Why $C_0^\infty$ is dense in $L^p$?

Solution 1:

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.