I am currently teaching an advanced undergraduate analysis class, and the following question came up.

Intuition suggests that "most" subsets of $[0,1]$ are not Lebesgue measurable. However, the power set $\mathcal{P}([0,1])$ has the same cardinality as the collection of measurable sets, so it is not clear how to make this statement precise.

One method is to view $\mathcal{P}([0,1])$ as a vector space over $\mathbb{Z}_2$, with addition corresponding to symmetric difference of sets. Then the measurable sets $\mathcal{M}$ form a subspace, and the quotient $\mathcal{P}([0,1])/\mathcal{M}$ is clearly uncountable.

So my question is: what is the cardinality of $\mathcal{P}([0,1])/\mathcal{M}$? It seems like it should be $2^c$, just like $\mathcal{P}([0,1])$, but I don't know how to prove it.

Also, in what other senses are "most" subsets of $[0,1]$ non-measurable?