Let $X$ be a measure space with finite $|X|=\int_X1$ and $f:X\rightarrow \mathbb{R}$ be a function. Under what condition on $X$ and $f$ does there exist a subset $Y \subset X$ satisfying the following?
To prove this kind of statement, we need some completeness of the set $\{Z \subset X \ | \ |Z|=|X|/2\}$, but I don't know how to get such a completion.
Look at the least real r such that the measure of the set Y of those points in X that map to something below r is one half the measure of X.
The "completeness" you ask for is essentially compactness in the weak* topology. Let $S\subset L^\infty(X)$ consist of all functions $g$ such that $\int g=|X|/2$ and $0\leq g\leq 1$. This set is weak*-compact, so for any $f\in L^1(X)$, the functional $g\mapsto \int fg$ is must have a minimum on $S$. Furthermore, by Krein-Milman, this minimum must be achieved at some extreme point of $S$. If $X$ is atomless, it is easy to see that any extreme point of $S$ must be a characteristic function.
(The hypothesis of atomlessness is necessary, of course, since otherwise there might not be any subset of $X$ at all with half measure.)
