For subsets $A$ and $B$ of $[0,1)$, say $A\sim B$ iff $\lambda(A\Delta B)=0$ where $\lambda$ is Lebesgue measure. 


**Question:** How many equivalence classes of subsets of $[0,1)$ are there given AC?


I would guess the answer is $2^c$ given AC, but I haven't got a proof. 


What got me thinking about this was trying to find a way to say that there are more nonmeasurable sets than measurable ones. There are, of course, $2^c$ of each, but modulo null sets there are only $c$ measurable ones (at least given AC), so if there were more than $c$ subsets modulo null sets, we could say that modulo null sets there are more nonmeasurable sets than measurable ones.



How many subsets of $[0,1)$ are there modulo null sets?
Yes, there are $2^\mathfrak{c}$ equivalence classes. In fact, I claim that there is a collection $S$ of $\mathfrak{c}$ disjoint non-null subsets of $[0,1)$; taking all unions of subcollections of $S$ gives $2^\mathfrak{c}$ inequivalent subsets of $[0,1)$.


To construct this $S$, let $N$ be the set of all null Borel sets and enumerate $N\times\mathfrak{c}$ with order type $\mathfrak{c}$. Using this enumeration, define a function $f:[0,1)\to\mathfrak{c}$ by induction such that for each $(n,\alpha)\in N\times\mathfrak{c}$, $f^{-1}(\{\alpha\})\not\subseteq n$. We can do this because at each stage of the induction, we have defined $f$ at fewer than $\mathfrak{c}$ points, and the complement of $n$ has cardinality $\mathfrak{c}$. The collection $S=\{f^{-1}(\{\alpha\})\}\_{\alpha\in\mathfrak{c}}$ then consists of $\mathfrak{c}$ disjoint non-null sets.




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Here is an earlier version of my answer, which is a slight variant on a standard argument that nonmeasurable sets exist and shows there must be more than $\mathfrak{c}$ equivalence classes, but does not show there must be $2^\mathfrak{c}$ of them. Let $S$ be any collection of $\mathfrak{c}$ subsets of $[0,1)$; we will find a subset $A\subseteq[0,1)$ that is not equivalent to any element of $S$. Let $N$ be the set of all Borel null sets and enumerate $N\times S$ with order type $\mathfrak{c}$. Define the characteristic function of $A$ inductively such that for each $(n,B)\in N\times S$, $A\Delta B\not\subseteq n$; we can do this because at each stage of the induction we have defined the characteristic function at fewer than $\mathfrak{c}$ points and the complement of $n$ has cardinality $\mathfrak{c}$.


