Measure Theory + Algebra + AOC = Vitali Set

Perhaps the least intuitive concept one first encounters in the study of Abstract Measure Spaces, or even the Lebesgue Measure on $\mathbb{R}^n$, is the concept of an unmeasurable set. What the heck does this even mean? Well it really just means that there’s no good way to assign a “size” to this set. A -algebra is really just trying to answer the question: “To what can we reasonably assign some size?”. Thus, if a set isn’t in the $\sigma$-algebra, we simply mean that it does not have a well-defined size. Luckily the Lebesgue measure does a pretty darn good job at handling any reasonable set that one would encounter in the wild.

In fact, to find an example of a set that is not measurable, we have to be quite clever and artificial; we even have to invoke the Axiom of Choice. Many of the constructions of the archetypical unmeasurable set, the Vitali Set, seem rather opaque at first glance. However, one construction, in my opinion, trumps all the others in both intuition and elegance.

Using simple algebra and the axiom of choice, we can construct our exotic set. So, on that note, let us begin. Let $\mathfrak{M} (\mu)$ be the set of Lebesgue-Measurable Sets in $\mathbb{R}^n$.

Construction of the Vitali Set:

Assume that every group below is equipped with standard addition. ($\coprod$ represents a disjoint union)

Begin with the quotient group $\mathbb{R} / \mathbb{Z} \simeq [0,1)$ Since $\mathbb{R} / \mathbb{Z}$ is abelian and $\mathbb{Q} / \mathbb{Z}$ is a subgroup it is a normal subgroup. Consider the group
$(\mathbb{R} / \mathbb{Z})/(\mathbb{Q} / \mathbb{Z}) := \{ r+ (\mathbb{Q}/ \mathbb{Z}) : r \in \mathbb{R} / \mathbb{Z} \},$

i.e. the set of all cosets of the group. By the Axiom of Choice, there exists a set $P$ such that, for each $p \in \mathbb{R} / \mathbb{Z}$, $|p \cap (r + \mathbb{Q} / \mathbb{Z})| = 1,$ that is, we may pick a single representative from each coset. Moreover, given an enumeration of $\mathbb{Q}$,
$[0,1) = \coprod \limits_{q_i \in \mathbb{Q}} P + q_i$

where $(P+q_i) \cap (P+q_j)$ are disjoint since if not, we would have two representatives in the same equivalence class. Also, $\mu(P+q_i) = \mu(P+q_j)$. If $\mu(P+q_i)>0$, then $\mu ([0,1)) = \mu ( \coprod \limits_{q_i \in \mathbb{Q}} P + q_i ) = \sum \limits_{q_i \in \mathbb{Q}} \mu(P+q_i) = \infty.$ If not, it is $0$. Either way, $\mu([0,1)) \neq 0 \neq \infty$. Thus, $P \notin \mathfrak{M}(\mu)$.

Let’s reflect on what we’ve actually done here. We take the real line, quotient out by the integers, to get something that can be visualized as $S^1$, the unit circle. Then we take the rationals and quotient out by the integers, which can be visualized as “the rationals on $S^1"$ in an odd sort of way. Another way to think of it is taking the rationals and gluing them together in such a way that this group now looks something like $\mathbb{Q} \cap [0,1)$. So in this group, $\frac{1}{2} + \frac{3}{4} = \frac{5}{4} \simeq \frac{1}{4}$. Alright cool, so let’s take it a step further and make a new group, $G$ by saying that two elements of $G$ are equal if their difference is a rational number as identified in $\mathbb{Q} / \mathbb{Z}$. So in our new group, for example, $((\pi+1) +\frac{3}{2}) - (\pi) = \frac{1}{2}$ since $\frac{3}{2} \simeq \frac{1}{2}$  in $\mathbb{Q} / \mathbb{Z}$ and $(\pi+1) \simeq \pi$ in $\mathbb{R} / \mathbb{Z}$.

I know. A bit odd and convoluted, but take a second to think about it and you’ll see what’s going on.

Now, by the Axiom of Choice, and since $G$ is defined by its equivalence classes, we may take a set whose elements are representatives for each equivalence class, call this set $P$. We could have picked say $\frac{1}{2}$ to represent all real numbers such that if you consider $x - \frac{1}{2}$, then you get something of the form $n+q_1$ with $n \in \mathbb{Z}$ and $q_1 \in \mathbb{Q}$. In fact, since $\frac{1}{2}$ is rational, it would be the representative for all of $\mathbb{Z}+\mathbb{Q} := \{n+q : n \in \mathbb{Z}, q \in \mathbb{Q}\}$!

In any case, we recognize that, just as cosets always do, the cosets partition $[0,1)$ and are disjoint, and all cosets have the same measure in this case. Clearly they can’t all have measure $0$ because, by the countable sub-additivity of measures, this would imply $\mu([0,1)) = 0$, a contradiction. Thus we are led to believe each has non-zero measure. Though, by simple facts about infinite series, since the elements of the sum do not converge to $0$, our sum must diverge. But this implies $\mu([0,1)) = \infty$! So, either way, we’re screwed.

Thus, we conclude that $P$ must not have been in the $\sigma$-algebra to begin with.

So there you have it. A set that cannot be measurable, constructed from purely algebraic concepts. If that doesn’t pique your interest, I don’t know what will. Yet another beautiful marriage on areas of mathematics.

Cheers,

J.T.