違うかも知れないが、検索ヒットと他にめぼしいヒットがないので下記を貼る （下記だと、cは連続濃度の意味だね） https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable (抜粋) Is a random subset of the real numbers non-measurable? Is the set of measurable sets measurable? edited Nov 29 '12 at 22:06
19 answered Jul 16 '12 The answer to your second question (assuming the axiom of choice, to dodge Asaf's comment) is that 2^R/Σ has dimension 2^c, where c=2^?0 is the cardinality of the continuum. The main ingredient of the proof is a partition of [0,1] into c subsets, each of which intersects every uncountable closed subset of [0,1]. To get such a partition, first note that there are only c closed subsets of [0,1], so you can list them in a sequence of length (the initial ordinal of cardinality) c in such a way that each closed set is listed c times. Second, recall that every uncountable closed subset of [0,1] has cardinality c. Finally, do a transfinite inductive construction of c sets in c steps as follows: At any step, if the closed set at that position in your list is C and if this is its α-th occurrence in the list, then put an element of C into the α-th of the sets under construction, being careful to use an element of C that hasn't already been put into another of the sets under construction. You can be this careful, because fewer than c points have been put into any of your sets in the fewer than c preceding stages, while C has c points to choose from. At the end, if some points in [0,1] remain unassigned to any of the sets under construction, put them into some of these sets arbitrarily, to get a partition of [0,1].
