The development of much of the earliest mathematics in human civilization was driven by the need to measure certain quantities: the amount of food harvested, the distance between cities, the area of land in a plot. These measurements, grounded as they were in the physical world, were entirely finite in character: the practical result of a measurement is a rational number, expressible as a ratio of finite integers. And, being entirely finite, they posed no serious foundational problems. But what happens when we remove measurement from the practical world of the finite? As we will see, when we start venturing into the realm of the infinite (which we can’t help ourselves from doing…) things become more complicated and more interesting.

In this intermittent series, we will look at various instances of “measuring” in mathematics and the sciences, paying particular attention to their interactions with the infinite. We will visit fin de siècle France, Poland in the 1930s, and the edges of a young universe. We will consider the development of the Lebesgue measure, climb the hierarchy of large cardinals, and look into the measure problem of cosmology. Today, though, we will simply try to count things.

Let us look at a particularly simple type of measure, known suggestively as the counting measure. Suppose we have a large but finite collection of objects: 10,000 potatoes, say. And suppose we want to describe a way to measure subsets of this collection. And suppose also that, for either practical or aesthetic reasons, we want the measure of the entire collection to be $1$. There is a very natural candidate for this. Let $P$ denote the set of all 10,000 potatoes. Given a subset $A \subseteq P$, let its measure, which we denote $m(A)$, simply be $\frac{|A|}{10,000}$. In other words, $m(A)$ is simply the proportion of all potatoes in $P$ that happen to lie in $A$.

This measure $m$ has the following attractive properties.

• For all $A \subseteq P$, $0 \leq m(A) \leq 1$.
• $m(\emptyset) = 0$.
• $m(P) = 1$.
• If $A$ and $B$ are disjoint subsets of $P$, i.e. $A, B \subseteq P$ and $A \cap B = \emptyset$, then $m(A \cup B) = m(A) + m(B)$.

This is all well and good, but, being the sophisticated mathematicians that we are, we might get bored with this and want to measure subsets of an infinite set. So, suppose we now have a countably infinite set $P$ and we want a measure $m$ on all subsets of $P$ satisfying the following.

1. For all $A \subseteq P$, $0 \leq m(A) \leq 1$.
2. $m(\emptyset) = 0$.
3. $m(P) = 1$.
4. For all $p,q \in P$, $m(\{p\}) = m(\{q\})$ (i.e. all one-element subsets of $P$ get the same measure).
5. If $\{A_n \mid n \in \mathbb{N}\}$ is a collection of pairwise disjoint subsets of $P$ (i.e., for all $m < n$ in $\mathbb{N}$, we have $A_m \cap A_n = \emptyset$), then $m(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}} m(A_n)$.

These seem like pretty natural requirements for a measure, and they generalize the attractive properties that do hold of the finite counting measure, but now we have a problem! For, suppose that we actually had such a measure, $m$. Let $r$ be the unique real number, which exists by requirement 4, such that $m(\{p\}) = r$ for all $p \in P$. Note that, by requirements 3 and 5 and the fact that $P$ is countable, we must have:

$1 = m(P) = m(\bigcup_{p \in P} \{p\}) = \sum_{p \in P} m(\{p\}) = \sum_{p \in P} r$.

If $r=0$, then this sum is simply $0$, so we obtain $0 = 1$, which is a contradiction. If $r > 0$, though, things aren’t any better. In that case, the sum is $r+r+r+\ldots = \infty$, so we obtain $\infty = 1$, another contradiction.

The situation can be somewhat salvaged, though, if we weaken requirement 5 to only deal with finite collections of sets:

5*. If $A$ and $B$ are disjoint subsets of $P$, then $m(A \cup B) = m(A) + m(B)$.

We first note that, if $m$ is any measure satisfying requirements 1-4 and 5*, and $r$ is the unique number such that $m(\{p\}) = r$ for all $p \in P$, then in must be the case that $r = 0$. To see this, suppose it were not the case. Then there is some natural number $n \in \mathbb{N}$ such that $n \times r > 1$. Let $A$ be a subset of $P$ of size $n$. Then, by repeated applications of requirement 5*, we obtain:

$m(A) = m(\bigcup_{a \in A}\{a\}) = \sum_{a \in A}m(\{a\}) = n \times r > 1$,

This should make sense; in the context of an infinite set, any single element should seem exceedingly small. Negligible, even.

Ok, fine, but have we really gained anything by replacing requirement 5 with 5*. Is there a measure satisfying these new requirements? Yes, there is, and there’s even one that’s quite easy for us to describe. Simply let $\mathcal{U}$ be a non-principal ultrafilter on $P$, and define a measure $m$ by

$m(A) = \begin{cases} 1 & \mbox{ if } A \in \mathcal{U} \\ 0 & \mbox{ if } A \not\in \mathcal{U}. \end{cases}$

It is immediate that $m$ satisfies requirements 1-3. Also, as $\mathcal{U}$ is non-principal, $\{p\} \not\in \mathcal{U}$ for all $p \in P$, so $m(\{p\}) = 0$ for all $p \in P$ and $m$ thus satisfies requirement 4.

It remains to verify requirement 5*. To do this, suppose $A$ and $B$ are disjoint subsets of $P$. Note first that it cannot be the case that $A$ and $B$ are both members of $\mathcal{U}$, for, if they were, then, by the definition of ultrafilter, we would also have $A \cap B \in \mathcal{U}$. But, as $A$ and $B$ are disjoint, $A \cap B = \emptyset$, and $\emptyset \not\in \mathcal{U}$ (again by the definition of ultrafilter). Therefore, one or both of $A$ and $B$ is not in $\mathcal{U}$.

Suppose first that neither $A$ nor $B$ is in $\mathcal{U}$. Then $P \setminus A$ and $P \setminus B$ are both in $\mathcal{U}$, since $\mathcal{U}$ is an ultrafilter. Therefore, $(P \setminus A) \cap (P \setminus B) = (P \setminus (A \cup B)) \in \mathcal{U}$, so $A \cup B \not\in \mathcal{U}$. Therefore, $m(A \cup B) = 0 = 0+0 = m(A) + m(B)$.

Suppose next that one of the sets is in $\mathcal{U}$ and the other is not. Since the two situations are symmetric, we may assume that $A \in \mathcal{U}$ and $B \not\in \mathcal{U}$. Then, since $A \subseteq A \cup B$, we have $A \cup B \in \mathcal{U}$, so $m(A \cup B) = 1 = 1+0 = m(A) + m(B)$. Therefore, we have verified requirement 5* in all possible situations.

Exercise for the reader: Suppose $P = \mathbb{N}$, $E$ is the set of even natural numbers, and $O$ is the set of odd natural numbers. Construct a measure $m$ on the subsets of $P$ satisfying requirements 1-4 and 5* and such that $m(E) = m(O) = \frac{1}{2}$.