Why $C_0^\infty$ is dense in $L^p$? Why $C_0^\infty$ is dense in $L^p$?
Would you give me a simple proof or the outline of the proof?
 A: The outline (the proof isn't simple, at least not according to my understanding of simple): 
Let $f \in L^p$. Then there is a sequence of simple functions $s_n \in L^p$ that converges pointwise to $f$. Then you show that they converge in norm.
These $s_n$ have finite support. You can therefore apply Lusin's theorem:
(Lusin) Let $\Omega$ be a locally compact Hausdorff space and $\mu$ a Radon measure and $f: \Omega \to \mathbb R$ a $\mu$-measurable function with finite support $E$. Then for every $\delta > 0$ there exists a closed set $K \subset E$ such that $\mu(E \setminus K) < \delta$ and $f$ is continuous on $K$.
to get a sequence of continuous functions.
Then you apply Tietze's theorem:
(Tietze) If $\Omega$ is a locally compact Hausdorff space and $K \subset \Omega$ compact then any $f \in C(K, \mathbb R)$ can be extended to a bounded function in $C(\Omega, \mathbb R)$.
to get a sequence of compactly supported functions.
I did this here for $p=1$.
Once you have $C_c$ dense in $L^p$ you mollify. Which I did here. Hope this helps.
A: Maybe I am missing something, but rather than using convolution and Lusins Lemma to approximate continuous finitely supported functions I would use the functions
$$
f(x) :=
\begin{cases}
e^{1-\frac{1}{1-x^2}} & \text{ for } \vert x\vert <1 \\
0,& otherwise
\end{cases}
$$
and
$$
g_n(x) := \frac{1}{n}\tan(\frac{x ~\pi }{2})
$$
to approximate indicator functions $\chi_{[a,b]}(x)$ by the functions $\frac{a+b}{2}+ \frac{b-a}{2} f\circ g_n (x)$.
