Consistency strength of Turing measurability

Solution 1:

What you mean is that all Turing-invariant sets of reals (that is, if $x\in A$ and $x\equiv_T y$, then $y\in A$) are Turing measurable.

This is equiconsistent with $\mathsf{AD}$, that is, with $\mathsf{ZFC}+$ "There are infinitely many Woodin cardinals". The equiconsistency was proved by Woodin in the $80$s, but remains unpublished -- in the last few years, most of the results from that period have appeared in one way or another. This is one of the few that did not.

He showed that, in $L(\mathbb R)$, under $\mathsf{ZF}+\mathsf{DC}$, Turing determinacy (that is, "all Turing invariant sets of reals are determined", which is easily seen to be equivalent to Turing measurability) implies full determinacy.

Woodin has improved the result over the years. The strongest version I know of, from the early $2000$s, I believe, is that in $\mathsf{ZF}$, Turing determinacy implies that every Suslin set is determined. This gives the result in $L(\mathbb R)$ by reflection.

I looked at this a few years ago, when thinking about a different problem, on Galvin's multiboard determinacy. Richard Ketchersid and I noted that in $\mathsf{ZF}$, $\omega$-board determinacy is incompatible with choice, and equiconsistent with full determinacy, because it implies Turing measurability. I do not know of a direct proof that avoids the detour through Martin's result. (In the $\kappa$-board game, a set $A\subseteq\omega^\omega$ is fixed, and players I and II play the usual game on $A$ simultaneously on $\kappa$ many boards. Player I wins iff they win the usual game in at least one of the boards, else II wins.)