As topic says, more precisely, is there set $E \subseteq [0,1]^2$ such that for every $(x,y) \in [0,1]^2$ sets $$E \cap ([0,1]\times \{y\}), \;E \cap (\{x\} \times [0,1])$$ have measure zero in $\mathbb{R}$ but $E$ is not measureable in $\mathbb{R}^2$?

I tried some sets as $V^2$ but it's clearly has measure zero, so $E$ should be more chaotic as set of $(x,y)$ s.t. $\frac{x}{y}\in \mathbb{Q}$. I think set of like this is should be candidate for answer of my question.

  • $\begingroup$ A slightly more general question would be if there are measurable subsets $E_0,E_1\subseteq [0,1]$ such that $E_0\times E_1$ is not even measurable in $[0,1]^2$. $\endgroup$ Mar 6 '16 at 19:49
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    $\begingroup$ @GyroGearloose, Cartesian product of two measurable set is always measurable. $\endgroup$ Mar 6 '16 at 20:19

If $V$ denotes the Vitali set, it's clear to me that $V^2$ is not measurable. And set $\{ (x,y) \in [0,1]^2 : {x \over y} \in \mathbb Q \}$ is hardly chaotic: it's just union of countable many lines.

In fact, there is non-measurable set $E \subseteq [0,1]^2$, such that every its vertical or horizontal section contains at most one point (and hence measurable). This $E$, in other words, is graph of injective function, defined on some subset of $[0,1]$.

$E$ can be constructed by transfinite induction using the fact that every set of positive Lebesgue measure has cardinality of continuum.


Let $c$ be the initial ordinal of continuum cardinality. And let $\{F_\alpha\ : \alpha < c\}$ be a well-ordering of all closed subsets of $[0,1]^2$ with positive measure. We will write $F^x$ for $F \cap (\{x\}\times[0,1])$ and $\lambda$ for the Lebesgue measure.

On each inductive step $\alpha < c$ we add one element to set of pairs $\{(x_\beta, y_\beta): \beta < \alpha\}$ constructed on previous steps. Note that cardinality of this set is strictly less than continuum, since $c$ is initial ordinal.

Step. By Fubini's theorem set $\{x \in [0,1] : \lambda(F_\alpha^x) > 0\}$ has positive measure and hence contains continuum points. So we can pick $x_\alpha \in [0,1] \setminus \{x_\beta : \beta < \alpha\}$ such that $\lambda(F^{x_\alpha}_\alpha) > 0$ and then take $y_\alpha \in F^{x_\alpha}_\alpha \setminus \{y_\beta : \beta < \alpha\}$.

Finally, $E := \{(x_\alpha, y_\alpha) : \alpha < c\}$. One can check that the sections of $E$ possess at most one point. If $E$ were measurable, its measure by Fubini's theorem would be equal zero. But this is nonsense, because $E$ intersects every closed subset of $[0,1]^2$ of positive measure and hence has full outer measure.

This example may seem too complicated, but I think those constructions of specific non-measurable set always are. Because to build such a set, you need set-theoretic tools, like the axiom of choice (or well-ordering principle, which we have used) whose results can be counter-intuitive.

Another remark: this example can be extended to $E$ being a graph of bijective function on $[0,1]$ by adding extra points at each inductive step.

UPD. The example can even be made universal for all non-atomic Borel measures on $[0,1]$, by taking all uncountable closed subsets of $[0,1]^2$ for $\{F_\alpha\}$. This is due to regularity of Borel measures and «continuum hypothesis» for metric compacta (that is, every uncountable metric compact has continuum cardinality).

  • $\begingroup$ +1 for pointing out that Fubini's theorem requires the function to be measurable as a hypothesis—thus both teaching me something that I should have known, and keeping me from writing a "proof" that your counterexample is impossible :) $\endgroup$ Mar 6 '16 at 20:18

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