Let $X$ be a metric space endowed with a measure $\mu$ satisfying the following condition: all the open and closed sets of $X$ are measurable and for any measurable set $M\subset X$ and any $\varepsilon>0$ there exists an open set $G\supset M$ such that $\mu(G\setminus M)<\varepsilon$.
Kolmogorov-Fomin's Элементы теории функций и функционального анализа proves (p. 378 here) that the set of all continuous functions [belonging to $L_1$, I would say; not all continuous functions on $X$ are in $L_1$] on $X$ [with the above said property] is everywhere dense in $L_1(X,\mu)$. (The English translation Introductory Real Analysis (p. 381 here) proves that the set of all continuous functions on $X$ is everywhere dense if $L_1(X,\mu)$ has the following property: all the open and closed sets of $X$ are measurable and $\mu(M)=\inf_{M\subset G}\mu(G)$, which is a similar property: I would say that it is equivalent to the property above if $\mu(X)< \infty$ but not equivalent, though implied, if $\mu(X)=\infty$.)
From Kolmgorov-Fomin's proof, since I think that $\varphi_\varepsilon$ is measurable because of its continuity and the above said property of $X$ and $|\varphi_\varepsilon|^p$ is integrable because it is bounded by the integrable function taking the values 1 for $x\in G_M$, where $\mu(G_M)<\infty$ (if I correctly understand the book) and 0 elsewhere, I would say that the same holds of any $L_p(X,\mu)$, $p\geq 1$, where $X$ has the above property, i.e. the set of all continuous functions belonging to $L_p(X,\mu)$ is everywhere dense in $L_p(X,\mu)$.
Am I right? I thank anybody for any answer!