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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?

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    $\begingroup$ I notice an answer just disappeared. (hopefully that wasn't a bug ...) $\endgroup$
    – user5810
    Commented Feb 13, 2011 at 10:29
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    $\begingroup$ I had posted an answer, but then realized it was flawed, so I deleted it. $\endgroup$
    – Joel David Hamkins
    Commented Feb 13, 2011 at 10:38
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    $\begingroup$ Fix a non-measurable set $A$. Then for any measurable set $B$, the set $A \Delta B$ is not measurable. This way you get a one-to-one function from the set of measurable sets to the set of non-measurable sets, which shows that there are at least as many non-measurables as measurables. $\endgroup$
    – Nick S
    Commented Feb 13, 2011 at 18:15

2 Answers 2

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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.

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    $\begingroup$ Thanks! I didn't know the result of Lusin and Sierpinski that you mentioned, and it seems potentially very helpful. Unfortunately, the argument you gave isn't correct as written: it's not strictly true that the sets $S_J$, $J\subseteq I$ are distinct modulo $\mathcal{M}$. For example, $S_I = [0,1]$ is measurable, so $S_J$ and $S_{I-J}$ are equivalent modulo $\mathcal{M}$ for each $J$. Is there some reason that there should be $2^c$ different equivalence classes of $S_J$'s? $\endgroup$
    – Jim Belk
    Commented Feb 13, 2011 at 17:02
  • $\begingroup$ Jim, the argument I suggest in my answer for forming the $2^c$ sets avoids that (probably very minor) problem but at the expense of knowing some infinite combinatorics. $\endgroup$
    – Bill Johnson
    Commented Feb 13, 2011 at 17:21
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    $\begingroup$ @Simon: You didn't fix it you still have the two equivalent sets Jim described. You should rather look at all subsets $J \subset I$ such that $i\notin J$ for some fixed $i\in I$. $\endgroup$
    – Thomas Kragh
    Commented Feb 13, 2011 at 17:42
  • $\begingroup$ It seems to be right now. Since each $S_J$ has outer measure $1$, and $[0,1] - S_J = S_{I-J}$, every $S_J$ is non-measurable as long as $J\ne\emptyset$ and $J\ne I$. Since $S_J + S_K = S_{J+K}$, it follows that $S_J$ and $S_K$ are distinct modulo $\mathcal{M}$ as long as $J\ne K$ and $J \ne I-K$. $\endgroup$
    – Jim Belk
    Commented Feb 13, 2011 at 18:11
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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 $2^c$.

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    $\begingroup$ I'm a bit confused about this answer. First, note that not every Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ is non-measurable, so we must be careful to pick $A$ originally to have positive outer measure, and to be decomposable into $B$ and $C$ as you have indicated. However, once we have succeeded at this, how do we know that the sets $B+D$ are inequivalent modulo $\mathcal{M}$? Indeed, $B+D_1$ and $B+D_2$ are certainly equivalent whenever $D_1$ and $D_2$ are countable. $\endgroup$
    – Jim Belk
    Commented Feb 13, 2011 at 17:27
  • $\begingroup$ You start with $A\subset [0,1]$ of outer measure one, as in one of the usual construction of non measurable sets (Folland's book contains this one, IIRC). $B$ (or $C$) is the intersection of $A$ with some interval $(0,t)$. All the sets $D$ have cardinality $c$. But I don't see that this gives all of these translates are inequivalent mod $\mathcal{M}$. $\endgroup$
    – Bill Johnson
    Commented Feb 13, 2011 at 18:26
  • $\begingroup$ Wait a minute: Aren't you using "$S$ inequivalent to $T$ mod $\mathcal{M}$" wrongly; i.e., to mean "$T$ is not in the sigma algebra generate by $S$ and $\mathcal{M}$"? For disjoint sets, inequivalence just means that the union is not measurable, yes? Am I missing something (yet once again)? $\endgroup$
    – Bill Johnson
    Commented Feb 13, 2011 at 18:47
  • $\begingroup$ Oh, I see you said essentially the same thing as I was writing my last comment. $\endgroup$
    – Bill Johnson
    Commented Feb 13, 2011 at 18:49

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