In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
That's a fine argument assuming the function is measurable. But what if it's not?
So there is an extension P′ of P such that P′-almost surely the dumb strategy works. Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives.
http://www.mdpi.com/2073-8994/3/3/636 Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis by Paul Bartha Symmetry 2011, 3(3), 636-652; 0554現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金) 12:00:58.15ID:+qJdNaLm>>553 DR Pruss氏は下記で、conglomerabilityの正確な意味がいまいち分からんけど 要するに”nonmeasurable”で、測度論的確率から外れているということでしょう （＾＾；
https://en.wikipedia.org/wiki/Alexander_Pruss Alexander Robert Pruss (born January 5, 1973) is a Canadian mathematician, philosopher, Professor of Philosophy and the Co-Director of Graduate Studies in Philosophy at Baylor University in Waco, Texas.
Pruss graduated from the University of Western Ontario in 1991 with a Bachelor of Science degree in Mathematics and Physics. After earning a Ph.D. in Mathematics at the University of British Columbia in 1996 and publishing several papers in Proceedings of the American Mathematical Society and other mathematical journals,[4] he began graduate work in philosophy at the University of Pittsburgh.
https://books.google.co.jp/books?id=RXBoDwAAQBAJ&;pg=PA77&dq=Pruss+conglomerability&hl=ja&sa=X&ved=0ahUKEwjplc_N8qfoAhWJ7WEKHXwVDuoQ6AEIKDAA#v=onepage&q=Pruss%20conglomerability&f=false Infinity, Causation, and Paradox 著者: Alexander R. Pruss （P76-77 に conglomerabilityの説明があるが、正確な定義は分からないが、 P76に”But typically, where there is no coutable additibity, there is lack of conglomerability(Scervish,Seidenfeld,and Kanade 1984).” と記されているので、”coutable additibity ”即ち σ-加法性 と密接に関連した（多分”σ-加法性”を拡張した）概念だと思う） （更に附言すれば、現代の測度論的確率が、σ-加法性をベースに成立っているとすれば、DR Pruss氏の指摘は、要するに”nonmeasurable”で、測度論的確率から外れているということでしょう （＾＾； ） 0555現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金) 12:52:30.11ID:+qJdNaLm>>554 （補足） 本 「Infinity, Causation, and Paradox」 著者: Alexander R. Pruss は、2018発行な
さて、 conglomerabilityは、下記の Reliability and Risk: A Bayesian Perspective 2006 では、ベイズ理論で、σ-加法性の代わりに使われる概念みたいだな
https://books.google.co.jp/books?id=szLqBTXsyJQC&;pg=PA23&dq=Pruss+conglomerability&hl=ja&sa=X&ved=0ahUKEwjplc_N8qfoAhWJ7WEKHXwVDuoQ6AEIMzAB#v=onepage&q=conglomerability&f=false Reliability and Risk: A Bayesian Perspective 2006 著者: Nozer D. Singpurwalla
P13 conglomerabilityの公理というのがあって コルモゴロフの確率公理のσ-加法性の代わりに 導入されたものらしい。 詳しくは、2.5, 5.2.3, 5.3.2を見ろってことらしい 0556現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/20(金) 14:16:36.13ID:+qJdNaLm>>554 （補足） 本 「Infinity, Causation, and Paradox」 著者: Alexander R. Pruss は、2018発行な
conglomerabilityは、下記の Reliability and Risk: A Bayesian Perspective 2006 では、ベイズ理論で、σ-加法性の代わりに使われる概念みたいだな（下記）
（参考） https://books.google.co.jp/books?id=szLqBTXsyJQC&;pg=PA23&dq=Pruss+conglomerability&hl=ja&sa=X&ved=0ahUKEwjplc_N8qfoAhWJ7WEKHXwVDuoQ6AEIMzAB#v=onepage&q=conglomerability&f=false Reliability and Risk: A Bayesian Perspective 2006 著者: Nozer D. Singpurwalla P13 conglomerabilityの公理というのがあって コルモゴロフの確率公理のσ-加法性の代わりに 導入されたものらしい。 詳しくは、2.5, 5.2.3, 5.3.2を見ろってことらしい（本を買う気がないのでスルー（大学生なら図書に入れてもらえば良い））
（なお、検索で見つけた論文では、下記が一番 conglomerabilityについて詳しい） http://www-dimat.unipv.it/~rigo/ Pietro Rigo Dipartimento di Matematica “F. Casorati”Universita di Pavia - Italia http://www-dimat.unipv.it/~rigo/cong.pdf International Journal of Approximate Reasoning, 88, 387-400. Basic ideas underlying conglomerability and disintegrability June 16, 2017 Abstract The basic mathematical theory underlying the notions of conglomerability and disintegrability is reviewed. Both the precise and the imprecise cases are concerned. 0557１３２人目の素数さん2020/03/21(土) 00:35:09.51ID:6p6Apyjd>>550 > ”決定番号d が存在して、d+1番以降のしっぽの箱から同値類E→代表数列rのrd=Xd”のところが怪しいと分かる > ・つまり、”そのような有限の決定番号dが存在する”というところが、数学的に怪しい雰囲気だってことです
どうしようもないな おサル 0572１３２人目の素数さん2020/03/21(土) 20:25:18.44ID:XWnhFsyt>>571 ＞底辺卒 Sランクということになってますね https://school-navi.org/university/rank/#SSSSS Sの個数については言及しません 0573現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/21(土) 20:28:59.97ID:gPebnXHG>>568 おサルさー、おまえ mathoverflowの DR Pruss氏議論が分かっていない 質問者 Denis氏そっくりの理解じゃんかｗ(゜ロ゜； DR Pruss氏は、”That's a fine argument assuming the function is measurable. But what if it's not?”ってあるよね
で、質問者 Denis氏は、この議論には、全く入れなかった ただ、壊れたレコードのように ”Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. ? Denis Dec 17 '13 at 15:21” を繰返したのだった（＾＾；
In general, Mj will be nonmeasurable (one can prove this in at least some cases). We likewise have no reason to think that M is measurable. But without measurability, we can't make sense of talk of the probability that the guess will be correct.
That's a fine argument assuming the function is measurable. But what if it's not?
So there is an extension P′ of P such that P′-almost surely the dumb strategy works. Just let P′ be an extension on which the set of representatives has measure 1 and note that the dumb strategy works on the set of representatives.
http://www.mdpi.com/2073-8994/3/3/636 Symmetry and the Brown-Freiling Refutation of the Continuum Hypothesis by Paul Bartha Symmetry 2011, 3(3), 636-652; 0574現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/21(土) 20:38:49.81ID:gPebnXHG>>572 Sランク？ サイテーの”S”だろｗ
（>>553より参考） https://mathoverflow.net/questions/151286/probabilities-in-a-riddle-involving-axiom-of-choice Probabilities in a riddle involving axiom of choice Denis氏 Dec 9 '13 DR Pruss氏 (抜粋) show 6 more comments 1)Our choice of index i is made randomly, but for this we only need the uniform distribution on {0,…,n}. It is made independently of the opponent's choice. ? Denis Dec 17 '13 at 15:21 2)I was assuming that "independently" has the meaning it does in probability theory (P(AB)=P(A)P(B) and generalizations for σ-fields). But that does require a probabilistic description of the opponent's choice. Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf ) Still, mixing the probabilistic with nonprobabilistic concepts might lead to some difficulties, though. ? Alexander Pruss Dec 18 '13 at 15:21 3)ah ok I see where the misunderstanding comes from, it's true that "independently" is ambiguous, because only one random variable is involved here. But I think it still has a mathematical meaning in the sense "it does not depend on the opponent's choice", namely we have ∃x∀y where x is our strategy and y is our opponent's strategy (i.e. the sequence), and we still win this game because we can choose devise a (probabilistic) strategy that works on all sequences. ? Denis Dec 19 '13 at 11:54
4)What we have then is this: For each fixed opponent strategy, if i is chosen uniformly independently of that strategy (where the "independently" here isn't in the probabilistic sense), we win with probability at least (n?1)/n. That's right. But now the question is whether we can translate this to a statement without the conditional "For each fixed opponent strategy". ? Alexander Pruss Dec 19 '13 at 15:05 5)How about describing the riddle as this game, where we have to first explicit our strategy, then an opponent can choose any sequence. then it is obvious than our strategy cannot depend on the sequence. The riddle is "find how to win this game with proba (n-1)/n, for any n." ? Denis Dec 19 '13 at 19:43 6)But the opponent can win by foreseeing what which value of i we're going to choose and which choice of representatives we'll make. I suppose we would ban foresight of i? ? Alexander Pruss Dec 19 '13 at 21:25 7)yes the order would be: 1)describe the probabilistic strategy 2)opponent choses a sequence 3)probabilistic variable i is instanciated ? Denis Dec 19 '13 at 23:02
(引用終り) 0588現代数学の系譜 雑談 古典ガロア理論も読む ◆e.a0E5TtKE 2020/03/22(日) 10:02:49.23ID:TMbOZsnt>>586 >Of course, one could mean "independently" here in some non-mathematical causal sense. (And there may be philosophical reason for doing this: fitelson.org/doi.pdf ) (補足) http://fitelson.org/doi.pdf Synthese ・ September 2014?137 (3), 273-323 Declarations of Independence Branden Fitelson and Alan Hajek Abstract According to orthodox (Kolmogorovian) probability theory, conditional probabilities are by definition certain ratios of unconditional probabilities. As a result, orthodox conditional probabilities are regarded as undefined whenever their antecedents have zero unconditional probability. This has important ramifications for the notion of probabilistic independence. Traditionally, independence is defined in terms of unconditional probabilities (the factorization of the relevant joint unconditional probabilities). Various “equivalent” formulations of independence can be given using conditional probabilities. But these “equivalences” break down if conditional probabilities are permitted to have conditions with zero unconditional probability. We reconsider probabilistic independence in this more general setting. We argue that a less orthodox but more general (Popperian) theory of conditional probability should be used, and that much of the conventional wisdom about probabilistic independence needs to be rethought. https://www.researchgate.net/publication/266136441_Declarations_of_Independence 同上 http://fitelson.org/ Branden Fitelson is Distinguished Professor of Philosophy at Northeastern University. Before teaching at Northeastern, Branden held teaching positions at Rutgers, UC-Berkeley, San Jose State, and Stanford and visiting positions at the Munich Center for Mathematical Philosophy at LMU-Munich (MCMP @ LMU) and the Institute for Logic, Language and Computation at the University of Amsterdam (ILLC @ UvA). 以上
