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.)


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.