Suppose that $f$ is Riemann integrable on $[a,b]$. Then there exists a sequence $f_{k}$ of continuous functions on $[a,b]$ so that $$\lim_{k\rightarrow \infty }\int_{a}^{b}\left | f(x)-f_k(x) \right |dx=0 $$
(My attempt) Suppose that $f$ is Riemann integrable on $[a,b]$ and for every sequence $f_{k}$ of continuous functions on $[a,b]$ , $\lim_{k\rightarrow \infty }\int_{a}^{b}\left | f(x)-f_k(x) \right |dx\neq 0 $.
I want to show contradiction. $\lim_{k\rightarrow \infty }\int_{a}^{b}\left | f(x)-f_k(x) \right |dx\neq 0 $ means that there exists $\epsilon > 0 $ such that for every $N$ ,
$n\geq N \Rightarrow \int_{a}^{b}\left | f(x)-f_k(x) \right |dx \geq \int_{a}^{b}\left (| f(x)|-|f_k(x) \right |)dx=\int_{a}^{b}\left | f(x) \right |dx-\int_{a}^{b}\left |f_k(x) \right |dx\geq \epsilon $
I don't know how to proceed from here.
There is a step function $g_k$ s.t. $$\lim_{k\to \infty }\int_a^b |g_k-f|=0.$$ Set $$g_k(x)=\sum_{i=0}^{n_k-1}a_i\boldsymbol 1_{[s_i,s_{i+1}[}$$ where $$a=s_0<s_1<...<s_{n_k}=b.$$ Let $f_k$ the function that connect linearly all $(s_i,a_i)$. It's a continuous function s.t. $$\lim_{k\to \infty }\int_a^b |f_k-f|=0.$$
This is almost the tropezoidal rule.