\in W^{1,p}(\Omega)$, the factor $1/r$ would blow up the error term. I haven't found a way to make it works yet. $\endgroup$
    – BigbearZzz
    Commented Jan 5, 2021 at 2:17
  • $\begingroup$ Oh, I see. It certainly looks like no good. $\endgroup$
    – sharpe
    Commented Jan 5, 2021 at 3:43

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