solved, thank you all
Suppose, for a contradiction, that there is a set of positive measure on which f is not zero. Then by Egorov's theorem there is a (maybe smaller) set A on which $\displaystyle f_n(x)\to f(x)$ uniformly. Let g be the function that is equal to f on A and 0 outside A. Then $\displaystyle \int_A|g|^2d\mu>0$.
On the other hand, it follows from Parseval's equality that $\displaystyle \langle g,f_n\rangle\to0$ as n→∞. But $\displaystyle \langle g,f_n\rangle = \int_X g\bar{f_n}\,d\mu = \int_A g\bar{f_n}\,d\mu$ since g is zero off A. This converges to $\displaystyle \int_A |g|^2d\mu$ as n→∞, by the uniform convergence. That gives your contradiction.