TITLE: How to split an integral exactly in two parts
QUESTION [15 upvotes]: This question is a by-product of a conversation with Theo Buehler in comments to this answer. Let's settle definitions.
Definition Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. We say that $X\in \mathcal{F}$ is an atom if $\mu(X) > 0$ and its only subset of strictly positive measure is $X$ itself. $\Omega$ is said to be non-atomic if no atoms exist.
So in a non-atomic measure space we can always split a measurable subset into smaller ones. Now the question is: have we got some control over this splitting? Can we split something in exactly two parts? I'm especially interested in the following.
Question Let $(\Omega, \mathcal{F}, \mu)$ be a non-atomic measure space and $f \in L^1(\Omega), f \ge 0$. Are there measurable and disjoint $A, B \subset \Omega$ such that $A \cup B = \Omega$ and 
$$\int_A f(x)\, d\mu= \int_B f(x)\, d\mu=\frac{1}{2}\int_\Omega f(x)\, d\mu?$$

REPLY [5 votes]: Here's a sketch of the proof that any set of finite measure can be cut into two halves of equal measure. This proof works in the setting of a non-atomic Radon measure defined over a locally-compact Hausdorff space.
We call a (measurable) set proper if it has a positive, finite measure.

Show that every proper set has a proper subset of smaller measure.
Deduce that every proper set has a proper subset of arbitrarily small measure.
Conclude that every proper set $A$ has a subset whose measure lies in $(\mu(A)/3, \mu(A)/2]$.
Deduce that $A$ can be cut into two halves of equal measure.

As Byron shows above, an easy corollary is that there are subsets of $A$ of arbitrary measure in $[0,\mu(A)]$.
There's a nice generalization: let $\mu_1,\ldots,\mu_k$ be measures as above, let $A$ be a set of finite measure (under all of these measures), and let
$$ M = \{(\mu_1(B),\ldots,\mu_k(B)) : B \subset A \text{ is measurable wrt } \mu_1,\ldots,\mu_k\}. $$
Then $M$ is a convex subset of $\mathbb{R}^k$.