Does this hold for product Lebesgue measure? Let $\mathscr M_n$ denote the set of lebesgue measurable subsets of $\Bbb R^n$. Given a set $K\subset \Bbb R^{n+m}$ with the property: 
$\forall x\in \Bbb R^n:  K_x:=\{y\in\Bbb R^m\mid(x,y)\in K\} \in \mathscr M_m$, 
$\forall y\in \Bbb R^m:  K_y:=\{x\in\Bbb R^n\mid(x,y)\in K\} \in \mathscr M_n$,
Is it then the case that $K$ is in $\mathscr M_n\times \mathscr M_m$ ? Or at least in $\mathscr M_{n+m}$ ?
Thanks in advance

Set $m$ and $n$ to $1$.  In the $\mathscr M_n\times \mathscr M_m$ case a counterexample is
$$
K:=\{(x,x)\mid x\in A\}, \qquad A \text{ a non-measurable subset of } \Bbb R.$$
Since the diagonal map $\Delta:{\mathscr M_1}\to {\mathscr M_1}\times{\mathscr M_1}$ is measurable as a map into the product $\sigma$-algebra, $K$ cannot be measurable.  However this set has measure $0$ in $\mathscr M_2$.  
In the $\mathscr M_{n+m}=\mathscr M_2$ case, there is a paper of Sierpiński (Sur un problème concernant les ensembles mesurables superficiellement, Fund. Math. ${\bf 1}$ $(1920)$, $112–115$, here) which gives a counterexample.  This paper describes a subset of ${\Bbb R}^2$ which is not Lebesgue measurable but which intersects every line in at most two points.
