For $u\in H^1(\mathbb{R}^n)$ such that every point is a Lebesgue point, we define a version of $u$ to be $$u(x)=\lim\limits_{r\rightarrow 0}\frac{1}{|B_r(x)|}\int_{B_r(x)}u(z)dz$$
Now we define a function $$f(x)= \frac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)}u(z)d\mathcal{H}^{n-1}$$
My question is : Is $f(x)$ continuous？
I came across this conclusion in the book "Regularity of the one-phase free boundaries" remark 3.6, so it is probably true. But it obviously doesn't hold for general $H^1$ functions for one can change the values of a Sobolev function on any measure zero sets, so I don't see how every point being a Lebesgue point changes that.
Any help would be appreciated!
This is a consequence of properties of the trace operator; more specifically the following integration by parts/divergence theorem formula:
Let $U\in H^1(\Omega)^n$, where $\Omega$ is a Lipschitz bounded domain; then $$ \int_\Omega \text{div}U(y)\, dy = \int_{\partial\Omega} U(y)\cdot n(y)\, d\sigma(y), $$ where $n$ denotes the outward unit normal to $\Omega$, and $\sigma$ is short for $\mathcal{H}^{n-1}$. Notice that when $\Omega=B(x,r)$ then $n(y)= \frac{y-x}{r}$.
Define, for fixed $u\in H^1(\mathbb{R}^n)$, $r>0$, and $x\in \mathbb{R}^n$, $U_{x,r}(y):= u(y)\frac{y-x}{r}$. Since continuity of $f$ is a local property, we may as well assume that $u$ has compact support, ensuring that $U_{x,r}\in H^1(\mathbb{R}^n)$. By the above formula we have \begin{equation} \begin{split} |\partial B(x,r)|f(x) & = \int_{\partial B(x,r)} U_{x,r}(y)\cdot n(y)\, d\sigma(y)= \int_{B(x,r)}[\nabla u(y)\cdot \frac{y-x}{r} + \frac{n}{r}u(y)]\, dy\\ & = \int_{\mathbb{R}^n} \mathbf{1}_{B(x,r)}(y)[\nabla u(y)\cdot \frac{y-x}{r} + \frac{n}{r}u(y)]\, dy. \end{split} \end{equation} At this stage the conclusion follows from the dominated convergence theorem since $\nabla u, u\in L^2$ and $\mathbf{1}_{B(x,r)}\frac{y-x}{r}$ is uniformly bounded (in $x$ and $y$) in compact subsets of $\mathbb{R}^n\times\mathbb{R}^n$. In fact, by the same argument, we see that $f$ is also continuous in $r>0$.
As for how to prove the above divergence theorem, use the extension operator to reduce to the case when $U\in H^1(\mathbb{R}^n)$, then convolve to get a smooth function for which the usual divergence theorem applies and then pass to the limit using continuity of the trace operator.