Theorem 1

— Let T be a topological space, and I a mapping of $\mathscr{K}(T)$ into $\mathbf{R}_+$. In order that there exist a measure $\mu$ on T such that $I(K) = \mu^\bullet(K)$ for all $K \in \mathscr{K}(T)$, it is necessary and sufficient that I satisfy the following conditions:

1) If K and L are compact subsets of T such that $K \subset L$, then $I(K) \leq I(L)$ ('I is increasing').

2) If K and L are compact subsets of T, then $I(K \cup L) \leq I(K) + I(L)$.

3) If K and L are disjoint compact subsets of T, then $I(K \cup L) = I(K) + I(L)$ ('I is additive').

4) For every decreasing directed family $(K_\alpha)_{\alpha \in A}$ of compact subsets of T, one has $I(\bigcap_{\alpha \in A} K_\alpha) = \inf_{\alpha \in A} I(K_\alpha)$.

5) For every $x \in T$, there exists a neighborhood V of x such that

('I is locally bounded').

The measure $\mu$ is then unique.

The uniqueness of $\mu$ follows from the Cor. of Prop. 2 of § 1, No. 2. The above conditions are necessary, the first three in obvious fashion, the last from the fact that $\mu^\bullet$ is a locally bounded encumbrance, and condition 4) by the Cor. of Prop. 5 of § 1, No. 6.

To show that these conditions are sufficient, we begin by treating the case that T is compact.