Is it provable in ZFC that there is a subset of the plane all of whose vertical cross sections have Lebesgue measure zero and all of whose horizontal cross sections are complements of sets of Lebesgue measure zero?


There are such sets in models in which every set of reals of cardinality less than the continuum is Lebesgue measurable.



Witness to a failure of Fubini/Tonelli
No. Suppose $m:\mathcal{P}([0, 1]) \to [0, 1]$ is a total extension of Lebesgue measure. Let $A \subseteq [0, 1]^2$ be such that every vertical section is Lebesgue null.


Claim: $A$ is $m \otimes \mu$-null.


Proof: For each rational $e>0$, let $U\_{e, x} \subseteq [0, 1]$ be an open set of measure less than $e$ that contain the vertical section $A\_x$. Let $G = \{(x, y) : (\forall e > 0)(y \in U\_{e, x})\}$. For each rational interval $J$, let $X\_{e, J} = \{x : J \subseteq U\_{e, x}\}$. Then $G$ belongs to the sigma algebra generated by rectangles of the form $X\_{e, J} \times J$ so $A$ is $m \otimes \mu$-null.


It follows that $\mu$-almost every horizontal section of $[0, 1]^2 \setminus A$ is not Lebesgue null. Note that this argument only requires the following: Every countably generated sigma algebra extending the Borel algebra admits a measure extending Lebesgue measure. A theorem of Carlson says that this holds in the random real model. 


