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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?

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up vote 3 down vote accepted

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.

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oh i hadn't seen that, thanks. – YN Chew Sep 8 '13 at 3:13
@YNChew Sure. It's a really useful counterexample to have in mind, since it breaks a lot of things which are obviously true. – user61527 Sep 8 '13 at 3:13
wait, are indicator functions continuous on $[0,1]$? – YN Chew Sep 8 '13 at 3:15
@YNChew You're right, I missed that assumption. An appropriate fix can be made easily by drawing very skinny triangles on each end of the indicator function, and requiring that the area of the triangle is very small. – user61527 Sep 8 '13 at 3:17

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., Corollary 3 for proof.

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