The approximation theorem states that if $U$ is bounded, $u \in W^{1,p}(U)$ for some $1 \leq p < \infty$ then there are functions $u_m \in C^\infty(U) \cap W^{k,p}(U)$, such that $u_m \rightarrow u$ in $W^{k,p}(U)$.
This is probably very simple, but I'm trying to follow along with an example in a book I am reading. I'm looking at the set $\Omega = (-1,0) \cup (0,1)$ and the function $u : \Omega \rightarrow \mathbb{R}$ defined by
$u(x) = \left\{ \begin{array}{ll} 1, & x > 0 \\ 0, & x < 0 \end{array} \right.$.
It's clear to me that this function is in $W^{1,p}(\Omega)$ for $1 \leq p \leq \Omega$; however, I'm having trouble proving that it cannot be approximated by functions in $C^\infty(\overline{\Omega})$. It seems that any sequence approaching $u$ would approach the heaviside function which I know is not in $L^p$ and therefore not in $W^{1,p}$; however, actually proving this rigorously has eluded me for a while now.
I also had a question about a global form of the approximation theorem. If we also add that $\partial U$ is $C^1$ to the hypothesis, then we have functions $u_m \in C^\infty(\overline{U})$ such that $u_m \rightarrow u$ in $W^{k,p}(U)$. I know that $W^{k,p}$ is Banach so I was wondering if in this case we are able to assume that $\lim\limits_{m \rightarrow \infty} u_m$ exists in $W^{1,p}$. I'm just not sure if the $C^\infty$ functions intersected with the $W^{k,p}$ functions are complete. Anyway, thanks a lot for looking. I appreciate any help you guys can provide.