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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!

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  • $\begingroup$ What is the Lebesgue measure on $\omega^\omega$? Simply any (finite?) measure on the Borel sets of $\omega^\omega$? $\endgroup$ – Dieter Kadelka Sep 26 at 14:54
  • $\begingroup$ 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 $X$. $\endgroup$ – Nate Eldredge Sep 26 at 15:07
  • $\begingroup$ @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 $\sigma\in\omega^{<\omega}$. $\endgroup$ – Noah Schweber Sep 26 at 15:09
  • $\begingroup$ To the OP, I presume "partial Borel function" means "partial function whose graph is Borel"? $\endgroup$ – Noah Schweber Sep 26 at 15:10
  • $\begingroup$ In the last sentence, do you mean "range" instead of "domain"? $\endgroup$ – Nate Eldredge Sep 26 at 15:11
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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.

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  • $\begingroup$ That's perfect thanks! $\endgroup$ – Corey Switzer Sep 27 at 12:47

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