If E is measurable and f a nonnegative measurable function on E, then for every $0 \leq t \leq \int_E f < \infty$, there exists a measurable set $A \subseteq E$ with $\int_A f = t$.
I can't figure out the right way to start this problem. It seems like I could work from constant functions to simple functions and use some kind of limiting argument on those to get general nonnegative functions if I knew a similar property held for the Lebesgue measure (specifically, if $|E| = M$ then for every $0 \leq t \leq M$ there exists a subset A of E with $|A| = t$), and I think that such a property does hold, but it's not given and I can't figure out how to prove that, either.
Is there an easier way to do this than trying to work up from case of simple functions? Or, is there a way to handle the simple-function case without assuming the property holds for the Lebesgue measure?
Consider the function
$$ F: \Bbb{R}\to [0,\int_E f\, dt], t\mapsto \int_{E \cap (-\infty,t)} f(s)\, ds. $$
I leave it to you (as a very nice exercise in using convergence theorems) to show that this map is increasing and continuous with $F(x)\to 0$ for $t\to-\infty$ and $F(t)\to \int_E f \, dx$ as $t\to\infty$.
Now apply the intermediate value theorem.