$f_n, f$ RI and continuous and $\int|f_n(x)-f(x)|dx \to 0$ as $n \to \infty$ implies $f_n \to f$ uniformly Let $f_n, f$ be Riemann Integral and continuous functions on $[a,b]$ and $\lim_{n \to \infty} \int_{a}^{b}|f_n(x)-f(x)|dx = 0$, then $f_n \to f$ uniformly on $[a,b]$. Prove or give a counterexample if false.
My attempt: I think this is true because $\lim_{n \to \infty} \int_{a}^{b}|f_n(x)-f(x)|dx = 0$ implies that $\forall \epsilon > 0$, can choose $N \in \mathbb{N}$ s.t. $n > N \implies |f_n(x)-f(x)|(b-a) < \epsilon \implies |f_n(x)-f(x)| < \frac{\epsilon}{b-a}$. Since $\epsilon$ is arbitrary, we have $f_n \to f$ uniformly. But I failed to use the fact that $f_n$ and $f$ are continuous. Any suggestions?
 A: Take $[a,b]=[0,1]$, $f_n(x)=x^n$, and $f\equiv 0$ for a counterexample.
A: Take the functions on $[0,1]$ $f_n(x)= \begin{cases}\
                                       0  & 0 \leq x \leq \frac{1}{2}\\ 
                                       n(t-\frac{1}{2}) & \frac{1}{2}<x<\frac{1}{2}+\frac{1}{n}\\
                                      1 & \frac{1}{n}+\frac{1}{2} \leq x \leq 1\
                                        \end{cases}$ 
We have that $f_n \to f(x)=0$ if $x\leq \frac{1}{2}$ and $f(x)=1$ if $x>\frac{1}{2}$
$\int_0^1 |f_n-f| \to 0$ but $f_n$ does not converge uniformly because then $f$ would be continuous as a uniform limit of continuous functions.

It would also help you to draw this sequence of functions to remember it easily,because it provides a counterexample which proves that the space $(C[0,1],d)$ where $d(f,g)=\int_0^1|f-g|$ is not complete.

A: Call $\Phi(x)=\begin{cases}\exp\frac1{x^2-x}&\text{if }0<x<1\\ 0&\text{if }x\le 0\lor x\ge1\end{cases}$ and $f_n(x)=\Phi(nx)$. Then, $f_n\to 0$ pointwise on $\Bbb R$ and $$\int_0^1\lvert f_n(x)\rvert\,dx\le\int_{-\infty}^\infty \lvert f_n(x)\rvert\,dx=\frac1n\int_0^1\Phi(x)\,dx\to 0$$
Yet, $f_n(\frac1{2n})=e^{-4}$, so $f_n$ does not converge uniformly to $0$ on any interval $[0,\delta]$.
