general topology - Show that a set of functions is dense in $L^2(0,2)$

Show that a set of functions
$$A=\{f\in C^0([0,2]):\ f(0)=f(1)=f(2)=0\}$$
is dense in $L^2(0,2)$.

I know the following theorem: Let $1\le p<\infty$, $\ \Omega\ $ be an open set of $\mathbb{R}^n$. Then continuous functions with compact support are dense in $L^p(\Omega)$.

In my case the functions are defined on $[0,2]$ that is closed. The definition of support is
$$supp\ f:=\overline{\{x\in X : f(x)\ne 0\}}$$
that is a closed set.

Now for a function in $A$ I can write the support like this (tell me any mistakes)
$$supp\ f=\overline{(0,1)\cup (1,2)}$$
that I think it's equal to $[0,2]$. Can I apply the previous theorem being my domain closed? Or maybe there is a different way to solve this?

Answer

It's enough to show $A$ is dense in $C[0,2]$ in the $L^2$ metric. So let $f\in C[0,2],$ with $M=\max_{[0,2]}|f|.$ For $n=1,2,\dots,$ set

$$f_n(x) = f(x)d(x,\{0,1,2\})^{1/n}.$$

Then each $f_n \in A.$ Furthermore $f_n \to f$ pointwise on $[0,2]\setminus \{0,1,2\}.$ We want to show

$$\tag 1\int_0^2 |f_n-f|^2\to 0.$$

But this is easy: First, we know $f_n\to f$ a.e. on $[0,2].$ Second, $|f_n|\le M$ for all $n.$ Hence the integrands in $(1)$ are bounded above by $4M^2.$ By the dominated convergence theorem, $(1)$ holds as desired.