In this supplemental note to the previous lecture notes, I would like to give an alternate proof of a (weak form of the) Carathéodory extension theorem.  This argument is restricted to the $\sigma$-finite case, and does not extend the measure to quite as large a $\sigma$-algebra as is provided by the standard proof of this theorem, but I find it conceptually clearer (in particular, hewing quite closely to Littlewood’s principles, and the general Lebesgue philosophy of treating sets of small measure as negligible), and suffices for many standard applications of this theorem, in particular the construction of Lebesgue measure.

Let us first state the precise statement of the theorem:

Theorem 1. (Weak Carathéodory extension theorem)  Let ${\mathcal A}$ be a Boolean algebra of subsets of a set X, and let $\mu: {\mathcal A} \to [0,+\infty]$ be a function obeying the following three properties:

1. $\mu(\emptyset) = 0$.
2. (Pre-countable additivity) If $A_1,A_2,\ldots \in {\mathcal A}$ are disjoint and such that $\bigcup_{n=1}^\infty A_n$ also lies in ${\mathcal A}$, then $\mu(\bigcup_{n=1}^\infty A_n) = \sum_{n=1}^\infty \mu(A_n)$.
3. ( $\sigma$-finiteness) X can be covered by at most countably many sets in ${\mathcal A}$, each of which has finite $\mu$-measure.

Let ${\mathcal X}$ be the $\sigma$-algebra generated by ${\mathcal A}$.  Then $\mu$ can be uniquely extended to a countably additive measure on ${\mathcal X}$.

We will refer to sets in ${\mathcal A}$ as elementary sets and sets in ${\mathcal X}$ as measurable sets. A typical example is when X=[0,1] and ${\mathcal A}$ is the collection of all sets that are unions of finitely many intervals; in this case, ${\mathcal X}$ are the Borel-measurable sets.

— Some basics —

Let us first observe that the hypotheses on the premeasure $\mu$ imply some other basic and useful properties:

1. From property 1. and 2. we see that $\mu$ is finitely additive (thus $\mu(A_1 \cup \ldots \cup A_n) = \mu(A_1) + \ldots + \mu(A_n)$ whenever $A_1,\ldots,A_n$ are disjoint elementary sets).
2. As particular consequences of finite additivity, we have monotonicity ( $\mu(A) \leq \mu(B)$ whenever $A \subset B$ are elementary sets) and finite subadditivity ( $\mu(A_1 \cup \ldots \cup A_n) \leq \mu(A_1) + \ldots + \mu(A_n)$ for all elementary $A_1,\ldots,A_n$, not necessarily disjoint).
3. We also have pre-countable subadditivity: $\mu(A) \leq \sum_{n=1}^\infty \mu(A_n)$ whenever the elementary sets $A_1,A_2,\ldots$ cover the elementary set A.  To see this, first observe by replacing $A_n$ with $A_n \backslash \bigcup_{i=1}^{n-1} A_i$ and using monotonicity that we may take the $A_i$ to be disjoint; next, by restricting all the $A_i$ to $A$ and using monotonicity we may assume that A is the union of the $A_i$, and now the claim is immediate from pre-countable additivity.

— Existence —

Let us first verify existence.  As is standard in measure-theoretic proofs for $\sigma$-finite spaces, we first handle the finite case (when $\mu(X) < \infty$), and then rely on countable additivity or sub-additivity to recover the $\sigma$-finite case.

The basic idea, following Littlewood’s principles, is to view the measurable sets as lying in the “completion” of the elementary sets, or in other words to exploit the fact that measurable sets can be approximated to arbitrarily high accuracy by elementary sets.

Define the outer measure $\mu_*(A)$ of a set $A \subset X$ to be the infimum of $\sum_{n=1}^\infty \mu(A_n)$, where $A_1,A_2,\ldots$ range over all at most countable collections of elementary sets that cover A.  It is clear that outer measure is monotone and countably subadditive.  Also, since $\mu$ is pre-countably subadditive, we see that $\mu_*(A) \geq \mu(A)$ for all elementary A.  Since we also have the trivial inequality $\mu_*(A) \leq \mu(A)$, we conclude that $\mu_*$ and $\mu$ agree on elementary sets.

The outer measure naturally defines a pseudometric (and thus a topology) on the space of subsets of X, with the distance between A and B being defined as $\mu_*(A \Delta B)$, where $\Delta$ denotes symmetric difference.  (The subadditivity of $\mu_*$ ensures the triangle inequality; furthermore, we see that the Boolean operations (union, intersection, complement, etc.) are all continuous with respect to this pseudometric.) With this pseudometric, we claim that the measurable sets lie in the closure of the elementary sets.  Indeed, it is not difficult to see (using subadditivity and monotonicity properties of $\mu_*$) that the closure of the elementary sets are closed under finite unions, under complements, and under countable disjoint unions (here we need finiteness of $\mu(X)$ to keep the measure of all the pieces absolutely summable), and thus form a $\sigma$-algebra.  Since this $\sigma$-algebra clearly contains the elementary sets, it must contain the measurable sets also.

By subadditivity of $\mu_*$, the function $A \mapsto \mu_*(A)$ is Lipschitz continuous.  Since this function is finitely additive on elementary sets, we see on taking limits (using subadditivity to control error terms) that it must be finitely additive on measurable sets also.  Since $\mu_*$ is finitely additive, monotone, and countably sub-additive, it must be countably additive, and so $\mu_*$ is the desired extension of $\mu$ to the measurable sets.  This completes the proof of the theorem in the finite measure case.

To handle the $\sigma$-finite case, we partition X into countably many elementary sets of finite measure, and use the above argument to extend $\mu$ to measurable subsets of each such elementary set.  It is then a routine matter to sum together these localised measures to recover a measure on all measurable sets; the pre-countable additivity property ensures that this sum still agrees with $\mu$ on elementary sets.

— Uniqueness —

Now we verify uniqueness.  Again, we begin with the finite measure case.

Suppose first that $\mu(X) < \infty$, and that we have two different extensions $\mu_1, \mu_2: {\mathcal X} \to [0,+\infty]$ of $\mu$ to ${\mathcal X}$ that are countably additive.  Observe that $\mu_1, \mu_2$ must both be continuous with respect to the $\mu_*$ pseudometric used in the existence argument, from countable subadditivity; since every measurable set is a limit of elementary sets in this pseudometric, we obtain uniqueness in the finite measure case.

When instead X is $\sigma$-finite, we cover X by elementary sets of finite measure.  The previous argument shows that any two extensions $\mu_1, \mu_2$ of $\mu$ agree when restricted to each of these sets, and the claim then follows by countable additivity. $\Box$

Remark 1. The uniqueness claim fails when the $\sigma$-finiteness condition is dropped.  Consider for instance the rational numbers $X = {\Bbb Q}$, and let the elementary sets be the finite unions of intervals ${}[a,b) \cap {\Bbb Q}$.   Define the measure $\mu(A)$ of an elementary set to be zero if A is empty, and $+\infty$ otherwise.  As the rationals are countable, we easily see that every set of rationals is measurable.  One easily verifies the pre-countable additivity condition (though the $\sigma$-finiteness condition fails horribly).  However, $\mu$ has multiple extensions to the measurable sets; for instance, any positive scalar multiple of counting measure is such an extension. $\diamond$

Remark 2. It is not difficult to show that the measure completion $\overline{{\mathcal X}}$ of ${\mathcal X}$ with respect to $\mu$ is the same as the topological closure of ${\mathcal X}$ (or of ${\mathcal A}$) with respect to the above pseudometric.  Thus, for instance, a subset of [0,1] is Lebesgue measurable if and only if it can be approximated to arbitrary accuracy (with respect to outer measure) by a finite union of intervals. $\diamond$

A particularly simple case of Theorem 1 occurs when X is a metrisable compact Hausdorff totally disconnected space (i.e. a metrisable Stone space), such as the infinite discrete cube $\{0,1\}^{\Bbb N}$ or any other Cantor space.  Then (see forthcoming lecture notes) the Borel $\sigma$-algebra ${\mathcal X}$ is generated by the Boolean algebra ${\mathcal A}$ of clopen sets.  Also, as clopen sets here are simultaneously compact and open, we see that any infinite cover of one clopen set by others automatically has a finite subcover.  From this, we conclude

Corollary 1. Let X be a compact Hausdorff totally disconnected space.  Then any finitely additive $\sigma$-finite measure on the clopen sets uniquely extends to a countably additive measure on the Borel sets.

By identifying $\{0,1\}^{\Bbb N}$ with ${}[0,1]$ up to a countable set, this provides one means to construct Lebesgue measure on ${}[0,1]$; similar constructions are available for ${\Bbb R}$ or ${\Bbb R}^n$.