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Equip $\mathbb R_{\geq 0} \times \Omega$ with the product measure structure of Borel measure for $\mathbb R_{\geq 0}$ and a probability measure for $\Omega$.

Suppose we have a set $S=\cup_{t \geq 0}(t,E_t)$ all of whose sections $E_t$ are measurable and have probability zero.

Unfortunately, $S$ is potentially nonmeasurable. Can we at least prove that it is contained in a measurable set of zero product measure?

The usual example of a nonmeasurable set on the diagonal of the Borel unit-square fails to form a counterexample since the diagonal is measurable and null.

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No. For instance, let us consider Lebesgue measure on $\mathbb{R}^2$. It can be shown that if $A\subseteq\mathbb{R}^2$ is measurable with positive measure, then there are $2^{\aleph_0}$ different values of $x$ such that $A$ contains a point of the form $(x,y)$. Also, there are only $2^{\aleph_0}$ different Borel subsets of $\mathbb{R}^2$, and any measurable set of positive measure contains a Borel set of positive measure. Using this, by a transfinite recursion of length $2^{\aleph_0}$, you can construct a set $S\subseteq\mathbb{R}^2$ such that $S$ intersects every Borel set of positive measure (and thus every measurable set of positive measure), but for each $x\in\mathbb{R}$, $S$ contains at most one point of the form $(x,y)$ (just one by one choose points to put in $S$ to make it intersect each Borel set of positive measure, avoiding any $x$-values you have already chosen). Then every section of $S$ is measurable with measure $0$. However, $S$ is not contained in any measurable set of measure $0$, or indeed in any measurable set which does not have full measure, since $S$ intersects every measurable set of positive measure.

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    $\begingroup$ Mathematicians are really a whole different breed of humans. Thank you! $\endgroup$ Jan 24 at 0:19
  • $\begingroup$ @YousefKaddoura Constructions like this are a big part of what makes mathematics fun (for me). $\endgroup$ Jan 24 at 1:53

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