The Caratheodory-Hahn Extension Theorem

From a Premeasure to a Measure

A premeasure is defined only on a ring, where countable additivity is required when the disjoint union remains in that ring. The extension problem is to produce a measure on the sigma-algebra generated by the ring. The covering construction provides a route: first define an outer measure on every subset of the underlying space, then restrict it to sets that satisfy the Caratheodory measurability criterion.

Let \(X\) be a set and \(\mathcal{R}\) a ring of subsets of \(X\). A ring is closed under finite unions and set differences; it need not contain \(X\). A premeasure is a function \(\mu_0:\mathcal{R}\to[0,\infty]\) with \(\mu_0(\varnothing)=0\) such that, whenever pairwise disjoint \(R_n\in\mathcal{R}\) have union \(R\in\mathcal{R}\), \(\mu_0(R)=\sum_{n=1}^{\infty}\mu_0(R_n)\). Values of \(+\infty\) are allowed. The measure eventually obtained need not be finite, and no sigma-finiteness assumption is needed for existence.

For \(A\subseteq X\), define the induced outer measure by

A finite cover is allowed by padding it with empty sets. If there is no countable cover, the infimum is \(+\infty\). An outer measure is a function on all subsets that vanishes on the empty set, is monotone, and is countably subadditive. The covering construction has these properties; its verification is part of the earlier material on induced outer measures. We will use those properties, in particular monotonicity and countable subadditivity.

Agreement on the Ring

The first essential point is that the covering infimum does not change the original values on \(\mathcal{R}\). The upper bound comes from covering a ring set by itself. For the reverse bound, any countable ring cover can be disjointified inside the set being covered.

Proof. Covering \(R\) by \(R\) itself, followed by empty sets, gives \(\mu^*(R)\leq\mu_0(R)\). For the reverse inequality, consider any countable cover \(R\subseteq\bigcup_{n=1}^{\infty}R_n\) with \(R_n\in\mathcal{R}\). Define \(S_1=R\cap R_1\) and, for \(n\geq2\), define \(S_n=(R\cap R_n)\setminus\bigcup_{j=1}^{n-1}R_j\). Each \(S_n\) is in \(\mathcal{R}\), since a ring is closed under finite intersections and differences. The sets \(S_n\) are pairwise disjoint, their union is \(R\), and \(S_n\subseteq R_n\). Countable additivity of the premeasure, applied to this disjoint union in the ring, and monotonicity of \(\mu_0\) give \(\mu_0(R)=\sum_n\mu_0(S_n)\leq\sum_n\mu_0(R_n)\). This holds for every cover; taking the infimum gives \(\mu_0(R)\leq\mu^*(R)\). The two inequalities prove the assertion. \(\square\)

Here monotonicity of \(\mu_0\) on the ring follows from its definition: if \(S\subseteq R\) are ring sets, then \(R=S\cup(R\setminus S)\) is a disjoint union in the ring, so \(\mu_0(R)=\mu_0(S)+\mu_0(R\setminus S)\geq\mu_0(S)\). Thus the agreement argument does not assume a separate extension of \(\mu_0\).

Caratheodory-Measurable Sets

The criterion says that \(E\) splits every test set \(A\) without losing outer measure. Every ring set passes this test. Indeed, fix \(E\in\mathcal{R}\). Subadditivity applied to \(A=(A\cap E)\cup(A\setminus E)\) gives \(\mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\setminus E)\). For the other direction, take any countable ring cover \(A\subseteq\bigcup_nR_n\). The ring sets \(R_n\cap E\) and \(R_n\setminus E\) cover \(A\cap E\) and \(A\setminus E\), respectively. Finite additivity of \(\mu_0\) on the disjoint pieces of \(R_n\) gives \(\mu_0(R_n)=\mu_0(R_n\cap E)+\mu_0(R_n\setminus E)\). Consequently, \(\mu^*(A\cap E)+\mu^*(A\setminus E)\leq\sum_n\mu_0(R_n)\). Taking the infimum over covers proves the reverse inequality. If \(A\) has no such cover, then \(\mu^*(A)=+\infty\), and the required lower bound is automatic. Thus \(E\) is Caratheodory-measurable.

For the extension theorem, we also need that the Caratheodory-measurable sets form a sigma-algebra and that \(\mu^*\) is countably additive on them. The following argument supplies these facts. Write \(\mathcal{C}\) for the collection of all Caratheodory-measurable subsets of \(X\).

Proof. The empty set belongs to \(\mathcal{C}\), since splitting by \(\varnothing\) gives \(\mu^*(A)=\mu^*(A)+0\). If \(E\in\mathcal{C}\), then \(E^c\in\mathcal{C}\), because the two terms in the defining split are merely exchanged.

To check finite unions, let \(E,F\in\mathcal{C}\). Apply the criterion first to \(E\), and then to \(F\) with test set \(A\setminus E\). This gives \[ \mu^*(A)=\mu^*(A\cap E)+\mu^*((A\setminus E)\cap F) +\mu^*(A\setminus(E\cup F)). \] The first two sets are disjoint and their union is \(A\cap(E\cup F)\). Subadditivity therefore gives \(\mu^*(A\cap(E\cup F))\leq\mu^*(A\cap E)+\mu^*((A\setminus E)\cap F)\). Together, these inequalities show \(\mu^*(A)\geq\mu^*(A\cap(E\cup F))+\mu^*(A\setminus(E\cup F))\). Subadditivity gives the reverse inequality, so \(E\cup F\in\mathcal{C}\). This argument uses the sum of the two pieces in the union; it does not assert that either piece alone has outer measure as large as a larger set.

Now let \(E_1,E_2,\ldots\in\mathcal{C}\) be pairwise disjoint and put \(U=\bigcup_nE_n\). Repeatedly applying the defining criterion gives, for every \(A\subseteq X\) and every positive integer \(N\), \[ \mu^*(A)=\sum_{j=1}^{N}\mu^*(A\cap E_j) +\mu^*\!\left(A\setminus\bigcup_{j=1}^{N}E_j\right). \] The remaining set contains \(A\setminus U\), so monotonicity bounds its outer measure below by \(\mu^*(A\setminus U)\). Letting \(N\) increase yields \(\mu^*(A)\geq\sum_{j=1}^{\infty}\mu^*(A\cap E_j)+\mu^*(A\setminus U)\). Countable subadditivity gives \(\mu^*(A\cap U)\leq\sum_j\mu^*(A\cap E_j)\). Hence \(\mu^*(A)\geq\mu^*(A\cap U)+\mu^*(A\setminus U)\); the reverse inequality follows from subadditivity. Thus \(U\in\mathcal{C}\). Any countable union of members of \(\mathcal{C}\) can be rewritten as a disjoint countable union by subtracting the preceding finite union from each set. Finite-union closure and the disjoint-union result therefore prove that \(\mathcal{C}\) is a sigma-algebra.

Finally, take pairwise disjoint \(E_n\in\mathcal{C}\) and set \(U=\bigcup_nE_n\). The preceding finite splitting with \(A=U\) gives \(\mu^*(U)\geq\sum_{j=1}^{N}\mu^*(E_j)\) for every \(N\). Taking the supremum over \(N\) gives \(\mu^*(U)\geq\sum_j\mu^*(E_j)\), while countable subadditivity gives the reverse inequality. Thus \(\mu^*|_{\mathcal{C}}\) is countably additive on disjoint sequences and vanishes at the empty set, so it is a measure. \(\square\)

Since every \(R\in\mathcal{R}\) belongs to \(\mathcal{C}\), the sigma-algebra \(\mathcal{C}\) contains \(\sigma(\mathcal{R})\). Restricting the measure \(\mu^*|_{\mathcal{C}}\) to \(\sigma(\mathcal{R})\) gives a measure \(\mu\) there. Agreement on the ring was proved above, so \(\mu(R)=\mu_0(R)\) for every \(R\in\mathcal{R}\). This proves the stated existence theorem. It makes no claim of uniqueness; uniqueness questions require additional hypotheses and are treated separately.

Worked Constructions

A Common Covering Error

A countable cover is essential. In the weighted example on \(\mathbb{N}\), a tempting but incorrect calculation is: “The odd integers have outer measure \(+\infty\), because no single finite ring set covers them.” The definition takes an infimum over countable covers, not just one-set covers. Covering each odd integer by its singleton gives total cost \(1/2+1/8+1/32+\cdots=2/3\), and the lower-bound argument in that example proves no cover has smaller total cost. Thus the induced outer measure of the odd integers is exactly \(2/3\), not \(+\infty\).

The construction separates three roles. The premeasure supplies costs on ring sets; the covering infimum defines an outer measure on all subsets; and the Caratheodory criterion identifies a sigma-algebra on which that outer measure is a measure. Ring agreement then ensures that restricting to \(\sigma(\mathcal{R})\) really extends the original premeasure. This existence result does not require the ring to contain \(X\), nor does it require the premeasure to be finite or sigma-finite.