Let be a measure space. Prove that if is a Lebesgue integrable function, then for each there exists a such that for each with .
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's Measure Theory) saying that absolutely continuous measures have this "continuity" property. I just took Halmos's proof and re-wrote it for this situation.