20/03/12 18:04:25 FZfOcjPG.net
>>445 補足
DR Pruss氏は、mathoverflowの回答で、下記を述べている
即ち、「the function is measurable.」ならば 良いが、そうでないときは、ダメだという
実際、コイントス（＝coin flips）で、Ω={0,1}^Nなのに、実数の数列の同値類と代表なら、”guess π”とかなって
それって、”Intuitively this seems a really dumb strategy. ”じゃんと、DR Pruss氏は いう （＾＾；
（参考）
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13
DR Pruss氏
(抜粋)
Here's an amusing thing that may help see how measurability enters into these things.
Consider a single sequence of infinitely many independent fair coin flips.
Our state space is Ω={0,1}^N, corresponding to an infinite sequence (Xi)^∞ i=0 of i.i.d.r.v.s with P(Xi=1)=P(Xi=0)=1/2.
That's a fine argument assuming the function is measurable.
If so, then guess according to the representative.
If not, then guess π. (Yes, I realize that π not∈{0,1}.)
Intuitively this seems a really dumb strategy.