Probability of selecting a non-measurable set

Solution 1:

The following topological smallness (rather than measure smallness) result may be of interest.

Let ${\mathbb P}[0,1]$ be the collection of all subsets of $[0,1]$ modulo the equivalence relation $\sim$ defined by $E \sim F \Leftrightarrow {\lambda^{*}(E \Delta F)} = 0,$ where $\lambda^{*}$ is Lebesgue outer measure and $\Delta$ is the symmetric difference operation on sets. The set ${\mathbb P}[0,1]$ can be made into a complete metric space by defining the distance function $d,$ where $d(E,F) = {\lambda^{*} (E \Delta F)}.$ In the paper cited below it is proved that the collection of measurable subsets of $[0,1]$ is a perfect nowhere dense set in ${\mathbb P}[0,1].$

Thus, in this setting, the collection of measurable subsets of $[0,1]$ makes up a very tiny part of the collection of all the subsets of $[0,1].$ Note that in the space ${\mathbb P}[0,1]$ the collection of measurable subsets is not just a first category subset of ${\mathbb P}[0,1]$ (this alone would make the collection a tiny subset of ${\mathbb P}[0,1]$), but in fact the collection of measurable subsets is actually a nowhere dense subset of ${\mathbb P}[0,1]$ (hence my saying the collection is a very tiny subset of ${\mathbb P}[0,1]$).

Nobuyuki Kato, Tadashi Kanzo, and Oharu Shinnosuke, A note on the measure problem, International Journal of Mathematical Education in Science and Technology 19 (1988), 315-318.

Solution 2:

Almost the same Question was asked on mathoverflow: https://mathoverflow.net/questions/102386/is-a-random-subset-of-the-real-numbers-non-measurable-is-the-set-of-measurable

The conclusion was that the set of all measurable sets was not measurable; Therefore, no Probability can be provided.