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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

$$\int_A f(x)\, d\mu= \int_B f(x)\, d\mu=\frac{1}{2}\int_\Omega f(x)\, d\mu?$$

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My answer below was deleted; it runs into annoying problems with subsets of measure zero and fixing it seems to basically require doing what Byron Schmuland suggests. –  Qiaochu Yuan Apr 9 '11 at 20:05
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2 Answers

up vote 10 down vote accepted

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.

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@Jonas T: Thanks, I've fixed it now. –  Byron Schmuland Apr 9 '11 at 19:53
    
+1, very nice answer. –  Willie Wong Apr 9 '11 at 22:52
    
@Byron: This is very nice indeed and I'm thinking at it. However, I'm having some problems with the proof of (2). I understand that the class of measurable $B$ such that $\int_B f d\mu \le \frac{1}{2} \lVert f \rVert$ doesn't reduce itself to $\{\varnothing\}$ (this is because of non-atomic property), but I can't prove that it has some maximal element. This must be a consequence of Zorn's lemma, but how to show that every chain has an upper bound...? –  Giuseppe Negro Apr 10 '11 at 1:07
    
@dissonance I'm not surprised that my hint wasn't too helpful. After posting the answer I realized that the problem is not quite that easy. I have a new answer which might be more useful. In particular, the case with a non-negative integrable $f$ reduces to the original case, by defining a new measure. –  Byron Schmuland Apr 10 '11 at 1:18
    
@dissonance I forgot to say: My apologies! –  Byron Schmuland Apr 10 '11 at 1:19
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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 compact subset of $\mathbb{R}^k$.

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+1 for a different approach. Anyway, I'm curious: are topological properties really necessary? I'm just reading Bogachev's book as Byron suggests and, if I see well, I think we could do without them. Am I right? Maybe topology simplifies the proof? –  Giuseppe Negro Apr 10 '11 at 9:15
    
I think the point of this exercise was that this setting is more general. Anyway, this background is only needed for part (1) of this argument. –  Yuval Filmus Apr 10 '11 at 15:33
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