# Proof: $C^\infty[0,1]$ is dense in $L^1[0,1]$

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

• "As we know, $g_\varepsilon(x)$ only have a finite and countable number of discontinuity points" -- How so? May 20 at 22:35
• @BrianMoehring Thk you for your comment. Because $g_ε(x)$ is choose as in (i) that means as a finite sum (finite linear combination) of indicator functions . See here: proofwiki.org/wiki/… the complete proof. May 20 at 22:50
• That doesn't relate to the number of discontinuity points at all. For instance, the indicator of a fat cantor set in $[0,1]$ is itself an indicator function but its set of discontinuity points is the entirety of $[0,1]$. May 20 at 22:52
• Right, there are only finitely many indicator functions in the sum of finitely many indicator functions. However, even with only one indicator function, the set of discontinuities can be the whole space. To proceed as you seem to want to proceed, you'd need to first approximate the indicator of a measurable set with the indicator of an open set from above, and then with the sum of the indicators of finitely many open intervals from below. May 20 at 22:58
• May 20 at 23:16