Friday, March 20, 2015

Are there more nonmeasurable sets than measurable ones?

Assume the Axiom of Choice.

Consider the subsets of the interval [0,1]. There 2c subsets of [0,1], where c is the cardinality of the continuum. Since the Cantor set C has measure zero, and every subset of a measure zero set is (Lebesgue) measurable, there are at least as many measurable subsets of [0,1] as subsets of the Cantor set. But the Cantor set has cardinality c, so there are 2c measurable subsets of [0,1].

On the other hand, however, if A is any nonmeasurable set, then adding and/or subtracting a set of measure zero to A won't change the nonmeasurability, so there are 2c nonmeasurable sets as well, since there are 2c ways of varying A by adding and/or substracting a subset of C.

But notice the curious role that the Cantor set played in my above arguments. The reason there are 2c measurable sets and 2c nonmeasurable sets is because there are 2c sets of measure zero, and we can vary any set by a measure zero set and preserve measurability / nonmeasurability.

This suggests another question. Say that two sets are equivalent provided that they differ by a set of measure zero. Are there more equivalence classes of nonmeasurable sets than of measurable sets?

Assuming the Axiom of Choice, the answer is yes. In fact:

  • Theorem: There are 2c equivalence classes of nonmeasurable sets while there are only c classes of measurable sets.

Why? Well, any measurable set differs from a Borel set by a set of measure zero, so there are no more equivalence classes of measurable sets than there are Borel sets, and there are c Borel sets. (And of course there are no fewer than c equivalence classes measurable sets: just consider the intervals [0,x].) Responding to a question I had asked on MathOverflow, Eric Wofsey yesterday showed that there are 2c equivalence classes of subsets of [0,1]. Since there are only c equivalence classes of measurable ones, there must be 2c equivalence classes of nonmeasurable ones.

In fact, we can say something stronger. Say that a subset A of [0,1] is saturated nonmeasurable (I used to call this "maximally nonmeasurable") provided that it is not only nonmeasurable but that every measurable subset of A has measure zero and every measurable superset of A has measure one. In other words, there is nothing measurable about A: we can't even give a non-trivial range of Lebesgue measures for A. (Technically: the inner measure is zero and the outer measure is one.) Then:

  • Theorem: There are 2c equivalence classes of saturated nonmeasurable sets.

Why? Well, this uses the same trick that Wofsey's proof did. Sierpiński showed in 1938 that you can find a disjoint family of perfect, and hence of cardinality c, subsets of the square [0,1]2 with the property that any set obtained by choosing one member from each element in the family is saturated nonmeasurable. Since as measure spaces [0,1]2 is isomorphic to [0,1], there will be a disjoint family F of subsets of [0,1], each of cardinality c, such that choosing one member of each element in F yields a saturated nonmeasurable set. For any AF, let φA be a one-to-one function from [0,1] to A. Define Ux={φA(x):AF} for x∈[0,1]. Then each Ux has inner measure zero and outer measure one, and the sets Ux are disjoint for different values of x. Any union of the sets Ux that (a) contains at least one of the sets (i.e., isn't empty) and (b) doesn't contain all the sets then also has inner measure zero and outer measure one. Why? Well, since it contains at least one of the sets, its outer measure is one. And since there is at least one that it doesn't contain, its complement has outer measure one and hence it itself has inner measure zero. But there are 2c such unions (since there are 2c possible unions given c nonempty disjoint sets, and throwing out the union that has no sets and the one that has all sets doesn't change that). Moreover, any two distinct such unions differ by at least one of the sets Ux, and hence differ by a set with inner measure one (and outer measure one for that matter), and so are not equivalent up to sets of measure zero.

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