If $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$, are there points $x_0\in[0,1]$ such that $\lim_{n\to\infty}f_n(x_0)=0$? This is part of an old qual problem at my school. 

Assume $\{f_n\}$ is a sequence of nonnegative continuous functions on $[0,1]$ such that $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$. Is it necessarily true that there are points $x_0\in[0,1]$ such that $\lim_{n\to\infty}f_n(x_0)=0$?

I think that there should be some $x_0$. My intuition is that if the integrals converge to $0$, then the $f_n$ should start to be close to zero in most places in $[0,1]$. If $\lim_{n\to\infty}f_n(x_0)\neq 0$ for any $x_0$, then the sequences $\{f_n(x_0)\}$ for each fixed $x_0$ have to have positive terms of arbitrarily large index. Since there are only countably many functions, I don't think it's possible to do this without making $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$.
Is there a proof or counterexample to the question? 

No. The standard counterexample would be indicator functions of $[0, 1]$, $[0, 1/2]$, $[1/2, 1]$, $[0, 1/3]$, $[1/3, 2/3]$, and so on. 
In order to make these continuous, add in line segments on either end with very large slope.
