$f,g: [0,1] \to \mathbb{R}$ with equal Lebesgue integrals

Question

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?

Answer 1

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 $X$."