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?


  • $\begingroup$ What is the Lebesgue measure on $\omega^\omega$? Simply any (finite?) measure on the Borel sets of $\omega^\omega$? $\endgroup$ Sep 26, 2020 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 $X$. $\endgroup$ Sep 26, 2020 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 $\sigma\in\omega^{<\omega}$. $\endgroup$ Sep 26, 2020 at 15:09
  • $\begingroup$ To the OP, I presume "partial Borel function" means "partial function whose graph is Borel"? $\endgroup$ Sep 26, 2020 at 15:10
  • $\begingroup$ In the last sentence, do you mean "range" instead of "domain"? $\endgroup$ Sep 26, 2020 at 15:11

1 Answer 1


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.

  • $\begingroup$ That's perfect thanks! $\endgroup$ Sep 27, 2020 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.