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 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 in that (†) implies a negative answer to the latter.)