Nonatomic measures and the nonstandard reals, for analysts

This argument has been useful to me in the past, as a way of constructing the nonstandard real numbers (real numbers with infinitesimals). Of course, infinitesimals are useful for cheesing epsilon-delta arguments where the explicit values of the constants that appear aren’t important, but one often feels “guilty” about using them because they come off as “less rigorous.” This is probably because the proof that nonstandard reals exist is usually thought of as “difficult to explain” to a non-logician, but the proof that I will give is meant to be “measure-theoretic” in nature, so much friendlier to someone with a background in analysis.

First, an incomprehensible definition. We will define the nonstandard reals to be any ultrapower $\mathbb R^\omega/\mathcal U$ , where $\omega$ is the set of natural numbers and $\mathcal U$ is a free ultrafilter on the powerset $2^\omega$ of $\omega$ .

An ultrafilter is nothing more than a finitely additive measure valued in {0, 1}, which is said to be free if it is non-atomic. (Throughout this post, we will take measures to be finitely additive, rather than countably additive.) If a set X is sent to 1, we write $X \in \mathcal U$ (so we identify the measure with the set of all sets which it sends to 1). Of course atomic ultrafilters on $2^\omega$ exist (just take any Dirac mass.) But as we will see, atomic ultrafilters are not very interesting for our purposes here.

Let $\lambda$ be an infinite set, k an ordered field, and $\mathcal U$ an ultrafilter on $2^\lambda$ . (This can be defined in much greater generality than ordered fields, but that’s all we’ll need for our purposes.) As usual $k^\lambda$ is the set of all maps $\lambda \to k$ (so if $\lambda = \omega$ , it is the sequence space of k). Now $\mathcal U$ determines an equivalence relation on $k^\lambda$ , by declaring that two maps x and y are equivalent if they are equal almost everywhere. The resulting quotient is known as the ultrapower $k^\lambda/\mathcal U$ of k.

We give $k^\lambda/\mathcal U$ the structure of an ordered field, as follows. First, 0 and 1 are nothing more than the equivalence classes of 0 and 1. We now observe that + and $\cdot$ lift to pointwise operations on $k^\lambda$ , and then descend to operations on $k^\lambda/\mathcal U$ , just as we can define operations on the space $L^p$ of Lebesgue-almost everywhere equivalence classes p-integrable functions by taking quotients of operations on the space $\mathcal L^p$ of p-integrable functions. Finally, we declare $x < y$ if $x(n) < y(n)$ for almost every $n \in \lambda$ .

We have to use some logical machinery somewhere in this proof, and so now we will. Specifically, we need Łoś’s theorem (also known as the fundamental theorem of ultrapowers, or the transfer principle.) To do this, we need the notion of a first-order formula in the language of ordered fields (or I will say “formula” for short). This is any mathematical statement about a given ordered field k which can be expressed using only finitely many copies of the symbols $\exists$ , $\forall$ , $\implies$ , $\neg$ (i.e. “not”), +, $\cdot$ , $<$ , =, 0, 1, and parentheses and variables, where $\exists x$ means that there exists an element $x$ of k (and not a subset of k, or a set of subsets of k, or so on) and $\neg$ is negation. For example, the statement that $\sqrt 2$ exists is a formula, namely $\exists x x\cdot x = 1 +1$ . A more sophisticated example of a formula is the quadratic formula, as would be a nice exercise to verify. We need infinitely many formulae to express that k is characteristic 0, namely $\neg((1 + 1 + \dots + 1) = 0)$ for every length of 1’s. And we cannot express that k is archimedean even with infinitely many formulae.

Łoś’s theorem says that if k is an ordered field and $k^\lambda/\mathcal U = F$ is any ultrapower of k, then for every formula $\varphi$ that is true in k, $\varphi$ is true in F as well (and if this conclusion holds, then we say that k and F are elementarily equivalent). Note carefully that elementary equivalence is weaker than isomorphism (indeed, by Łoś’s theorem, $\mathbb R$ is elementarily equivalent to the nonstandard reals, which are nonarchimedean).

To prove Łoś’s theorem, we use induction on the complexity of $\varphi$ . For so-called “atomic formulae”, those that can be expressed without use of $\neg$ , $\implies$ , or $\exists$ , this is an easy “almost everywhere” argument. For negation, notice that if $\varphi = \neg\psi$ , then $\psi$ is true almost everywhere iff $\varphi$ is false almost everywhere by the inductive hypothesis. Similarly for implication. For quantification, assume $\varphi = \exists x \psi$ . If $\varphi$ is true in k, then there is an $a \in k$ be such that a satisfies $\psi$ . So the map which is constantly a satisfies $\psi$ by the inductive hypothesis. On the other hand, if $\varphi$ is true in $k^\lambda/\mathcal U$ , then find a map $A \in k^\lambda/\mathcal U$ satisfying $\psi$ . Then there is a $a$ in the image of $A$ which satisfies $\psi$ , since $A$ satisfies $\psi$ almost everywhere. This proves Łoś’s theorem.

We are almost done. The problem is that if we choose any ultrapower of $\mathbb R$ , we have not accomplished anything. In fact if we take $\mathbb R^\omega/\delta_0$ , then the map $x \mapsto (x, x, \dots)$ is an isomorphism $\mathbb R \to \mathbb R^\omega/\delta_0$ , so we have not shown the existence of infinitesimals. This is why we assumed that the ultrafilter $\mathcal U$ was free.

To prove that a free ultrafilter exists, we define a filter $\mathcal F$ to be a measure on a subalgebra $\Sigma_{\mathcal F}$ of $2^\omega$ valued in {0, 1}, and declare that a filter is free if it is not atomic. Clearly the filter which sends finite sets to 0 and cofinite sets to 1 is non-atomic, so one exists. We order filters by saying that $\mathcal F \leq \mathcal G$ if there is an inclusion of algebras $\Sigma_{\mathcal F} \subseteq \Sigma_{\mathcal G}$ , and $\mathcal F$ agrees with $\mathcal G$ if they agree whenever both are defined, just like in the proof of the Hanh-Banach theorem. By taking limits we see that the supremum of filters is a filter. Now we apply Zorn’s lemma to the space of free filters, which is easily seen to be an ultrafilter.

(Unimportant aside: The existence of a $\omega$ -additive free ultrafilter on $2^\omega$ is a very special property of $\omega$ . It is not true for $\mathbb R$ or most other infinite sets. In fact, if $\lambda$ is uncountable and has a $\lambda$ -additive free ultrafilter on $2^\lambda$ , we say that $\lambda$ is a measurable cardinal. The existence of a measurable cardinal cannot be proven from the axioms of set theory, and in fact is much stronger than the consistency of said axioms.)

Now if $\mathcal U$ is a free ultrafilter, then the ultrapower $\mathbb R^\omega/\mathcal U$ is by definition the ordered field of nonstandard real numbers. By Łoś’s theorem, any statement which does not require quantifying over infinitely many reals or over sets of reals and is true in $\mathbb R$ is still true in $\mathbb R^\omega/\mathcal U$ . Moreover, the reals embed in $\mathbb R^\omega/\mathcal U$ by the map $x \mapsto (x, x, \dots)$ . On the other hand, the sequence $(1, 1/2, 1/3, 1/4, \dots)$ , descends to an positive element $\varepsilon$ of $\mathbb R^\omega/\mathcal U$ which is smaller than any standard real number, since it is smaller than any standard real number almost everywhere. So we do have infinitesimals, and the inclusion $\mathbb R \to \mathbb R^\omega/\mathcal U$ is not an isomorphism, as desired.