Are nonmeasurable sets whose sections are null always contained in a null measurable set?

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.

1 Answer

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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