How to split an integral exactly in two parts

Solution 1:

I've revised my answer, as I felt the original did a disservice to the OP and the reader by downplaying the difficulty of the question. It is a nice problem!

The OP is correct that a non-atomic probability space is wonderfully flexible, and can be split into pieces with specified probabilities. The crucial fact needed is that we can always split such a space exactly in two. That is, there exists $B\in{\cal F}$ so that $\mu(B)=1/2$. Finding $B$ is not trivial, but not terribly difficult. It is often assigned as an exercise in a measure theory class. An argument using Zorn's Lemma can be found on Wikipedia. Also, it is Exercise 2.17 (page 31) in Probability and Measure (2nd edition) by Patrick Billingsley, and Corollary 1.12.10 (page 56) in Volume 1 of Bogachev's Measure Theory.

Once we have this result, we can create a uniform$(0,1)$ random variable on our space as follows. Start with a set $B(1/2)\in{\cal F}$ with $\mu(B(1/2))=1/2$. Splitting $B(1/2)$ and its complement exactly in two, we get a partition of four sets each with $\mu$ measure of $1/4$. By induction, you can create a collection $B(q)$ indexed by the dyadic rationals $q$ so that $q\leq r$ implies $B(q)\subseteq B(r)$ and $\mu(B(q))=q$. Now define $U(\omega)=\inf (q : \omega\in B(q))$. The random variable $U$ has a uniform$(0,1)$ distribution, i.e. $U$ is a measurable map to Lebesgue space $(\Omega,{\cal F},\mu) \to ((0,1),{\cal B},\lambda)$. Any splitting (or representation) on $(0,1)$ can now be pulled back to $\Omega$ via $U$.

A consequence is that $(\Omega,{\cal F},\mu)$ can serve as the "common probability space" for Skorokhod's representation theorem. In particular, for any probability measure $\nu$ on $\mathbb{R}$, there exists a random variable $Y:\Omega\to\mathbb{R}$ so that the law of $Y$ is $\nu$. Choosing $\nu$ to be a discrete measure shows that you can split $\Omega$ into a finite or countable collection $B_i$ of measurable sets with $\mu(B_i)=p_i$ for any $\sum_i\ p_i=1$ and $p_i\geq 0$.

My answer (and the comments) here Construction of "pathological" measures may also help.

Granted, you don't really need the full power of Skorokhod's result. But it is worth remembering that any non-atomic probability space is "universal" in this sense.

I cannot resist the temptation to also note that the representation extends to Polish spaces. For any probability measure $\nu$ on a Polish space $(S,{\cal B}(S))$, there exists a random variable $Y:\Omega\to S$ on your non-atomic space so that the law of $Y$ is $\nu$.

Finally, about question 2? We either have the trivial case $\int_\Omega f(x) \mu(dx)=0$, or $$\mu_f(B)={\int_B f(x) \mu(dx)\over \int_\Omega f(x) \mu(dx)}$$ defines a new non-atomic probability measure which can split $\Omega$ exactly in half.

Solution 2:

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.

1. Show that every proper set has a proper subset of smaller measure.
2. Deduce that every proper set has a proper subset of arbitrarily small measure.
3. Conclude that every proper set $A$ has a subset whose measure lies in $(\mu(A)/3, \mu(A)/2]$.
4. 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 $\mathbb{R}^k$.