Carathéodory's Measurability Criterion

Outer measure gives every subset of $\mathbb{R}$ a size, but it is only sub-additive. The problem is that some “bad” sets leak measure across boundaries in a way that makes additivity fail. Constantin Carathéodory’s insight (1914) was to characterise the “good” sets — the measurable ones — by a single geometric condition: a set $E$ is measurable precisely when it perfectly splits every test set $A$ into two non-overlapping pieces.

The criterion

Let $\mu^*$ be an outer measure on a set $X$ (see Outer Measure for the abstract definition). A set $E \subseteq X$ is Carathéodory-measurable (or simply measurable) with respect to $\mu^*$ if

$$\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c) \qquad \text{for every } A \subseteq X. \tag{1}$$

The set $A$ in $(1)$ is called the test set. You are asking: does $E$ divide $A$ into two pieces whose outer measures add up exactly to $\mu^*(A)$?

Because $A = (A \cap E) \cup (A \cap E^c)$ and the two pieces are disjoint, sub-additivity always gives

$$\mu^*(A) \leq \mu^*(A \cap E) + \mu^*(A \cap E^c).$$

So the content of criterion $(1)$ is really the reverse inequality:

$$\mu^*(A) \geq \mu^*(A \cap E) + \mu^*(A \cap E^c). \tag{1'}$$

In words: $E$ does not “create” extra measure when it cuts $A$ in two.

Remark. The criterion is symmetric in $E$ and $E^c$: if $E$ satisfies $(1)$, so does $E^c$ (just swap the roles). So measurability is preserved under complementation.

The measurable sets form a σ-algebra

Let $\mathcal{M}$ denote the collection of all $\mu^*$-measurable sets. You will now verify that $\mathcal{M}$ satisfies all three σ-algebra axioms.

Axiom 1: $X \in \mathcal{M}$. For any test set $A$: $A \cap X = A$ and $A \cap X^c = A \cap \emptyset = \emptyset$, so $\mu^*(A \cap X) + \mu^*(A \cap X^c) = \mu^*(A) + 0 = \mu^*(A)$. ✓

Axiom 2: Closed under complement. Already observed above: criterion $(1)$ is symmetric in $E$ and $E^c$. ✓

Axiom 3: Closed under countable union. This is the heart of the argument. Here is the key step.

Finite unions first. Suppose $E, F \in \mathcal{M}$. You want to show $E \cup F \in \mathcal{M}$. For any test set $A$, use the measurability of $E$ with test set $A$ and then the measurability of $F$ with test set $A \cap E^c$:

$$\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$$ $$\mu^*(A \cap E^c) = \mu^*(A \cap E^c \cap F) + \mu^*(A \cap E^c \cap F^c).$$

Note $E^c \cap F^c = (E \cup F)^c$, so

$$\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c \cap F) + \mu^*(A \cap (E \cup F)^c). \tag{2}$$

Since $A \cap (E \cup F) = (A \cap E) \cup (A \cap E^c \cap F)$ and these are disjoint, sub-additivity gives $\mu^*(A \cap (E \cup F)) \leq \mu^*(A \cap E) + \mu^*(A \cap E^c \cap F)$. Substituting into $(2)$:

$$\mu^*(A) \geq \mu^*(A \cap (E \cup F)) + \mu^*(A \cap (E \cup F)^c),$$

which (combined with sub-additivity in the other direction) proves $E \cup F \in \mathcal{M}$. By induction, $\mathcal{M}$ is closed under finite unions.

Countable unions. Let $E_1, E_2, \ldots \in \mathcal{M}$ be pairwise disjoint (the general case reduces to this by writing $\tilde{E}_k = E_k \setminus (E_1 \cup \cdots \cup E_{k-1})$, which remain in $\mathcal{M}$ by the closure properties already established). Set $S_n \coloneqq \bigsqcup_{k=1}^n E_k \in \mathcal{M}$. For any test set $A$ and each $n$:

$$\mu^*(A \cap S_n) = \sum_{k=1}^{n} \mu^*(A \cap E_k). \tag{3}$$

Equation $(3)$ follows by induction on $n$ using the measurability of $E_n$ with test set $A \cap S_n$. Now let $S \coloneqq \bigcup_k E_k$. Since $S_n \subseteq S$, monotonicity and $(3)$ give:

$$\mu^*(A \cap S) \geq \mu^*(A \cap S_n) = \sum_{k=1}^{n} \mu^*(A \cap E_k).$$

Taking $n \to \infty$ yields $\mu^*(A \cap S) \geq \sum_{k=1}^{\infty} \mu^*(A \cap E_k)$, so together with sub-additivity:

$$\mu^*(A \cap S) = \sum_{k=1}^{\infty} \mu^*(A \cap E_k). \tag{4}$$

Also $S^c \subseteq S_n^c$, so $\mu^*(A \cap S^c) \leq \mu^*(A \cap S_n^c)$. Using measurability of $S_n$:

$$\mu^*(A \cap S_n) + \mu^*(A \cap S_n^c) = \mu^*(A),$$

hence $\mu^*(A \cap S^c) \leq \mu^*(A) - \sum_{k=1}^n \mu^*(A \cap E_k)$ for every $n$. Combining with $(4)$:

$$\mu^*(A) \geq \mu^*(A \cap S) + \mu^*(A \cap S^c),$$

so $S \in \mathcal{M}$. Conclusion: $\mathcal{M}$ is a σ-algebra. ✓

Countable additivity on $\mathcal{M}$

The real payoff of the above argument is equation $(4)$ applied with $A = X$:

$$\mu^*\!\left(\bigsqcup_{k=1}^{\infty} E_k\right) = \sum_{k=1}^{\infty} \mu^*(E_k) \qquad \text{for pairwise disjoint } E_k \in \mathcal{M}. \tag{5}$$

This is countable additivity — the defining property of a true measure. The restriction $\mu^* \restriction_{\mathcal{M}}$ therefore turns the outer measure into a genuine measure on the σ-algebra $\mathcal{M}$.

Completeness

A measure space $(X, \mathcal{M}, \mu^*)$ is complete when every subset of a null set is measurable. Carathéodory’s construction automatically delivers completeness.

Proposition. If $\mu^*(N) = 0$ and $A \subseteq N$, then $A \in \mathcal{M}$.

Proof. For any test set $T$: $T \cap A \subseteq N$, so $\mu^*(T \cap A) \leq \mu^*(N) = 0$. Also $T \supseteq T \cap A^c$, so $\mu^*(T \cap A^c) \leq \mu^*(T)$ by monotonicity. Thus $\mu^*(T \cap A) + \mu^*(T \cap A^c) \leq 0 + \mu^*(T) = \mu^*(T)$, and the reverse inequality holds by sub-additivity. So $A \in \mathcal{M}$.

This completeness is important in practice: it means you never need to worry about sub-sets of measure-zero sets “falling outside” the σ-algebra.

Summary

• Carathéodory’s criterion $(1)$ says $E$ is measurable when it splits every test set $A$ additively; the non-trivial content is the reverse inequality $(1')$.
• The collection $\mathcal{M}$ of all measurable sets is a σ-algebra: it contains $X$, is closed under complements (symmetry of the criterion), and is closed under countable unions (the finite-union argument extends by the limit argument above).
• On pairwise disjoint measurable sets, the outer measure satisfies countable additivity — equation $(5)$. This makes $\mu^*\restriction_{\mathcal{M}}$ a genuine measure.
• The resulting measure space is complete: every subset of a null set is measurable.
• Next: Lebesgue Measure applies this machinery to $\mu^* = \lambda^*$ (the Lebesgue outer measure on $\mathbb{R}$) to produce the canonical notion of length.