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Suppose that $A$ is a non-determined set in Baire space (although perhaps my argument is a little easier to see for Cantor space, using the usual coin-flipping measure), and we interpret player I as trying to play into $A$. Let's build a new set $B$ from $A$ by inserting a $0$ into every other play by player I, and ignoring the next move by player II. That is, we insert a round of dummy moves between each real round of play, insisting in the dummy moves that I play $0$ and II can play anything. Thus, I wins the game for B by making sure to play a $0$ in the dummy rounds, and otherwise paying attention only to the real moves of player II. Player II wins by ignoring the dummy moves, and otherwise playing as in $A$. Thus, the game $B$ is not determined, since from a strategy for $B$ for either player we could make a strategy for $A$ by inventing appropriate dummy moves. But meanwhile, because of all those inserted $0$s, every sequence in $B$ has every forth digit $0$, and so the measure of $B$ is zero, and so $B$ is a measurable set.

This argument does not seem to use AC, but only the existence of a non-determined set.

(My argument is somewhat easier if you consider games where the players play $0$s and $1$s, so that the payoff set is in Cantor space $2^\omega$, and we use the usual coin-flipping probability measure; but an essentially similar idea works in Baire space.)

For any game with payoff set $A$, where player I wins if the play is in $A$, consider the following slightly modified game $A^\ast$, which is just like $A$, except we insert a pair of dummy moves between each pair of actual moves, and insist that player I play a $0$ in this dummy round, while player II can play anything. Thus, a sequence or play is in the payoff set $A^\ast$ if indeed that sequence shows that player I did play a $0$ in all the dummy rounds (so every fourth digit is $0$), and furthermore, if we omit the dummy rounds entirely from the sequence, we get a sequence in $A$. 

Thus, playing the game $A^*$ is just like playing $A$, except that the play is interrupted for these silly dummy rounds. Note that player I has no incentive not to play a $0$ on those rounds, and player II's plays in the dummy rounds are ignored entirely.

Thus, it is clear that a player has a winning strategy for $A$ if and only if he or she has a winning strategy for $A^\ast$, since we can translate the strategies from $A$ to $A^\ast$ and back again. The dummy rounds really don't change the difficulty of winning the game. 

But the point now is that because every fourth digit of $A^\ast$ is $0$, it follows that $A^\ast$ has measure $0$. (Every time you insist that a particular digit is $0$, it cuts the measure in half again.)

The conclusion, therefore, which does not use the axiom of choice, is that if there is a non-determined set, then there is a non-determined set with measure $0$. In particular, there is a non-determined set that is measurable.

Suppose that $A$ is a non-determined set in Baire space, and we interpret player I as trying to play into $A$. Let's build a new set $B$ from $A$ by inserting a $0$ into every other play by player I, and ignoring the next move by player II. Thus, I wins the game for B by making sure to play a $0$ every other time, and otherwise paying attention only to every other move of II. Player II wins by ignoring every other move, and otherwise playing as in $A$. Thus, the game $B$ is not determined, since from a stratwgy for $B$ we could make a strategy for $A$. But meanwhile, because of all those inserted $0$s, the measure of $B$ is zero, and so $B$ is a measurable set.

This argument does not seem to use AC, but only the existence of a non-determined set.

Suppose that $A$ is a non-determined set in Baire space (although perhaps my argument is a little easier to see for Cantor space, using the usual coin-flipping measure), and we interpret player I as trying to play into $A$. Let's build a new set $B$ from $A$ by inserting a $0$ into every other play by player I, and ignoring the next move by player II. That is, we insert a round of dummy moves between each real round of play, insisting in the dummy moves that I play $0$ and II can play anything. Thus, I wins the game for B by making sure to play a $0$ in the dummy rounds, and otherwise paying attention only to the real moves of player II. Player II wins by ignoring the dummy moves, and otherwise playing as in $A$. Thus, the game $B$ is not determined, since from a strategy for $B$ for either player we could make a strategy for $A$ by inventing appropriate dummy moves. But meanwhile, because of all those inserted $0$s, every sequence in $B$ has every forth digit $0$, and so the measure of $B$ is zero, and so $B$ is a measurable set.

This argument does not seem to use AC, but only the existence of a non-determined set.

The usual proof that there are non-determined sets is a diagonalization against all possible strategies, and in this proof one can ensure that the resulting non-determined set is contained within, say, the Cantor set, which has measure zero. Thus, one can find a non-determined set that has measure zero and in particular is measurable.

Suppose that $A$ is a non-determined set in Baire space, and we interpret player I as trying to play into $A$. Let's build a new set $B$ from $A$ by inserting a $0$ into every other play by player I, and ignoring the next move by player II. Thus, I wins the game for B by making sure to play a $0$ every other time, and otherwise paying attention only to every other move of II. Player II wins by ignoring every other move, and otherwise playing as in $A$. Thus, the game $B$ is not determined, since from a stratwgy for $B$ we could make a strategy for $A$. But meanwhile, because of all those inserted $0$s, the measure of $B$ is zero, and so $B$ is a measurable set.

This argument does not seem to use AC, but only the existence of a non-determined set.