Today I have the following question: Say $\Omega\subset\mathbb R^d$ is open, bounded, with Lipschitz boundary. Let $u\in W^{1,2}(\Omega)$. Then $u|_{\partial\Omega}$ is well-defined as the image of the trace operator. My main question is: if you have a sequence of $(d-1)$-dimensional "nice" sets $S_n\subset\Omega$ (nice in the Lipschitz sense) that approximate $\partial\Omega$ uniformly, does it follow that $$ u|_{S_n}\rightarrow u|_{\partial\Omega} $$ in the sense that $$ \int\limits_{S_n}u\phi\,dx\rightarrow\int\limits_{\partial\Omega}u\phi\,dx $$ for each test function $\phi\in C^{\infty}(\overline{\Omega})$, as $n\rightarrow\infty$?
Think of $S_n$ as approximations of the boundary. If these approximations do not occur uniformly, but say, only in an appropriate $L_2$ sense or in an a.e. sense, does the answer to the main question change? Finally/alternatively, please give a reference to a good book/paper that discusses boundary approximations in PDE applications, if it exists.
I would say that in general the answer is no. However something very close to what you want is true. Consider the case $d=2$ and $\Omega=(0,1)^2$. Then for every smooth function $u$, by the fundamental theorem of calculus you have $$u(x,y)-u(x,0)=\int_0^y\partial_y u(x,s)\,ds.$$ By Holder's inequality, $$|u(x,y)-u(x,0)|^2\le y\int_0^y|\partial_y u(x,s)|^2\,ds.$$ Integrate over $(0,1)\times (0,\varepsilon)$ to get $$\int_0^\varepsilon \int_0^1|u(x,y)-u(x,0)|^2 dxdy \le \varepsilon^2 \int_0^1\int_0^\varepsilon|\partial_y u(x,s)|^2\,dsdx.$$ By continuity of the trace operator the same inequality holds for $u$ in $H^1(\Omega)$. Set $y=\varepsilon t$. Then $$\int_0^1 \int_0^1|u(x,\varepsilon t)-u(x,0)|^2 dxdt \le \varepsilon\int_0^1\int_0^\varepsilon|\partial_y u(x,s)|^2\,dsdx.$$ Letting $\varepsilon \to 0$ you get $$\lim_{\varepsilon \to 0}\int_0^1 \Bigl(\int_0^1|u(x,\varepsilon t)-u(x,0)|^2 dx\Bigr)dt=0$$ Take $\varepsilon=1/n$. Then there exists a subsequence $1/n_k$ such that $$\lim_{k \to \infty}\int_0^1|u(x,\frac1{n_k} t)-u(x,0)|^2 dx=0,$$ for $\mathcal{L^1}$ a.e. $t$. This is telling you that the trace of $u$ on the segment $(0,1)\times \{\frac1{n_k} t\}$ converges in $L^2(0,1)$ to the trace of $u$ on $(0,1)\times \{0\}$ for $\mathcal{L^1}$ a.e. $t$. I think one can construct examples (the mystic bed of nails?) that show that this cannot happen for every $t$. For a general domain you can flatten the boundary locally and reduce to this case. I don't know a good reference for boundary approximations.