Here's a proof I'm struggling with:
Let $f\in L$ and $\epsilon > 0$. Show that there exists a step function $\varphi$ on $[a,b]$ such that
$$\int |f(x)−\varphi(x)|\,dx < \epsilon\ .$$
I figured I would need to use something about $L^+$ but I'm really not sure. Any advice?
A possible idea of proof :
Consider the step function
$$\phi_n(x) = \sum_{k=-n^2}^{n^2} \frac{k}{n} \chi_{\{ f(x) \in [ \frac{k}{n}, \frac{k+1}{n}[ \}} $$
Now,
$$ \int |f(x) - \phi_n(x) | dx = \underbrace{\int_{ |f| > n } |f(x) - \phi_n(x) | dx}_{R_n} + \sum_{k=-n^2}^{n^2} \underbrace{\int_{ f(x) \in [\frac{k}{n},\frac{k+1}{n}[ }|f(x) - \phi_n(x) |dx }_{I_{n,k}}$$
We have
$$R_n = \int_{ |f| > n } |f(x)| dx$$
And this converge to $0$ as $n \to \infty$ , because $f\in L^1$
We also have
$$I_{k,n} = \int_{ f(x) \in [\frac{k}{n},\frac{k+1}{n}[ }|f(x) - \phi_n(x) |dx \leq \frac{1}{n} \mu( \{f(x) \in [\frac{k}{n},\frac{k+1}{n}[\} )$$
Hence,
$$\sum_{k=-n^2}^{n^2} I_{k,n} = \frac{1}{n} \sum_{k=-n^2}^{n^2} \mu( \{f(x) \in [\frac{k}{n},\frac{k+1}{n}[\} )$$
$$= \frac{1}{n} \mu( \{f(x) \in [-n,n[\} ) \leq \frac{1}{n} \mu([a,b])$$
So this also converge to $0$ as $n \to \infty$
This will give us the existence of a $n$ such that $\int | f(x) - \phi_n(x) | dx \leq \epsilon$