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

The answer is no.
An example pointed out by Gerald Edgar to Harvey Friedman in the late 70s, but probably noted by many before, is: There are permutations of $\mathbb R$ whose graph is non-measurable (but of course, each section -vertical or horizontal- of the graph is measurable). 
I gave another example in an answer a while ago, namely, that no well-ordering of a non-null set of reals is measurable. (So, to be completely explicit, for example under CH, there is a set with countable vertical sections and co-countable horizontal sections which is not measurable.)
A natural follow-up question is whether we can have a Fubini theorem when the function being integrated is not assumed measurable, that is, can we have $$\int\int f(x,y)\,dx\,dy=\int\int f(x,y)\,dy\,dx$$ provided that both integrals exist (but without requiring $f$ to be measurable). This is independent of the axioms of set theory, meaning that it is consistent that the answer is no, and it is consistent that the answer is yes. This was studied first by Harvey Friedman:

Harvey Friedman. A consistent Fubini-Tonelli theorem for nonmeasurable functions, Illinois J. Math. 24 (3), (1980), 390–395. MR0573474 (82b:03097).

In the context of Martin's axiom and in the presence of (atomlessly) real-valued measurabile cardinals, this was studied by J. Shipman. See, for example:

Joseph Shipman. Cardinal conditions for strong Fubini theorems, Trans. Amer. Math. Soc. 321 (2), (1990), 465–481. MR1025758 (91c:03041).

