# measure theory - Codimension of Measurable Sets

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?

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 $\mathcal{M}$.
I think this works to show that $\mathcal{P}/\mathcal{M}$ has cardinality $2^c$. It is a small variation on one of the usual constructions of Lebesgue non measurable sets.
Let $A\subset [0,1]$ be a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ and partition $A$ into two subsets $B$ and $C$ so that the outer measure of both is one-half and $C$ has cardinality $c$. For every subset $D$ of $C$ (maybe you need $C\sim D$ infinite), $B+D$ mod $[0,1]$ is not Lebesgue measurable. Now let $D$ vary over an almost disjoint family of subsets of $C$ that has cardinality $2^c$.