Does Vitali construction exhaust non-measurable sets?

Question

We can construct Vitali type of non-measurable sets. But is this the only type of non-measurable sets? That is, are there any other non-measurable sets constructed from a different approach?

Please provide some answers or references.

Thanks.

Answer 1

Yes, there are other ways.

For example, the existence of a non-principal ultrafilter on ${\mathbb N}$ is weaker than the existence of the choice function required in Vitali's construction, and suffices to ensure the existence of non-measurable sets. One way of seeing this is noticing that we can identify the measurable subsets of ${\mathbb R}$ and the measurable subsets of $2^{\mathbb N}$ (the product of infinitely many copies of a two point set, that we endow with the product topology and the product measure), which follows from the fact that we can identifying the Borel structures of these two spaces (see Kechris's "Classical descriptive set theory" for details).

So it suffices to argue that a non-principal ultrafilter gives us a non-measurable subset of $2^{\mathbb N}$. But general results from ergodic theory tell us that a "tail-invariant" set must have measure 1 or 0. Here, a set $C$ is invariant iff for any $a$ and $b$ infinite sequences of $0$s and $1$s that agree from some point on, either both $a$ and $b$ are in $C$, or neither is. A non-principal ultrafilter is easily seen to "be" tail-invariant, and one can easily argue that if measurable, it must have measure $1/2$ (because the map that sends $a$ to the sequence $\bar a$ that "flips" all $0$s and $1$s in $a$ maps the ultrafilter to its complement, and is measure-preserving; here we think of a sequence of 0s and 1s as the characteristic function of a set of natural numbers, so we can identify the ultrafilter with a set of sequences).

Another example comes from observing that no well-ordering of the reals can be measurable (as a subset of the plane). This is a result of Sierpiński that is essentially a refinement of the better known result that under CH a well-ordering of smallest order type is non-measurable. Rudin's book on "Real and complex analysis" has a proof of this last fact.

The one thing all examples have in common is that at some point they must make an appeal to a (somewhat significant) amount of the axiom of choice. This is essential: A famous result of Robert Solovay shows that it is consistent that all sets of reals are measurable, and the weakening of choice known as the axiom of Dependent choices holds.

Answer 2

John C. Oxtoby's "Measure and Category" has a chapter on non-measurable sets. He covers the following three examples of non-measurable sets:

• Vitali sets (group-theoretic)
• Bernstein sets (topological)
• A construction due to Ulam (set-theoretic)