TITLE: Intersection of two 'huge' sets in the plane
QUESTION [9 upvotes]: Consider two sets on the plane $A=\mathbb{Q}\times \mathbb{R}$ and $B=\mathbb{R}\times \mathbb{Q}$.
We know that $A\cap B=\mathbb{Q}\times \mathbb{Q}\neq\emptyset$. What about the general cases?
That is, will $A\cap B\neq\emptyset$ if $A,B\subset\mathbb{R}^2$ satisfy that 

each vertical fiber $A_y$ of $A$ is dense in $\mathbb{R}\times y$, 
each horizontal fiber $B_x$ of $B$ is dense in $x\times\mathbb{R}$?


What if we replace the assumption by
a. each vertical fiber $A_y$ of $A$ is of positive volume, 
b. each horizontal fiber $B_x$ of $B$ is of positive volume?
The only case that I know is if the positive volume assumption is replaced by full vulmue (Fubini theorem).
Thanks!

REPLY [2 votes]: A simple way to construct sets with the required properties is to first pick $U,V\subseteq\mathbb{R}$ and set
$$
\begin{align}
&A = \left\{ (x,y)\in\mathbb{R}^2\colon x+y\in U\right\},\\
&B = \left\{ (x,y)\in\mathbb{R}^2\colon x+y\in V\right\}.
\end{align}
$$
Then the vertical fibres $A_y$ are all translates of $U$ and the horizontal fibres $B_x$ are translates of $V$ (and, similarly, for the horizontal fibres of $A$ and vertical fibres of $B$). Choosing $U,V$ disjoint, then $A,B$ are also disjoint. You can take $U=\mathbb{Q}$ and $V=\mathbb{R}\setminus\mathbb{Q}$ for the first example and $U=(0,1)$, $V=(1,2)$ for the second. Or, if you prefer, take $U=\left((0,1)\setminus\mathbb{Q}\right)\cup\left(\mathbb{Q}\setminus(0,1)\right)$ and $V=\mathbb{R}\setminus U$ to give a simultaneous counterexample to both.
If you want to venture into non-measurable sets, which requires the axiom of choice, then taking $U$ to be any subset of $\mathbb{R}$ with full outer measure and zero inner measure (e.g., a Vitali set) and $V=\mathbb{R}\setminus U$ then $A$ and $B$ (as a subset of $\mathbb{R}^2$) together with their horizontal and vertical fibres (as subsets of $\mathbb{R}$) will have full outer measure, but have zero inner measure. In fact, again using the axiom of choice, its possible to find uncountably many pairwise disjoint sets whose horizontal and vertical fibres all have full outer measure.