# 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.
• wait, are indicator functions continuous on $[0,1]$? – YN Chew Sep 8 '13 at 3:15
As T. Bongers points out the answer is no. However, we can say that there is a subsequence $f_{n_k}$ such that $f_{n_k}(x) \to 0$ for almost every $x \in [0,1]$. The statement that $\int_0^1 f_n(x)dx \to 0$ exactly tells us $f_n \to 0$ in $L^1$ which implies the existence of a subsequence which converges to zero almost everywhere. See, e.g., http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ Corollary 3 for proof.