Is the following statement (†) consistent with ZFC?


* If $E \subseteq [0,1]^2$ is such that $E\_x := \{y\in[0,1] : (x,y)\in E\}$ has measure zero for almost all $x$, then $E^y := \{x\in[0,1] : (x,y)\in E\}$ has measure zero for almost all $y$.


Of course I'm not assuming $E$ to be measurable, because that case follows trivially from Fubini's theorem. I know that (†) is not *implied* by ZFC, because it contradicts CH: namely, if $\prec$ is a well-ordering of $[0,1]$ with order type $\omega\_1$ then $E := \{(x,y) \in [0,1]^2 : y\prec x\}$ is such that $E\_x$ is countable for all $x$, and in particular has measure zero, whereas $E^y$ is co-countable for all $y$, and in particular has measure $1$. The consistency of (†) does not seem to follow from strong Fubini theorems such as [this one by Harvey Friedman](https://doi.org/10.1215/ijm/1256047607) because they assume that both iterated integrals exist in order to conclude that they are equal.


(The consistency of (†) would imply a partial reply to [this other question](https://mathoverflow.net/questions/447925/if-a-b-is-a-non-trivial-partition-of-s1-is-it-possible-that-r-thetaa/447929) in that (†) implies a negative answer to the latter.)



Consistency of a strong Fubini type theorem for measure zero sets
ZFC refutes this principle. Let $\kappa=\text{non}(\mathcal{L}),$ i.e. the least cardinality of a set of reals of positive outer measure. Let $X \subset [0,1]$ be such that $|X|=\kappa$ and $\lambda^\*(X)>0,$ and let $\prec$ be a well-ordering of $X$ of order type $\kappa.$ Then $E:=\{(x,y) \in X^2: y \prec x\}$ is a counterexample.


