Problem: Does there exist a sequence of continuous functions $ f_n:[0,1] \to [0,\infty)$ such that $\lim_{n \to \infty} \int_0^1 f_n(x) dx=0$ but their doesn't exist any $x \in [0,1]$ for which the sequence $ f_n(x)$ converges?
My approach: Actually, I think the above assertion can be proven false. Rather we can show that for some $\alpha \in [0,1]$, the sequence converges to $0.$ If that doesn't happen, then, for each $\alpha \in [0,1]$, there exists an infinite subsequence of $f_n(\alpha)$ such that $\| f_n(\alpha) \| \geq 1/k$, then define $S_k=\{\alpha \in [0,1] | f_n(\alpha)| \geq 1/k$ for infinitely many $n$ $\}$, then $S_k$ is a nested sequence and I think as $\lim_{n \to \infty} \int_0^1 f_n(x)dx=0$ then closure of $S_k$ has an empty interior .
But the collection $S_k$ may not be closed . Though the union of these collection gives the whole space $[0,1]$ but it is Baire space, then some contradiction may arise. Any help is greatly appreciated .
Just take $$ f_n(x) = \sin(2\pi x + n) $$ Then all the $f_n$ are continuous, $∫_0^1 f_n = 0$ but $f_n(x)$ never converges for whatever $x\in[0,1]$.