I was told to give an example of compact $E\subset\mathbb{R}^2$ with $m_2(E)=0$ and a function $u\in W^{1, p}(\mathbb{R}^2\setminus E)$ such that $u\notin W^{1, p}(\mathbb{R}^2)$. I came up with the following example: Let $E=\partial B(0, 1)$ and define
$$ u(x)=\begin{cases} |x|^{\alpha} \quad x\in B(0, 1) \\ 0 \quad \ \ \ |x|>1 \end{cases} $$ where $\alpha>1-n/p$ so that $|x|^{\alpha}\in W^{1, p}(B(0, 1)$. Now clearly $u\in L^p(\mathbb{R}^2\setminus E)$, $D_i u\in L^p (\mathbb{R}^2\setminus E)$ and $u$ is ACL on $\mathbb{R^2}\setminus E$. It then follows that $u\in W^{1, p}(\mathbb{R}^2\setminus E)$.
Now if $u\in W^{1, p}(\mathbb{R}^2)$, then $u$ would have ACL representative $g$ such that $g=u$ almost everywhere in $\mathbb{R}^2$. But this is impossible, because $g$ cannot be absolutely continuous on any line that intersects the ball $B(0, 1)$, because $g=0$ in $\mathbb{R}^2\setminus\bar{B}(0, 1)$ and $g(x)\rightarrow 1$ when $|x|\rightarrow 1$. Contradiction.
Is my reasoning correct and does my counterexample also hold in general dimension $n$? I didn't use any special properties of $\mathbb{R}^2$ in my proof.