Witness to a failure of Fubini/Tonelli 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.
 A: In 1980 H. Friedman proved (in A consistent Fubini-Tonelli Theorem) that after adding $\mathfrak{c}^+$-many random reals the equality $\int f dx dy=\int f dy dx$ holds for any non-negative function for which the iterated integrals exist. In particular, a subset of the plane with the required properties does not exist in such model.
A: 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.   
