Non-measurable sets and Determinacy... - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:11:08Z http://mathoverflow.net/feeds/question/67366 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67366/non-measurable-sets-and-determinacy Non-measurable sets and Determinacy... George Lazou 2011-06-09T18:47:33Z 2011-06-10T08:30:51Z <p>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 $\mathbb{R}$?</p> http://mathoverflow.net/questions/67366/non-measurable-sets-and-determinacy/67368#67368 Answer by Joel David Hamkins for Non-measurable sets and Determinacy... Joel David Hamkins 2011-06-09T19:09:26Z 2011-06-09T21:12:08Z <p>(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.)</p> <p>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 $A$.</p> <p>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.</p> <p>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.</p> <p>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.)</p> <p>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.</p> http://mathoverflow.net/questions/67366/non-measurable-sets-and-determinacy/67414#67414 Answer by Matteo Mio for Non-measurable sets and Determinacy... Matteo Mio 2011-06-10T08:30:51Z 2011-06-10T08:30:51Z <p>I have two possible answers. The first is short and possibly not the one you want. The second is probably the right one.</p> <p><strong>First:</strong> 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).</p> <p>I guess this is a consistent proof of "no", to your question.</p> <p><strong>Second</strong> 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.</p> <p>This second example is mentioned in Martin's "Blackwell's determinacy", where it is credited to Greg Hjorth. </p> <p>This is a proof in ZFC of "no", to your quesiton.</p>