19/06/06 23:46:32.04 2NTuckfC.net
スレ64 ｽﾚﾘﾝｸ(math板:838番)
古典ガロア理論も読む 2019/05/05
>正しくは
>「数列の測度が必要だが、その場合
> 選択公理で（非可測集合ができるから）台無しになる」
ご指摘ありがとう
これか？（＾＾
URLﾘﾝｸ(mathoverflow.net)
Probabilities in a riddle involving axiom of choice edited Dec 9 '13 at 16:32 asked Dec 9 '13 at 16:16 Denis
3 Answers
11 edited Dec 12 '13 answered Dec 11 '13 Alexander Pruss
(抜粋)
But we have no reason to think the event of guessing correctly is measurable with respect to the probability measure induced by the random choice of sequence and index i, and we have no reason to think that the conglomerability assumption is appropriate.
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).
Assume CH. Let < be a well-order of [0,1].
Suppose X and Y are i.i.d. uniformly distributed on [0,1].
Consider the question of which variable is bigger. Fix a value y∈[0,1] for Y.
Then P(X<=y)=0, since there are only countably many points <- prior to y.
By a conglomerability assumption, we could then conclude that P(X<=Y)=0, which would be absurd as the same reasoning would also show that P(Y<=X)=0. The argument fallaciously assumes conglomerability.
We are neither justified in concluding that P(X<=Y)=0, nor that {X<=Y} is measurable (though for each fixed y, {X<=y} is measurable).
And indeed it's not measurable: for were it measurable, we could use Fubini to conclude that it has null probability.
Note that one can repeat the argument without CH but instead using an extension of Lebesgue measure that assigns null probability to every subset of cardinality <c, so clearly there is no refutation of CH here.
(引用終り)