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50(2): 現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/17(土)09:22 ID:cr30r3uy(3/5) AAS
>>49 追加

＜英文資料＞
外部ﾘﾝｸ:mathoverflow.net
mathoverflow
Probabilities in a riddle involving axiom of choice asked Dec 9 '13 at 16:16 Denis

Answers
12 answered Dec 11 '13 at 21:07 Alexander Pruss

A quick way to see that the conglomerability assumption is going to be dubious is to consider the analogy of the Brown-Freiling argument against the Continuum Hypothesis
(see here for a discussion 外部ﾘﾝｸ:www.mdpi.com Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis by Paul Bartha Symmetry 2011, 3(3), 636-652; ).

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. Start with P being the completion of the natural product measure on Ω.
省4

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51(2): 現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/17(土)09:22 ID:cr30r3uy(4/5) AAS
>>50
つづき

外部ﾘﾝｸ:www.ma.huji.ac.il
Sergiu Hart The Hebrew University of Jerusalem
Game Theory
Economic Theory

Some nice puzzles:
Choice Games
外部ﾘﾝｸ[pdf]:www.ma.huji.ac.il
PUZZLES ”Choice Games”Sergiu Hart November 4, 2013
省2

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56(2): 現代数学の系譜 雑談 ◆yH25M02vWFhP 2021/04/17(土)13:21 ID:cr30r3uy(5/5) AAS
>>52
Alexander Pruss氏の前振りの部分だけをつまみ食いするのはいかがか？
氏の結論部分は、はっきりと質問のstrategyを否決しています！（＾＾

なお、Alexander Pruss氏は
（>>50 の”the conglomerability assumption”）
2018年のInfinity, Causation, and Paradox (Oxford University Press, 2018)
で
conglomerability について
P75-202 に記載があります
どうぞ、お読みください
省5

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