Let $\Omega\subset\mathbb{R}^n$ be a bounded Lipschitz domain, $\delta>0$ and $x_0\in\partial\Omega$ fixed. Moreover let $Q_\delta(x_0)$ be a cube with sidelength $\delta$ around $x_0$ and define for $s>0$ the set $\Omega_s = \{x\in\Omega: \mathrm{dist}(x,\partial\Omega)>s\}$. Now I expect/hope that something like
$$ \lim_{s\to0} \,\frac{1}{\mathcal{L}^n(Q_\delta(x_0)\cap(\Omega\setminus\Omega_s)}\int_{Q_\delta(x_0)\cap(\Omega\setminus\Omega_s)}u(y)\, \mathrm{d}y = \frac{1}{\mathcal{H}^{n-1}(Q_\delta(x_0)\cap\partial\Omega)}\int_{Q_\delta(x_0)\cap\partial\Omega}\mathrm{tr}(u)\mathrm{d}\mathcal{H}^{n-1} $$
should hold for $u\in W^{1,1}(\Omega)$, but at the moment I'm not able to prove it. $\mathcal{L}^n$ and $\mathcal{H}^{n-1}$ denote the Lebesgue and the Hausdorff-measure respectively. First I want to sketch why I think this should be true:
- If I take a smooth function $g\in C^ \infty(\overline{\Omega})\cap W^{1,1}(\Omega)$ one can apply a consequence of the coarea-formula, namely the integration over level sets (e.g. Evans-Gariepy, Measure Theory and Fine Properties of Functions- Theorem 3.11) to get that $g|_{\mathrm{dist}^{-1}(\xi)}$ is $\mathcal{H}^{n-1}$-summable for $\mathcal{L}^1$-a.e. $\xi$ and $$ \int_{\Omega\setminus\Omega_s} g\,\mathrm{d}x = \int_0^s \int_{\mathrm{dist}^{-1}(\xi)}g(y)\,\mathrm{d}\mathcal{H}^{n-1}(y) \,\mathrm{d}\xi $$ where we have used that the distance function is 1-Lipschitz with Jakobian equals 1 a.e. (for details I have consulted Theorem 3.14 in the same book). In this situation I can apply Lebesgue-differentiation theorem to get the claimed result (For a smooth function every point is a Lebesgue-Point).
- Now my idea was use a approximation sequence $(u_k)_{k\in\mathbb{N}}$ in $C^ \infty(\overline{\Omega})\cap W^{1,1}(\Omega)$ for $u\in W^{1,1}(\Omega)$. For such a sequence we know by continuity of the trace operator that $\mathrm{tr}(u_k)=u_k|_{\partial\Omega} \to \mathrm{tr}(u)$ in $L^1(\partial\Omega,\mathcal{H}^{n-1})$ as $k\to\infty$. At this point I don't have a idea how to argue, that I can interchange limits (if this is even possible?), more precisely I stop at $$ \lim_{k\to\infty}\,\lim_{s\to0} \,\frac{1}{\mathcal{L}^n(Q_\delta(x_0)\cap(\Omega\setminus\Omega_s)}\int_{Q_\delta(x_0)\cap(\Omega\setminus\Omega_s)}u_k(y)\, \mathrm{d}y = \frac{1}{\mathcal{H}^{n-1}(Q_\delta(x_0)\cap\partial\Omega)}\int_{Q_\delta(x_0)\cap\partial\Omega}\mathrm{tr}(u)\mathrm{d}\mathcal{H}^{n-1}. $$
- Another idea: One could conclude as above if we would have better regularity of the map $$ \xi \mapsto \int_{\mathrm{dist}^{-1}(\xi)}u(y)\,\mathrm{d}\mathcal{H}^{n-1}(y) $$ I think continuity should be enough (or that 0 is a Lebesgue-point). What can we say about the regularity using e.g. the fact $u\in W^{1,1}(\Omega)$? If the mapping would be $W^{1,1}$ one could use the fact, that we have already absolute continuity.
I'm happy about every comment, idea, reference request (maybe such a question has been already treated in the literature) or even some counterexample. Thanks a lot in advance.