Non-measurable sets and Determinacy... 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 $\mathbb{R}$?
 A: (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.
A: I have two possible answers. The first is short and possibly not the one you want. The second is probably the right one.
First: 
take any (co)analytic subset $X$ of the Baire space $\omega^{\omega}$, which is not Borel.
Than it is consistent with ZFC that the game $G(\omega,X)$ is not determined (ZFC just proves the determinacy of Borel games), but $X$ is Lebesgue-measurable (in fact universally measurable).
I guess this is a consistent proof of "no", to your question.
Second
Assume AC. Then there exists a universally-null set (hence Lebesgue null, since the Lebesgue measure is atomless) $X$ of cardinality $\geq\aleph_{1}$.
Then one can show that such a set $X$ can not be a Perfect set. 
Now let us consider the so-called Perfect-set game $PSG(\omega, X)$, which is technically a Gale -Stewart game $G(\omega, Y)$, with $Y$ of about the same complexity of $X$ (in particular $Y$ is universally-null non-perfect set of cardinality $\geq\aleph_{1}$). This game is not determined since $Y$ is not Perfect nor countable by construction, and it is knonw that such a game is determined only if $Y$ is Perfect or countable. Yet $Y$ is universally null, hence Lebesgue measurable.
This second example is mentioned in Martin's "Blackwell's determinacy", where it is credited to Greg Hjorth. 
This is a proof in ZFC of "no", to your quesiton.
