# Is there non Lebesgue measurable set in $I^2$ that its every section has measure $0$?

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.

• 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 $[0,1]^2$. Mar 6 '16 at 19:49
• @GyroGearloose, Cartesian product of two measurable set is always measurable. 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.

Example.

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).

• +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 :) Mar 6 '16 at 20:18