A set of positive measure contains a product set of positive measure?

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?

-
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
Here is a related question: math.stackexchange.com/questions/42748/… – Jonas Meyer May 28 '12 at 3:19
@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

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.
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