Are "most" subsets of reals non-Lebesgue Measurable? I have an intuition that most subsets of reals are not Lebesgue measurable. The tricky part is defining "most". For that, we need a measure on the set of subsets of the powerset of reals, in other words, a measure on the powerset of the powerset of reals. Has anyone defined such a measure, and either confirmed or denied my intuition?
 A: The following isn't entirely an answer, since you specifically ask about measure and this is an observation about category, but I've come around to think that it's more than a comment (which is what it originally was).

Consider the function
$$d:\mathcal{P}(\mathbb{R})^2\rightarrow\mathbb{R}:(X,Y)\mapsto\mu^*(X\triangle Y),$$ where $\mu^*$ is Lebesgue outer measure and $\triangle$ is symmetric difference. This is a pseudometric since we always have $$(X\triangle Y)\cup (Y\triangle Z)\supseteq X\triangle Z$$
and outer measure is finitely subadditive. In the topology coming from this pseudometric, the set of measurable sets is nowhere dense. So "generic sets aren't measurable." (Separately it's worth noting that measurability is well-defined up to distance zero: if $d(X,Y)=0$ and $X$ is measurable then $Y$ is measurable, since all null sets are measurable. So if we "mod out by distance zero" to get a genuine metric space, we don't lose any information as far as measurability is concerned - which I think helps emphasize the naturality of this approach.)
Of course generic is not the same as random, and I don't see a similar way to put a measure on $\mathcal{P}(\mathbb{R})$ (or anything like a measure) which would help in this context.
