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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.

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  • $\begingroup$ @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. $\endgroup$
    – timur
    Commented Dec 1, 2018 at 0:52

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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.

This has nothing to do with measure theory, it is just basic set theory: given a well-ordering
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.

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  • $\begingroup$ A bit late here, but could you provide an argument or reference for the claims of the first paragraph? My set theory is lacking. $\endgroup$ Commented Jun 21, 2021 at 18:52
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    $\begingroup$ @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. $\endgroup$
    – bof
    Commented Jun 21, 2021 at 19:30
  • $\begingroup$ Great thanks for the quick reply! $\endgroup$ Commented Jun 21, 2021 at 19:32
  • $\begingroup$ This is a great example. I believe that the quite involved construction given by Joel David Hamkins here shows that just ZFC is sufficient to construct such a set (in fact one such that each horizontal and vertical section is a singleton). $\endgroup$ Commented Dec 10, 2023 at 22:52

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