Sobolev Spaces – Lebesgue Differentiation Theorem at Boundary Points

Question

$\newcommand{\R}{\mathbb R}$
Let $\Omega\subset \R^d$ be a smooth, bounded open set and fix $p\geq 1$.

• Fact 1: the usual Lebesgue differentiation theorem says that, if $u\in L^p(\Omega)$, then
  $$
  u(x)=\lim\limits_{r\to 0}\frac{1}{|B_r(x)|}\int_{B_r(x)}u(y)dy
  $$
  for Lebesgue-a.e. $x\in \Omega$.

• Fact 2: for $u\in W^{1,p}(\Omega)$ (the usual Sobolev space) it is well known that the boundary trace $v:= tr(u)$ is well-defined as an element of $W^{1-1/p,p}(\partial\Omega)$.

Question: is it true that
$$
v(x)=\lim \limits_{r\to 0}\frac{1}{|B_r(x)\cap \Omega|}\int_{B_r(x)\cap \Omega}u(y)dy
$$
for $\mathcal H^{d-1}$-a.e. $x$ in the boundary?

This seems a very natural question to ask and I hope that the answer is already written somewhere out there. I suspect that the notion of $p$-capacity should play a role here (being the precise representative of a $W^{1,p}$ function $p$-quasicontinuous), but I'm not too familiar with capacity theory so I'd rather avoid appealing to such raffinate machinery and reinventing the wheel, if possible.
(I am also aware that the trace can be recovered as the continuous limit along any transversal curve, but somehow I did not manage to conclude from this.)
Actually any reference to either a positive or negative statement would make me happy, I just need a black-box theorem that I could apply.

Answer 1

It is true - this is Theorem 5.7 in Evans and Gariepy’s Measure Theory and Fine Properties of Functions (2015 version).

Note that the theorem is stated for BV functions, but Sobolev functions are BV, so it holds also for your case.