Does there exist a sequence of continuous functions $f_{n}: [0,1] \rightarrow [0,\infty)$ such that $$\lim_{n \rightarrow \infty }\int_{0}^{1}f_{n}(x)dx = 0$$ but there does not exist any $x \in [0, 1]$ for which the sequence $f_{n}(x)$ converges?
Typewriter sequence is an example such type of sequence of function, but this is not a sequence of continuous function, I have no idea how to construct such type of sequence of function. Give some hints.
It is enough to modify the construction of the typerwriter sequence a bit. For each integer $n \geq 0$ and $0 \leq k < 2^n$, choose a smooth function $g_{n,k}$ such that
$$ \mathbf{1}_{[k/2^n, (k+1)/2^n]}(x) \leq g_{n,k}(x) \leq \mathbf{1}_{[(k-1)/2^n, (k+2)/2^n]}(x). $$
Then we re-enumerate $(g_{n,k})$ in lexicographical order to obtain a 1-D array $(f_n)$, that is,
$$ (f_1, f_2, f_3, f_4, f_5, f_6, f_7, \ldots) = (g_{0,0}, g_{1,0}, g_{1,1}, g_{2,0}, g_{2,1}, g_{2,2}, g_{2,3}, \ldots) $$
Then $(f_n)$ is smooth, and $(f_n(x))_{n\in\mathbb{N}}$ attains both $0$ and $1$ infinitely often for each $x \in [0, 1]$. Moreover, it is clear that the integral of $f_n$ converges to $0$.