Give an example of continuous functions $f_n$ for which $\lim_{n \to \infty} f_n(x)=0$, but $\int_0^1 f_n(x) \ dx$ doesn't have a limit. [duplicate]

Nope, not true, even for everywhere pointwise convergence: Consider $$ f_n(x) = \begin{cases} n^2x & 0\leq x\leq 1/n\\ 2n-n^2x & 1/n<x\leq 2/n\\ 0 & x>2/n \end{cases}. $$ (Note that all $f_n$ are continuous…). Also look up Lebesgue's dominated convergence theorem.

An easy counterexample in the case of almost everywhere convergence, since I've already written it. Dirk took care of the question better.

I also recommend to read something like this Tao's post about modes of convergence, to get a better idea of what's going on.

Take $f_n(x)=n\chi_{[0,\frac{1}{n}]}(x)$. Then $f_n \to f$ almost everywhere and $$ \int_0^1 |f_n(x)|\,dx=\int_0^1 f_n(x)\,dx=n\int_0^1 \chi_{[0,\frac{1}{n}]}(x)\, dx= n \mu\left(\left[0,\frac{1}{n}\right]\right)=1, $$ for any $n$ and where we denoted by $\mu$ the Lebesgue measure.