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 uniformly. Let g be the function that is equal to f on A and 0 outside A. Then .

On the other hand, it follows from Parseval's equality that as n→∞. But since g is zero off A. This converges to as n→∞, by the uniform convergence. That gives your contradiction.