Let $U \subseteq \mathbb{R}^n$ be an open set. Let $f \in W^{1,p}(U)$ and suppose all the weak derivatives of $f$ are continuous. Is $f$ itself continuous?
It is a classic fact that if we assume a-priori $f$ is also continuous, then it is in $C^1$.
Here, however I do not assume $f$ is continuous.
On
It depends on exactly what we mean by "$f$ is continuous". Taken literally, the answer is no: If $$f(x)=\begin{cases}1,&(x=0), \\0,&(x\ne0)\end{cases}$$then $f$ is not continuous, but the weak derivatives of $f$ equal $0$.
Typically in this sort of context when one says $f$ is continuous one means that $f=g$ almost everywhere, where $g$ is continuous. If that's what we mean then the answer is yes.
Note, inspired by a comment: One has to be careful about jumping to conclusions; the continuity condition above has more or less nothing to do with being continuous almost everywhere! For ease of reference let's label the two:
(i) $f=g$ almost everywhere for some continuous function $g$.
(ii) $f$ is continuous almost everywhere.
On $\Bbb R$, the function $\chi_{(0,\infty)}$ shows that (ii) does not imply (i), while $\chi_{\Bbb Q}$ shows that (i) does not imply (ii).
Exercise investigate the implications, if any, between (i), (ii) and a third version:
(iii) There exists a set $E$ of full measure such that $f_E$ is continuous (wrt to the subspace topology on $E$).
You can show the continuity of $f$ as follows:
If $p = \infty$, then $f$ is Lipschitz continuous and you are done.
In case $p < \infty$, you can consider an open subset $O$ with $\overline O \subset U$. Then, $\overline O$ is compact and the continuous functions $D f$ are bounded on $\overline O$. Now, you can use the Sobolev embedding theorem to get $f \in W^{1,\infty}(O)$. Again, $f$ is Lipschitz on $O$ and (since $O$ was arbitrary), $f$ is continuous on $U$.
Now you know that $f$ is continuous and you can use the technique from the other question to show that $f$ is actually $C^1$.