# Every section null implies null in product measure

Let $$A\subset[0,1]^2$$ be a set such that every section $$A_x=\{y:(x,y)\in A\}$$ is a null set in $$[0,1]$$. Can we conclude that $$A$$ is a null set in $$[0,1]^2$$?

Some context: It is a standard fact that if $$A$$ is measurable in the product measure, then each section $$A_x$$ is measurable. In the converse direction, it seems that each $$A_x$$ measurable does not imply that $$A$$ is measurable (even in the 2D Lebesgue sense). The question above is about what happens if we replace measurable by null.

• @bof: You are right, if we take $A$ to be Lebesgue measurable in $\mathbb{R}^2$. On the other hand, if we consider the product of 1-dimensional Lebesgue measures in $\mathbb{R}^2$, then every section would be measurable. Commented Dec 1, 2018 at 0:52

Assuming the continuum hypothesis, $$\mathbb R^2$$ can be partitioned into two (nonmeasurable) sets, one with all vertical sections countable, the other with all horizontal sections countable. If you just want the sections to be Lebesgue null sets instead of countable sets, the continuum hypothesis can be replaced by the weaker assumption, that every set of real numbers of cardinality less than $$2^{\aleph_0}$$ is a Lebesgue null set.
of $$\mathbb R$$, we can define a partition of $$\mathbb R^2$$ into two sets $$A$$ and $$B$$ such that every vertical section of $$A$$, and every horizontal section of $$B$$, has cardinality less than the continuum.
• @PhysicalMathematics Assuming the axiom of choice, any set $X$ admits a well-ordering $\lt_\text w$ such that, for each $x\in X$, $|\{y\in X:y\lt_\text wx\}|\lt|X|$. Given such a well-ordering $\lt_\text w$ of $\mathbb R$, let $A=\{(x,y):y\lt_wx\}$. The intersection of $A$ with every vertical line has cardinality $\lt\mathfrak c$ and is countable if the continuum hypothesis holds. Likewise the complement of $A$ has small intersection with every horizontal line.