>>295
>Pruss氏が論文であげた例
>したがって、∫∫[0,1]✕[0,1] f(x,y)dxdyを、
>∫[0,1](∫[0,1]f(x,y)dy）dxとして計算すると
>いかなるx∈[0,1]についてもf(x,y)＝0となるyは可算個だから
>∫[0,1]f(x,y)dyの値は1、したがって∫[0,1]1dx＝1　となる

スレ主です、戻る
補足：下記 >>1 より
https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice
Probabilities in a riddle involving axiom of choice
asked Dec 9 '13 at 16:16 Denis
（Pruss氏）
The probabilistic reasoning depends on a conglomerability assumption, ・・・and we have no reason to think that the conglomerability assumption is appropriate.

（追加引用）
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).
http://www.mdpi.com/2073-8994/3/3/636

（補足追加）
１）ここの”the Brown-Freiling argument”は、直接の箱入り無数目の説明ではないみたい
２）つまり、”the conglomerability assumption”の説明に引き合いに出したものです
３）”the conglomerability assumption”の数学的定義は、見つけられなかった
４）下記 Pruss著 ”Infinity, Causation, and Paradox (Oxford University Press, 2018)”に追加説明がありそう

https://en.wikipedia.org/wiki/Alexander_Pruss
Alexander Robert Pruss (born January 5, 1973) is a Canadian philosopher and mathematician.
Biography
After earning a Ph.D. in mathematics at the University of British Columbia with a dissertation on Symmetrization, Green’s Functions, Harmonic Measures and Difference Equations, under John J. F. Fournier in 1996, and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals, he began graduate work in philosophy at the University of Pittsburgh.
Bibliography
Infinity, Causation, and Paradox (Oxford University Press, 2018)