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 $0$s and $1$s 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 $\mathsf{CH}$ 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. For the general version, see here.

A classical example that is used frequently is due to Bernstein. A set $A$ is a Bernstein set iff $A$ meets every uncountable closed set but contains none. Note that the complement of $A$ is also a Bernstein set, and that any set of positive outer measure must meet both $A$ and its complement. These sets are described carefully in Oxtoby's Measure and category. Here is a question on MO on the relation between Vitali and Bernstein sets.

Another example comes from the study of Hamel bases: A Hamel basis is a basis for $\mathbb R$ understood as a vector space over $\mathbb Q$. The list continues...

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.

A nice reference on this matter is the book

Alexander B. Kharazishvili. Nonmeasurable sets and functions, North-Holland Mathematics Studies, 195. Elsevier Science B.V., Amsterdam, 2004. MR2067444 (2005d:28001).

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)

Answer 3

Here is an easy way to construct $|Pow(\mathbb{R})|$ new nonmeasurable subsets of $\mathbb{R}$ (starting from just a single one). Let $\mathcal{F}$ be any sigma algebra on $\mathbb{R}$ that satisfies:

i) $[0,1] \in \mathcal{F}$

ii) There is a set $A \subseteq [0,1]$ such that $A \notin \mathcal{F}$.

Both the Lebesgue sigma algebra on $\mathbb{R}$ and the standard Borel sigma algebra on $\mathbb{R}$ satisfy (i) and (ii). Fix $A$ as the set that satisfies point (ii).

Claim 1: For all $B \subseteq [2,3]$ we have $A \cup B \notin \mathcal{F}$.

Proof: Suppose $A \cup B \in \mathcal{F}$. Since $[0,1] \in \mathcal{F}$, we know $(A\cup B) \cap [0,1] \in \mathcal{F}$. But $(A\cup B)\cap [0,1]= A$, contradicting the fact that $A \notin \mathcal{F}$. $\Box$.

Claim 2: There are exactly $|Pow(\mathbb{R})|$ distinct subsets of $\mathbb{R}$ that are not in $\mathcal{F}$.

Proof: There are $|Pow([2,3])|$ distinct ways to choose set $B$, each leading to a distinct set $A \cup B$ that is not in $\mathcal{F}$. Since $|Pow([2,3])|=|Pow(\mathbb{R})|$, there are at least $|Pow(\mathbb{R})|$ subsets of $\mathbb{R}$ that are not in $\mathcal{F}$. On the other hand, the number of subsets of $\mathbb{R}$ that are not in $\mathcal{F}$ is less than or equal to the number of subsets of $\mathbb{R}$, which is $|Pow(\mathbb{R})|$. $\Box$