Existence of Borel uniformization for coanalytic set with non-$K_\sigma$ sections

Question

Suppose that $X$ is a Polish (or standard Borel) space and $\omega^\omega$ is the Baire space of all natural number sequences. My question is: If $A\subseteq X\times \omega^\omega$ is a coanalytic set such that for all $x\in X$, $A_x=\{y\in\omega^\omega:(x,y)\in A\}$ is not $K_\sigma$, must there exist a Borel uniformization for $A$? That is, does there exist a Borel measurable $f:X\to\omega^\omega$ such that for all $x\in X$, $f(x)\in A_x$? (Note that as I want a total function on $X$, being Borel measurable is equivalent to being $\mathbf{\Pi}^1_1$-measurable.)

In other words, I am looking for a $\mathbf{\Pi}^1_1$ "large-section" uniformization result, where the notion of "largeness" is being a non-$K_\sigma$ subset of $\omega^\omega$. (This is not to be confused with the Arsenin-Kunugui "small-section" uniformization result, for Borel sets with $K_\sigma$ sections.)

I believe this would follow from Theorem 36.23 in Kechris's Classical Descriptive Set Theory, if we knew that the $\sigma$-ideal of $K_\sigma$ subsets of $\omega^\omega$ was $\mathbf{\Pi}^1_1$-additive, meaning that whenever $(A_\alpha)_{\alpha<\eta}$ is an ordinal length sequence of $K_\sigma$ sets and the relation $\leq^*$ on $\bigcup_{\alpha<\eta}A_\alpha$ defined by $x\leq^* y$ iff the least $\alpha$ such that $x\in A_\alpha$ is $\leq$ the least $\beta$ such that $y\in A_\beta$, was $\mathbf{\Pi}^1_1$, then $\bigcup_{\alpha<\eta}A_\alpha$ is $K_\sigma$. This holds for both the null and meager ideals, yielding the corresponding uniformization results for coanalytic sets with non-null or non-meager sections (Corollary 36.24 in Kechris).

Any suggestions, references, or related (possibly weaker) results would be appreciated.

Answer 1

No. For let $A\subseteq{^\omega}\omega\times{^\omega}\omega$ be the set of pairs $(x,c)$ such that $c=(f,t)$ where $f\in{^\omega}\omega$ and $t\in{^\omega}2$ codes the theory of $L_\alpha[x]$ (in the language of set theory augmented with a constant symbol for $x$) for some ordinal $\alpha$ such that $\alpha>\omega_1^{\mathrm{ck},x}$, and $L_\alpha[x]$ is pointwise definable. Here let us do the coding $c=(f,t)$ by setting $c(2n)=f(n)$ and $c(2n+1)=t(n)$.

This is a $\Pi^1_1$ set (the main complexity is in saying that the model $M$ coded by $t$ is wellfounded; we can express that $\alpha>\omega_1^{\mathrm{ck},x}$ just by saying that $M$ satisfies "there is an ordinal $\beta$ such that $L_\beta[x]\models$ KP). And note that for every $x$, $A_x$ is non-$K_\sigma$, because given any $c_0\in{^\omega}\omega$, there is $c=(f,t)$ such that $(x,c)\in A$ but $c$ is not eventually dominated by $c_0$; in fact we can arrange that $f(n)=c(2n)=c_0(2n)+1$ for all $n$.

But there is no Borel measurable $F:{^\omega}\omega\to{^\omega}\omega$ which uniformizes $A$, since if $F$ were such, then fixing a real $z$ such that the graph of $F$ is $\Sigma^1_1(\{z\})$, then $F(z)\in\Delta^1_1(\{z\})$, and hence $F(z)\in L_{\omega_1^{\mathrm{ck},z}}[z]$, which clearly contradicts the definition of $A$ (that is, letting $F(z)=c=(f,t)$, we can't have $t\in L_{\omega_1^{\mathrm{ck},z}}[z]$, by construction).