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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?

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  • $\begingroup$ Do you know that the lebesgue integral is "absolutetly continuous"? Maybe this property or the proof of this property helps $\endgroup$ – Quickbeam2k1 Mar 5 '15 at 21:39
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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.

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