Given a set $A \subseteq \omega^\omega$, let $G\_A$ denote the Gale-Stewart game with payoff set $A$ (so player $I$ wants the real built over the course of play to be in $A$ and player $II$ wants it not to be). Notice that the set of (not necessarily winning) strategies for player $I$ is simply the set of functions mapping finite sequences of natural numbers of even length to $\omega$ and the set of (not necessarily winning) strategies for $II$ is the same with "even" replaced by "odd". Thus, via some coding, in both cases if we wanted to, we could think of these sets as homeomorphic copies of Baire space itself. Now suppose that $A$ is determined for one of the players, say $I$. Then in the space of strategies for $I$, let $W(A)$ be the set of winning strategies. My question is about the relationship between the topological complexity of $A$ and $W(A)$. 


Let me give a few naive examples to motivate what I mean. First, let $A = \omega^\omega$. In this case, not only does $I$ have a winning strategy for $G\_A$ but of course any strategy that $I$ plays is winning. Therefore in this case $A \cong W(A) \cong \omega^\omega$. A (very) slightly less trivial example is as follows. Let $n \in \omega$ and let $U\_n$ be the basic open set of all sequences whose first element is $n$. Then of course $I$ wins $G\_{U\_n}$ and the set $W(A)$ is the basic open set of all strategies whose first move is $n$. Therefore again $A$ and $W(A)$ have the same topological complexity. My main question is whether these naive examples are simply naive, or whether there is a general theorem to be mined from it.


**Question 1:** Is there a general theorem dictating the relationship between the topological complexity of $A$ and $W(A)$? What about if we restrict the possible $A$'s to "nice" sets (e.g. the Borels)?


Of course similar a similar question can be asked for measure:


**Question 2:** Is there a general theorem dictating the relationship between the measure theoretic properties of $A$ and $W(A)$? For instance, if $A$ is measurable, then is $W(A)$? What about the converse? Can one of $A$ and $W(A)$ be null and the other have positive measure?


I'm thinking of these questions in the context of ZFC but given the relationship between determinacy and inner model theory I would also be interested to hear if anything more interesting happens if we assume large cardinals or axioms related to games such as $AD^{L(\mathbb R)}$. 



On the topological complexity of the set of winning strategies for Gale-Stewart Games
Abstractly, $W(A)$ is of the form $\forall^{\mathbb R} A \vee \forall^{\mathbb R} \lnot A$; in fact, if $I$ has a winning strategy for $A$, then $W(A)$ is $\forall^{\mathbb R} A$, and if $II$ has a winning strategy for $A$, then $W(A)$ is $\forall^{\mathbb R} \lnot A$. As Joel points out, though, $W(A)$ may be of much simpler complexity than $A$, as $A$ may contain some irrelevant "noise," e.g. $A$ may contain a copy of a Vitali set even though $A$ is determined, so like in most other cases of descriptive set theory it should be a question concerning pointclasses rather than individual sets. For example we now have that if $A$ is Borel, then $W(A)$ is coanalytic, if $A$ is $\Pi^1\_n \cup \Sigma^1\_n$, then $W(A)$ is $\Pi^1\_{n+1}$ or even $\Pi^1\_n$, etc. Does this answer your Question 1?


Concerning your Question 2, all coanalytic sets are Lebesgue measurable, so we get that $W(A)$ is always Lebesgue measurable if $A$ is Borel; also, if all $\Pi^1\_n$ sets are determined, then all $\Pi^1\_{n+1}$ sets are Lebesgue measurable, so that from this hypothesis we get that $W(A)$ is always Lebesgue measurable if $A$ is $\Pi^1\_n$. But as above, $W(A)$ can be pretty simple even though $A$ itself is not contained in a model of determinacy, so there is no converse here. Finally, if $A$ is null, then $W(A)$ is null by Fubini; again, no converse here. 


