Assume AC. Suppose $X$ is a subset of the irrationals (Baire Space) for which neither player has a winning strategy (i.e. the game $G(\omega, X)$ is not determined). Is $X$ non-measurable in the Lebesgue sense as a subset of $\mathbb{R}$?



Non-measurable sets and Determinacy...
(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.


