Intro: I would like to know if my demonstration of $C^\infty[0;1]$ is dense in $L^1[0,1]$ is correct because I didn't find any complete demonstration of that statement.
-(i) As we know from here all functions in $f(x) \in L^{1}[0;1]$ are the point wise limit of a finite and countable linear combination of indicator functions.
-(ii) So from (i) we can write that there exists a function $g_\varepsilon(x)$ (wich is a finite and countable linear combinaison of indicator functions as explain in (i)) s.t it verifies pointwise convergence as follows: $\forall 0\leq x\leq 1, \; |f(x)-g_\varepsilon(x)|<\varepsilon/2 $ (such function exists according to (i) for any given $\varepsilon>0$).
Thus in this case we have: $\left \| f-g_\varepsilon \right \|_1=\int_{0}^{1}|f-g_\varepsilon|dx \leq \int_{0}^{1}|\varepsilon/2|dx = \varepsilon/2<\varepsilon$
We just prove that the set of all simple functions is dense in $L^{1}[0;1]$
-(iii) Now let define $\tilde{g}(x)=\begin{Bmatrix} g_\varepsilon(x), x \notin \mathbb{Q}\\ 0, x \in \mathbb{Q} \end{Bmatrix}$. Thus: $\Rightarrow \left \|g_\varepsilon(x)-\tilde{g}(x) \right \|_1=\int_{0\leq x \in \mathbb{Q} \leq 1}^{}|g_\varepsilon(x)|dx=0 < \varepsilon$
-(iv) Because we know that every function of type $\tilde{g}(x)$ can be approximated by a continuous function (in $L^1[0;1]$) in $C[0;1]$ so $C[0;1]$ is dense in $L^1[0;1]$.
-(v) Now by Weierstrass Theorem we know that the set of all polynomials is dense in C[0;1]. But we know too that all polynomials belong to $C^\infty[0;1]$.
So $C^\infty[0;1]$ is dense in $L^1[0,1]$.
Q.E.D