Definition: A function $f: 2^{\omega} \rightarrow 2^{\omega}$ is Turing invariant if $x \equiv_T y \rightarrow f(x)\equiv_T f(y)$.
Question I (under $ZFC$): Let $f: 2^{\omega} \rightarrow 2^{\omega}$ be a Borel function, is there necessarily a set $A$ of positive Lebesgue measure such that $f\restriction A$ is Turing invariant?
Remark: Assuming the axiom of choice, it's easy to construct a non-Borel counterexample, so I suspect that the answer to the question might have something to do with the interplay of choice-pathologies-definability, therefore I'm also interested in the following question:
Question II: Same as the previous question, but in choiceless models where we have regularity properties for all sets of reals (e.g. Solovay's model and $AD$ models).
Let $f:2^{\omega} \to 2^{\omega}$ be projection on even coordinates: $f(x) = x_0$ where $x = x_0 \oplus x_1$. Suppose $A \subseteq 2^{\omega}$ is compact and has positive measure. WLOG, we can assume that for every $x \in A$, $x_0$ is Turing incomparable with $x_1$. Using Lebesgue density, choose $\sigma, \tau \in 2^{k}$, $k < \omega$ such that $A$ has, say, more than $99$ percent measure in $[\sigma \oplus \tau]$. Since $A$ has high density in $\sigma \oplus \tau$, we can choose $x \in 2^{\omega}$ such that both $y = \sigma x_0 \oplus \tau x_1$ and $z = \sigma x_1 \oplus \tau x_0$ are in $A$. But $y, z$ have same Turing degrees while $f(y) = \sigma x_0$ and $f(z) = \sigma x_1$ are not even Turing comparable.