Q:

if $\int_{0}^{1}|f_n(x)-f(x)|dx\to 0$ then $f_n\to f$ uniformly on $[0,1]$: True/false

Let  $\{f_n\}$ be a sequence of continuous real valued functions on $[0, \infty)$. Suppose $f_n(x)\to f(x) ~~~\forall x\in [0,\infty)$ and $f$ is integrable. Then
if $\int_{0}^{1}|f_n(x)-f(x)|dx\to 0$ then $f_n\to f$ uniformly on $[0,1]$: True/false
Im  not  able   to  find  a  counter example ?
Pliz help me

A:

False .. $x^n$ is not uniformly convergent on $[0,1]$. Here the limit function is $f(x) = 0$ when $x \in [0,1)$ and $f(1) = 1$
But $\int_{0}^{1}|x^n-f(x)|dx\to 0$ 

A:

Here is an example where $f_n(x)$ does not converge at any point! Let $I_n=[\frac {i-1} {2^{n}},\frac i {2^{n}})$ for $1\leq i \leq 2^{n}$ and $n \geq 1$. Arrange the characteristic functions of these intervals in a sequence by first listing those with $n=1$, then the one's with $n=2$ etc. Then $\int_0^{1} |f_n(x)-0|\, dx \to 0$ but the functions do not converge even at one point. 

