I’m considering a $H^1$ function u on a open domain D. Is the integral:
$$ \int_{\partial B_r(x)} u \hspace{2pt}dH^{n-1}$$
continuous with respect to x?
I tried to prove that it’s differential by showing that the derivative can be written as the integration:
$$ \int_{\partial B_r(x)} Du \hspace{2pt}dH^{n-1}$$
But it’s just right a.e. From which we can deduce that it’s continuous a.e.. Could we prove that it’s continuous everywhere?
To show it’s right a.e., we start by considering a sequence of smooth functions approaching u in $H^1$. For each $u_n$, the derivative is:
$$ \int_{\partial B_r(x)} Du_n \hspace{2pt}dH^{n-1}$$
And it approaches
$$ \int_{\partial B_r(x)} Du \hspace{2pt}dH^{n-1}$$
Maybe I'm missing something, but it's not clear to me that, fixing $r>0$, you can conclude that $\int_{\partial B_r(x)} Du(y)\, dH^{n-1}(y)$ is well defined, even for a.e. $x$: Fubini (and polar coordinates) gives that this integral is finite, for fixed $x$, for a.e. $r>0$, but this is not the same.
On the other hand, the continuity of your function follows without such arguments. Notice that for $y\in \partial B_r(x)$, the vector $n(y)= \frac{y-x}{r}$ is the exterior unit normal to $B_r(x)$ at $y$. So we can write, using the divergence theorem, \begin{equation} \begin{split} \int_{\partial B_r(x)} u(y)dH^{n-1}(y) & = \int_{\partial B_r(x)} u(y) \dfrac{y-x}{r}\cdot n(y)\, dH^{n-1}(y)\\ & = \int_{B_r(x)} \left[ \nabla u(y)\cdot \dfrac{y-x}{r} + \dfrac{nu(y)}{r} \right]\, dH^{n-1}(y), \end{split} \end{equation} and this last term is continuous by the dominated convergence theorem.