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

Solution 1:

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).