FOM: Re: Elaboration on not-CH proof:
Martin Schlottmann
martin_schlottmann at math.ualberta.ca
Tue Sep 15 16:40:46 EDT 1998
To my understanding the analysis of the game is not
correct unless we also stipulate a strategy for player I.
Given the other rules of the game, it is completely
irrelevant that I has the first pick.
Let P_I be the probability measure of player I, P_II
the probability measure of player II. If we call
A(r):={r'|r'<r} and B(s):={s'|s'<=s},
then the event that II wins is:
{II wins}={(r,s)|r\in B(s)}={(r,s)|s\not\in A(r)}
which occurs with probability:
(P_I x P_II)( {(r,s)|r\in B(s)} )
= (P_I x P_II)( {(r,s)|s\not\in A(r)} ),
but only in the case that {II wins} is a measurable
set (then we could use Fubini in order to calculate
this probability to be 1, given certain properties of
P_I and P_II).
Therefore, the only thing shown is that certain sets
in the product space are not measurable, a rather
non-surprise.
--
Martin Schlottmann <martin_schlottmann at math.ualberta.ca>
Department of Mathematical Sciences, CAB 583
University of Alberta, Edmonton AB T6G 2G1, Canada
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