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The following question arose in my research on variations on Bell's theorem. I have tried to solve it on my own, but my weak background in measure theory apparently doesn't allow me to do so within a reasonable amount of time.

This is my first post on any SE site. Since the question is probably not research-level, I'm posting it here instead of on MO.

Let $(\Omega_1,\mathcal{F}_1,P_1)$ and $(\Omega_2,\mathcal{F}_2,P_2)$ be probability spaces. The product $\Omega_1\times\Omega_2$ comes equipped with the standard product $\sigma$-algebra and product measure.

If $A\subseteq \Omega_1\times\Omega_2$ is of positive measure, do there exist $B_1\subseteq\Omega_1$ and $B_2\subseteq\Omega_2$ of positive measure such that $B_1\times B_2\subseteq A$?

If this turns out to be false, then what about the same question with $B_1\times B_2\subseteq_{a.s.} A$ instead of exact containment?

Edit: I have accepted @leslie's answer as it resolves the original problem. I still hope for a positive answer to the revised question, where I allow $A$ to be modified by a set of measure zero. Can anyone say anything about this?

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Sorry, the first sentence was supposed to contain a link: en.wikipedia.org/wiki/Bell%27s_theorem –  Saibot May 26 '12 at 8:27
    
I've added the link. –  Zev Chonoles May 26 '12 at 8:29
    
Thanks! By now, I've also found the edit button ;) –  Saibot May 26 '12 at 8:51
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Here is a related question: math.stackexchange.com/questions/42748/… –  Jonas Meyer May 28 '12 at 3:19
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@Jonas: good find, thanks! I haven't read the references you gave there in detail, but now I fear that the answer to my revised question will also be negative. Luckily, what I can prove using the approximation lemma is that for every $\varepsilon>0$, there exists a rectangle $B_1\times B_2$ of positive measure for which at most an $\varepsilon$th part lies outside the original set $A$. This turns out to be enough for my application. –  Saibot May 28 '12 at 10:09

1 Answer 1

up vote 1 down vote accepted

One counterexample is the subset of $[0,1] \times [0,1]$ (with the usual Lebesgue $\sigma$-algebras on both copies of $[0,1]$) given by $$ E = \{(x,y) \in [0,1] \times [0,1]: y - x \not \in \mathbb{Q}\}. $$ It turns out that $E$ has planar measure $1$, but $E$ does not contain any cylinder set of the form $A \times B$ with $A,B$ Lebesgue measurable sets of positive measure. One way to see this is to appeal to the nontrivial but better known fact that if $A$ and $B$ are Lebesgue measurable subsets of $\mathbb{R}$ with positive measure, the difference set $A - B = \{a - b: a \in A, b \in B\}$ must contain a nontrivial open interval (so, in particular, rational numbers). The special case of this assertion when the sets $A$ and $B$ are the same is very well known and apparently originally due to Steinhaus.

I recalled this example from Falconer's The Geometry of Fractal Sets, where it is Exercise 5.4 (and the generalization of Steinhaus's observation is Exercise 1.7).

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Very cool! Yes, I've heard of Steinhaus' theorem before. Can you say anything about the case in which I would allow an enlargement of the original set by a set of measure zero? Then your counterexample fails, and I hope for the revised statement to be correct. –  Saibot May 26 '12 at 9:08
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How can you show that $E$ is measurable and has measure $1$? –  Michael Greinecker May 26 '12 at 11:29
    
The complement of $E$ is a countable disjoint union of lines, and therefore has measure $0$. –  Saibot May 26 '12 at 13:04
    
@Saibot I can't think of any counterexample for the more general problem; it seems genuinely "harder". My guess is that it would not be rejected from MO, and might be answered more quickly there. (For whatever reason, math.SE and MO seem to have increasingly disjoint user bases...) –  leslie townes May 26 '12 at 22:56

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