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 $\mathbb{R}$?
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2 Answers
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(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.
answered Jun 9, 2011 at 19:09
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$\begingroup$ I'm not convinced... Cantor Space is homeomorphic to the Cantor Set which has measure zero... Thus any payoff set in it will have measure zero when considered as a subset of $\mathbb{R}$ by virtue of being a subset of the Cantor Set... I'm not convinced that your argument about halving the measure carries through to non-measurable sets... further my guess (I should probably ask this as a question) is that the existence of a non-determined game in Cantor Space is stronger than AC... $\endgroup$ Jun 9, 2011 at 23:51
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$\begingroup$ I am not using the Cantor set as a subset of $\mathbb{R}$ and the Lebesgue measure on that, but rather, using the natural probability measure on $2^\omega$, for which $2^\omega$ is the whole space and has measure $1$; the measure of the basic open set determined by a finite binary sequence of length $n$ has measure $\frac{1}{2^n}$, like flipping a coin. A subset of $2^\omega$ which is $0$ in every fourth digit has measure $0$ (an easy calculation). The same idea works in Baire space, but you seem to be mapping spaces by homeomorphisms that may not be measure-preserving... $\endgroup$ Jun 10, 2011 at 0:24
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1$\begingroup$ As for your final question, I think you mean to ask whether the existence of a non-determined game is weaker than AC, rather than stronger, since ZFC proves the existence of such non-determined games for Cantor space in exactly the same way that it does for Baire space. I don't expect the existence of such a non-determined set to imply full AC, but I'd have to think a bit more to give a precise model showing this. If this is right, then the existence of non-determined sets would be a weak choice principle. $\endgroup$ Jun 10, 2011 at 0:29
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1$\begingroup$ But I would encourage you to ask that as a question, since perhaps someone knows a good model, and I would be interested to see it. $\endgroup$ Jun 10, 2011 at 2:16
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1$\begingroup$ There is an old beautiful argument of Sierpinski that shows that the axiom of choice for PAIRS $AC_2$ implies the existence of a nonmeasurable set; which in turn shows that $AD$ fails as soon as $AC_2$ is true. So a model in which $AC$ fails but $AC_2$ holds is yet another way of seeing that the negation of $AD$ does not imply $AC$. Cohen's so-called "first (symmetric) model" in which $AC$ fails is such a model (since every set has can be linearly ordered in that model, but the reals cannot be well-ordered there). $\endgroup$ Jun 10, 2011 at 19:14
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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.