I am given two functions $f, g: [0,1] \rightarrow \mathbb{R}$, with $$\int_{0}^{1} f(x) dx = \int_{0}^{1} g(x) dx = 1,$$ (where the previous integrals are Lebesgue integrals), and I am asked to prove that for every $\alpha \in (0,1)$, there exists a measurable $E \subset [0,1]$, such that $$\int_{E} f(x) dx = \int_{E} g(x) dx = \alpha.$$ I have already proved this, in the special case that $f$, $g$ are simple functions, but I don't know how to generalize it. Could anyone give me a hint?
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If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions. – Xander Henderson Nov 21 '18 at 15:56
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I have already tried to show it following these steps but it fails. Could you please be more specific? – A.M. Nov 22 '18 at 12:00
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@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits. – David C. Ullrich Nov 22 '18 at 17:12
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@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start. – Xander Henderson Nov 22 '18 at 17:23
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I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $\sigma$-field has compact and convex range if $\dim(X)<\infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
David C. Ullrich
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