Approximating an $L^2$ function in the Riemann sense $\newcommand{\R}{\Bbb R}$
Consider the Lebesgue measure in $\R$ and the following proposition:

P. For each representative of a function class $f\in L^2[0,1]$ there is a sequence of continuous functions $(f_n)_{n\in\Bbb N}$ such that:

*

*$|f_n-f|$ is Riemann integrable on $[0,1]$, for all $n\in\Bbb N$.

*$\lim\limits_{n\to\infty} \int\limits_0^1 |f_n(x)-f(x)|^2\ \mathrm d x=0$.


There is no reason why this proposition should be true, but I cannot find a counterexample.

This proposition is not true. 
Let $f:[0,1]\to\mathbb R$ be defined by $f(x)=1$ iff $x\in\mathbb Q$ and $f(x)=0$ otherwise.  $f$ is Lebesgue integrable but not Riemann integrable.
Suppose $|f_n-f|$ is Riemann integrable and $\int_0^1|f_n-f|dx<1/4$. 
Then there is a partition $0=t_0<t_1<\dots<t_n=1$ of the unit interval such that 
$$\sum_{i=1}^n(t_i-t_{i-1})\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4.$$
Now there is $i\in\{1,\dots,n\}$ such that $\sup\{|f_n(x)-f(x)|:x\in[t_{i-1},t_i]\}<1/4$. 
The function $f_n$ is not continuous on $[t_{i-1},t_i]$ since on a dense subset of that interval, $f_n<1/4$ and on a dense set, $f_n>3/4$.  
