# Do Borel subsets of the plane with null sections have Borel projections?

This might be a very easy question, and it might be better for mathstackexchange in which case I apologize. I'm stuck on something an anonymous referee wrote to me about a paper of mine and I'm hoping for some clarity.

Suppose $$X$$ and $$Y$$ are Polish spaces and $$A \subseteq X \times Y$$ is Borel. It's well known that if for each $$y \in Y$$, $$A_y =\{x \; | \; (x, y) \in A\}$$ is "small" in various sense, then in fact $$proj_X(A)$$ is Borel. For instance, in Kechris' Classical Descriptive Set Theory Theorem 35.46 proves that if $$A$$ has $$\mathcal K_\sigma$$ sections then $$proj_X(A)$$ is Borel. My question is simply, what if the sections are null for (some appropriate version of) Lebesgue measure?

For the specific case I'm interested in, let $$\mu$$ denote (any choice of formulating) the Lebesgue measure on $$\omega^\omega$$ and let $$f:\omega^\omega \to \omega^\omega$$ be a partial Borel function with $$\mu$$-null domain. Is the domain of $$f$$ Borel?

Thanks!

• What is the Lebesgue measure on $\omega^\omega$? Simply any (finite?) measure on the Borel sets of $\omega^\omega$? – Dieter Kadelka Sep 26 at 14:54
• What if I simply take an uncountable null closed subset $C \subset X$? Since $C$ is again an uncountable Polish space, I should be able to find a set $A \subset C \times Y$ whose projection onto $C$ is not Borel, and now by embedding this into $X \times Y$ I should have the desired counterexample. Every section of $A$ is contained in $C$ and therefore null in $X$. – Nate Eldredge Sep 26 at 15:07
• @DieterKadelka The one I learned is gotten by letting the measure of the basic open set $$\{f: \sigma\prec f\}$$ be $$\prod_{i<\vert\sigma\vert}2^{-\sigma(i)-1}$$ for each $\sigma\in\omega^{<\omega}$. – Noah Schweber Sep 26 at 15:09
• To the OP, I presume "partial Borel function" means "partial function whose graph is Borel"? – Noah Schweber Sep 26 at 15:10
• In the last sentence, do you mean "range" instead of "domain"? – Nate Eldredge Sep 26 at 15:11

Notice that $$\omega^\omega$$ can be embedded to a null subset of itself by sending any sequence $$(a_0,a_1,a_2,\dots)$$ to $$(a_0,0,a_1,0,a_2,0,\dots)$$. So any Borel phenomenon that can happen in $$\omega^\omega$$ can also happen in a null subset.