Why $\mathcal C^\infty (\Omega )\cap W^{k,p}(\Omega )$ dense in $W^{k,p}(\Omega )$ instead of $\mathcal C^\infty (\Omega )$.

Question

Let $\Omega \subset \mathbb R^n$ open. In my course, I have a theorem that says that $\mathcal C^\infty (\Omega )\cap W^{k,p}(\Omega )$ is dense in $W^{k,p}(\Omega )$. But don't we have that $\mathcal C^\infty (\Omega )\subset W^{k,p}(\Omega )$ ? and thus the theorem should be $\mathcal C^\infty (\Omega )$ is dense in $W^{k,p}(\Omega )$ ?

Answer 1

We do not have $C^\infty(\Omega) \subset W^{k,p}(\Omega)$. For example, if $\Omega = (0,1)$, the function $$u(x) = \frac{1}x, \,\,\,\,\,\, x \in \Omega$$ is in $C^\infty(\Omega)$ but is not even in $L^p(\Omega)$ for any $p \ge 1$.

EDIT: As a previous (now deleted) answer pointed out, sets of infinite measure also pose an issue since constant maps are $C^\infty$ but not $W^{k,p}$. I believe that we can only conclude $C^\infty(\Omega) \subset W^{k,p}(\Omega)$ when $\Omega$ is compact.