Dear ladies and gentlemen, I have written a proof (sketch) for the statement below and would like to know if it is the right approach.
It goes as follows:
Let $\Omega \subset \mathbb{R}$ be an open subset and define $C_p(\Omega) = \{ f \in C(\Omega) : |f|^p \in L(\Omega,\lambda)\}$.
Statement: $C_p(\Omega)$ is dense in $L^p(\Omega)$.
Proof (sketch)
Since the simple functions are dense in $L^1(\Omega)$ and for each simple function $\chi_{_\Omega}$ there exists $g\in C_1(\Omega)$, such that for all $\epsilon$>0 one gets $|\chi_{_\Omega} - g| < \frac{\epsilon}{2}$. Thus, for $f \in L^1(\Omega)$
$\begin{equation} |f - g| = |f - \chi_{_\Omega} +\chi_{_\Omega} - g| \end{equation} < |f - \chi_{_\Omega}| + |\chi_{_\Omega} - g| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} < \epsilon $.
To turn this into a proof, I'd additionaly show that the two assumptions hold. For the second assumption, namely that $\forall \chi_{_\Omega} \exists g\in C_1(\Omega) $ such that $g$ approximates $\chi_{_\Omega}$, I'd construct $g$ explicitly for $\Omega = \cup_{j\in \mathbb{N}}I_j$, $I_j$ open intervals in $\mathbb{R}$.
If you are allowed to use the theorem of density of continuous functions with compact support,i.e. $C_c(\Omega)$ then it is easy enough since $$C_c(\Omega) \subseteq C_p(\Omega)$$