Let $S=\mathbb{Q}\cap [0,1]$. Let $1_S$ be the characteristic function regarding S. I want to show that for all $\epsilon >0$ there exists a step function $f$ satisfying $\| f-1_S\|_1 <\epsilon$.
So, let $\epsilon >0$. I have to construct a peace-wise constant function, depending of $\epsilon$, on disjoint intervals with the desired property. Since the rationals are dense in $\mathbb{R}$, on each such intervall we will have that $1_S$ attains the value 1. How to construct such a step function? I struggle with the condition $\|f-1_S\|_1 <\epsilon$. Regards
$\|f-I_S\|_1=\|f\|_1$ because $I_S=0$ almost everywhere. Hence the consatnt function $f(x)=\epsilon /2$ for all $x$ will do.