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?



Codimension of Measurable Sets
In 1917, Lusin and Sierpinski showed that the unit interval $[0,1]$ can be partitioned into $2^{\aleph\_{0}}$ many pairwise disjoint sets each having Lebesgue outer measure 1; say, $X\_{i}$, $i \in I$. Fix $i\_{0} \in I$.
For each proper subset $J$ with $i\_{0} \notin J \subset I$, let $S\_{J} = \bigcup\_{j \in J} X\_{j}$. Then the sets $S\_{J}$, $i\_{0} \notin J \subset I$, are distinct modulo $\mathcal{M}$.


Nikolai N. Lusin and Waclaw Sierpinski, “Sur une décomposition d’un intervalle en une infinité non dénombrable d’ensembles non mesurables” [On a decomposition of an interval into a nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424.


