# Borel set without Borel uniformization

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!)

-
There is a $\Pi_1^0$ (light-face effectively closed) subset of $\omega^\omega \times \omega^\omega$ which can not be uniformized by $\mathbf{\Sigma}_1^1$ (bold-face analytic). This is Exercise 4D.11 in Moschovakis $\textit{Descriptive Set Theory}$ which comes with a hint. (Moschovakis' hints are practically the solution.) –  William Aug 5 '13 at 5:56