Suppose on the contrary that for some there exists, for each n=1,2,3,..., such that but . Let , where . Then for all n, and hence .

On the other hand it follows from Dominated Convergence or some such theorem that . That is a contradiction since clearly |f| cannot have a nonzero integral over a null set.

The idea of this proof comes from the fact that if we define a new measure on by , then . There is then a standard theorem (Theorem 30B in Halmos'sMeasure Theory) saying that absolutely continuous measures have this "continuity" property. I just took Halmos's proof and re-wrote it for this situation.