What is an example of a Borel subset $X$ of $\omega^\omega\times\omega^\omega$, such that $$ \forall f\in\omega^\omega\exists g\in\omega^\omega((f, g)\in X)$$ which has no Borel uniformization? That is, there is no Borel set $Y\subseteq\omega^\omega\times\omega^\omega$ such that $Y\subseteq X$ and $$\forall f\in\omega^\omega\exists!g\in\omega^\omega((f, g)\in Y).$$
I know of theorems showing that there do exist Borel uniformizations of $X$ whenever all sections of $X$ are "small" (=countable) or "large" (=containing a perfect set), but beyond that I don't know how to construct such an $X$.
(I suspect this is very easy, and I'm just not seeing something. Oh well!)