How many subsets of $[0,1)$ are there modulo null sets?

Question

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.

Answer 1

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 $\mathfrak{c}$.

Answer 2

Edit. I had posted this answer to complement Eric's original answer, which showed that the number of classes was at least ${\frak c}^+$, since at that time we didn't quite yet know whether there were $2^{\frak c}$ classes. Afterwards, however, Eric improved his answer to get $2^{\frak c}$ directly. Following the comments, though, I have left this answer up.

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Let me complement Eric's answer by showing that it is relatively consistent to have strictly more than ${\frak c}^+$ many equivalence classes. Indeed, it is relatively consistent with ZFC to have $2^{\frak c}$ many equivalence class, in a case where this is larger than ${\frak c}^+$.

Specifically, I claim that if the continuum hypothesis holds and there is a thick Kurepa tree (an $\omega_1$ tree with $2^{\omega_1}$ many branches), then there are $2^{\omega_1}=2^{\frak c}$ many equivalence classes. Indeed, I shall construct an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

To see this, let $T$ be a thick Kurepa tree, and let $\langle A_\alpha\mid\alpha<\omega_1\rangle$ enumerate the equivalence classes of reals under translation-by-a-rational. Label the $\alpha^{th}$ level of $T$ with the countably many elements of $A_\alpha$. For any path $s$ through $T$, the set $A_s$ of labels appearing on the nodes of $s$ will be a Vitali set, and therefore non-measurable. Further, any two distinct paths $s\neq t$ will have $A_s\cap A_t$ being countable, and so $A_s\not\sim A_t$. Since $T$ is a thick Kurepa tree, we therefore have $2^{\omega_1}$ many branches and thus this many equivalence classes modulo your relation. The collection $\{\ A_s\mid s\in[T]\ \}$ is an almost-disjoint family of $2^{\omega_1}$ many Vitali sets.

Finally, let me explain that it is relatively consistent from an inaccessible cardinal that there is a thick Kurepa tree, yet CH holds and $2^{\omega_1}$ is very large. One way to do this is as follows. Start with $\kappa$ inaccessible in $V$ and $2^\kappa$ very large (by forcing if necessary). Let $V[G]$ be the forcing extension by the Levy collapse, so that $\kappa=\omega_1^{V[G]}$. Consider the tree $T=(2^{<\kappa})^V$ in the model $V[G]$. Since every ordinal less than $\kappa$ was made countable, this has become an $\omega_1$-tree. Yet, since $2^\kappa$ was very large and cardinals $\kappa$ and above were preserved, we have $(2^\kappa)^V$ many branches through this tree. So it is thick.