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

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.

It is clear that $J$ is an increasing set function, extending $I$, and that $J(A) + J(\mathbf{C}A) \leq l$ for every $A \subset T$.

Let $K$ and $S$ be two compact sets in $T$; we are first going to show that

On considering the restrictions of $I$ to $\mathscr{K}(S)$ and of $J$ to $\mathfrak{P}(S)$, one reduces immediately to the case that $S = T$. Since $T$ is normal, $K$ is the intersection of the decreasing directed family of its compact neighborhoods, and condition 4) implies the existence, for every $\varepsilon > 0$, of a compact neighborhood $H$ of $K$ such that $I(H) \leq I(K) + \varepsilon$. Let $L$ be the closure of $T - H$; $L$ is compact, $L \cap K = \varnothing$ and $H \cup L = T$, therefore $l = I(H \cup L) \leq I(H) + I(L) \leq I(K) + I(L) + \varepsilon$ (condition 2)), whence the relation $J(K) + J(\mathbf{C}K) \geq I(K) + I(L) \geq l - \varepsilon$. Since $\varepsilon$ is arbitrary, $J(K) + J(\mathbf{C}K) = l$. This proves the formula (2), as well as the inclusion $\mathscr{K}(T) \subset \Phi$.

Let us now prove that $\Phi$ is a clan. Since $\Phi$ is obviously stable under passage to the complement, it suffices to show that if $A_1$ and $A_2$ denote elements of $\Phi$, then $A_1 \cup A_2 \in \Phi$, or again that

Denote by $\varepsilon$ a number $> 0$, and, for $i = 1, 2$, let $K_i$ be a compact set contained in $A_i$, and $L_i$ a compact set contained in $\mathbf{C}A_i$, such that

Set $M_1 = K_1 \cup L_1$; the relations $l = J(M_1) + J(\mathbf{C}M_1)$ and

imply that $J(\mathbf{C}M_1) \leq 2\varepsilon$. Then if $S$ is a compact subset of $T$, the relation (2) (applied to $K = M_1$) implies $J(S) \leq J(M_1 \cap S) + 2\varepsilon$, whence

Let us add the inequalities obtained by making $S = K_2$ and $S = L_2$, and take into account the inequality $J(K_2) + J(L_2) \geq l - 2\varepsilon$ and the fact that $K_1 \cap K_2$, $L_1 \cap K_2$ and $K_1 \cap L_2$ are three disjoint compact sets contained in $A_1 \cup A_2$. Denoting by $C$ the union of these three compact sets, it follows that

whence immediately the desired formula (3), in view of the arbitrariness of $\varepsilon$. This having been established, the preceding inequalities imply that $J(C) \geq J(A_1 \cup A_2) - 6\varepsilon$; if $A_1$ and $A_2$ are disjoint, then $C$ is the union of $K_1 \cap L_2 \subset A_1$ and $K_2 \cap L_1 \subset A_2$, from which one deduces that $J(A_1 \cup A_2) \leq J(A_1) + J(A_2)$. The reverse inequality being obvious, $J$ is indeed additive on $\Phi$, and the lemma is established.

Let us complete the proof of the theorem for the case that $T$ is compact. Let $\mathscr{E}(\Phi)$ be the vector space of $\Phi$-step functions on $T$, equipped with the topology of uniform convergence (Ch. IV, §4, No. 9, Def. 4); we shall again denote by $J$ the positive linear form on $\mathscr{E}(\Phi)$ associated with the additive function $J$ (loc. cit., Prop. 18). Since $J(T) = l$, $J$ is continuous and has norm $l$. Let then $\mathscr{H}$ be the closure of $\mathscr{E}(\Phi)$ for the topology of uniform convergence; one verifies at once that $J$ may be extended by continuity to a positive linear form on $\mathscr{H}$, again denoted $J$. Since $\mathscr{H}$ contains $\mathscr{C}(T)$ (loc. cit., No. 10, Prop. 19) the restriction of $J$ to $\mathscr{C}(T)$ is a positive measure $\mu$. It remains to show that $\mu^\bullet(K) = I(K)$ for every compact subset $K$ of $T$. Now, we have $\mu^\bullet(K) = \inf_{f \in S_K} \mu^\bullet(f)$, where $S_K$ denotes the set of elements of $\mathscr{C}(T)$ that are $\geq \varphi_K$ (§ 1, No. 6, Prop. 5). Since $J(f) = \mu^\bullet(f)$ for $f \in \mathscr{C}(T)$, it clearly suffices to show that $J(K) \geq \inf_{f \in S_K} J(f)$. As in the proof of Lemma 1, let $H$ be a compact neighborhood of $K$ such that $J(H) \leq J(K) + \varepsilon$, and let $f$ be a continuous function on $T$, between 0 and 1, equal to 1 on $K$ and to 0 outside $H$ (GT, IX, §4, No. 1, Prop. 1). Then

$\varepsilon$ being arbitrary, the desired inequality is proved, and the theorem is thus established when $T$ is compact.

Let us now pass to the general case. For every compact set $L$ in $T$, let $I_L$ be the restriction of $I$ to $\mathfrak{K}(L)$. By the special case just treated, there exists a measure $\mu_L$ on $L$, unique, such that $\mu_L(K) = I_L(K)$ for every compact set $K \subset L$. Let then $L'$ be a compact set contained in $L$; we have $(\mu_L)_{L'}^\bullet(K) = \mu_{L'}^\bullet(K) = \mu_{L'}^\bullet(K)$ for every compact set $K \subset L'$, therefore $\mu_{L'} = (\mu_L)_{L'}$; the mapping $\mu : L \mapsto \mu_L$ is a premeasure. The condition 5) expresses that $\mu$ is a measure, and the relation $I(K) = \mu^*(K)$ for all compact $K \subset T$ is obvious.

These three conditions are necessary: for, the mapping $A \mapsto \mu^*(A)$ on $\mathscr{B}(T)$ is countably additive (§ 1, No. 5, Cor. of Prop. 4), locally bounded by the definition of measures (§ 1, No. 2, Def. 5), and inner regular by Remark 3 of § 1, No. 2.

We pass to existence. It is clear that the restriction of I to $\mathscr{K}(T)$ satisfies conditions 1), 2), 3) and 5) of the statement of Th. 1; let us show that 4) is satisfied as well. Let K be a compact subset of T, the intersection of a decreasing directed family $(K_\alpha)_{\alpha \in A}$ of compact sets, and let $\varepsilon$ be a number $> 0$; I being locally bounded, there exists an open (hence Borel) neighborhood V of K such that $I(V) < +\infty$, and then there exists an index $\alpha$ such that $K_\alpha \subset V$; changing notation if necessary, we can suppose that $K_\alpha \subset V$ for all $\alpha \in A$. By the inner regularity of I, there exists a compact set $L \subset V - K$ such that $I(L) \geq I(V - K) - \varepsilon$; since L does not intersect K, there exists an index $\alpha$ such that $L \cap K_\alpha = \varnothing$, and one then has $I(V - K_\alpha) \geq I(L) \geq I(V - K) - \varepsilon$. Since $K_\alpha \subset V$, it follows that $I(K_\alpha) \leq I(K) + \varepsilon$ and the condition 4) is verified.

By Th. 1, there exists a measure $\mu$ such that $\mu^\bullet(K) = I(K)$ for all $K \in \mathscr{K}(T)$. The inner regularity of the set functions $\mu^\bullet$ and $I$ on $\mathscr{B}(T)$ then implies that $\mu^\bullet(A) = I(A)$ for all $A \in \mathscr{B}(T)$, and existence is proved. The uniqueness of $\mu$ follows from the uniqueness assertion of Th. 1.

Let T be a Lindelöf space that is Radon, and let I be a set function on the tribe $\mathscr{B}(T)$ that is positive, countably additive and locally bounded. The open sets V such that $I(V) < +\infty$ form a covering of T, from which one can extract a countable covering $(V_n)_{n \in \mathbf{N}}$. Set $G_n = V_0 \cup V_1 \cup \cdots \cup V_n$ for every $n \in \mathbf{N}$; set $H_0 = G_0$ and $H_n = G_n - G_{n-1}$ for $n \geq 1$; finally, denote by $I_n$ the set function $A \mapsto I(A \cap H_n)$ on $\mathscr{B}(T)$, which is obviously countably additive and bounded. Since the sets $H_n$ form a partition of T, we have $I = \sum_n I_n$. The space T being Radon, for every $n \in \mathbf{N}$ there exists a bounded measure $\mu_n$ on T such that $\mu_n^\bullet(A) = I_n(A)$ for all $A \in \mathscr{B}(T)$; therefore also $\sum_n \mu_n^\bullet(A) = I(A)$. Since I is locally bounded, the family $(\mu_n)$ is summable (§ 1, No. 7, Prop. 7); if $\mu$ denotes $\sum_n \mu_n$, we have $\mu^\bullet(A) = I(A)$ for all $A \in \mathscr{B}(T)$, and it follows that I is inner regular. In other words, T is strongly Radon.

Recall that a subset A of a topological space T is said to be universally measurable if A is $\mu$-measurable for every measure $\mu$ on T. This is equivalent to saying that A is $\mu$-measurable for every measure $\mu$ on T with compact support (§ 1, No. 8, Prop. 9).

(1) Recall (GT, I, §9, Exer. 14; TG, IX, Appendix I) that a Lindelöf space is a topological space T such that every open covering of T contains a countable covering.

a) Suppose that T is a Radon space. Then, for every function I defined on $\mathscr{B}(X)$ that is positive, countably additive and bounded, one has

Moreover, T is universally measurable in X.

b) Conversely, suppose that X is a Radon space and that T is universally measurable in X; then T is a Radon space.

Let us prove a). We denote by $\alpha$ the right side of (6); for every $n \in \mathbf{N}$, there exists a set $C_n \in \mathscr{B}(X)$ containing T, such that $I(C_n) \leq \alpha + 2^{-n}$. Setting $C = \bigcap_n C_n$, we then have $T \subset C$, $I(C) = \alpha$. If $A \in \mathscr{B}(T)$, let us choose a Borel set B of X such that $A = B \cap T$ (GT, IX, §6, No. 3) and set $J(A) = I(B \cap C)$. This number does not depend on the choice of B, for if $B'$ is a second Borel set in X such that $A = B' \cap T$, then $B \cap C$ and $B' \cap C$ differ only by a Borel set M contained in $C - T$, and $I(M) = 0$ by the construction of C. Clearly $J(K) = I(K)$ for every compact set $K \subset T$. Let $(A_n)$ be a sequence of Borel sets of T, pairwise disjoint, and, for each $n$, let $B_n$ be a Borel set of X such that $B_n \cap T = A_n$. Replacing $B_n$ by $B_n - \left( \bigcup_{k < n} B_k \right)$ if necessary, we can suppose that the sets $B_n$ are pairwise disjoint. Set $A = \bigcup_n A_n$ and $B = \bigcup_n B_n$; then

J is thus a countably additive and bounded function on $\mathscr{B}(T)$. Since T is by hypothesis a Radon space, there exists a bounded measure $\mu$ on T such that $J(A) = \mu^\bullet(A)$ for every $A \in \mathscr{B}(T)$; consequently

by the definition of $\mu^\bullet$. Formula (6) is thus established.

Let us now show that T is universally measurable. Let $\lambda$ be a bounded measure on X; the preceding argument may be applied to the set function $I : A \mapsto \lambda^\bullet(A)$ on $\mathscr{B}(X)$, thus there exists a sequence $(K_n)$ of compact subsets of T such that (with the above notations)

Set $K' = \bigcup_{n \in \mathbf{N}} K_n$; $K'$ is Borel in $X$, $K' \subset T \subset C$, $\lambda^*(K') = \lambda^*(C)$, therefore these three sets differ only by $\lambda$-negligible sets, and so $T$ is $\lambda$-measurable. This completes the proof of a).

Let us pass to b). Suppose that $X$ is a Radon space, and that $T$ is universally measurable in $X$. Let $I$ be a positive function on $\mathscr{B}(T)$ that is countably additive and bounded; the function $A \mapsto I(A \cap T)$ on $\mathscr{B}(X)$ is then positive, countably additive and bounded, therefore there exists a bounded measure $\nu$ on $X$ such that $I(A \cap T) = \nu^*(A)$ for all $A \in \mathscr{B}(X)$. Now, $T$ is $\nu$-measurable; the preceding relation shows that $\nu^*(K) = 0$ for every compact subset $K$ of $X$ that is disjoint from $T$, therefore $\nu$ is concentrated on $T$. Consequently, for every Borel set $A$ of $X$, we have $I(A \cap T) = \nu^*(A \cap T) = \mu^*(A \cap T)$, where $\mu$ is the measure induced by $\nu$ on $T$. Finally, it follows that $I(B) = \mu^*(B)$ for every set $B \in \mathscr{B}(T)$ (GT, IX, §6, No. 3, Remark 2), and $I$ is indeed inner regular.

For, $T$ is universally measurable in $X$.

Let $T$ be a Souslin space; since $T$ is a Lindelöf space (TG, IX, Appendix I, Cor. of Prop. 1),(2) it suffices to show that $T$ is Radon (Prop. 1). Let $I$ be a function defined on $\mathscr{B}(T)$, positive, countably additive and bounded. We extend $I$ to $\mathscr{P}(T)$ by setting, for every subset $A$ of $T$,

Let us show that this extension is a capacity on $T$ (GT, IX, §6, No. 9). It is clear that the relation $A \subset A'$ implies $I(A) \leq I(A')$. Let $(A_n)$ be an increasing sequence of subsets of $T$, and let $A = \bigcup_n A_n$. The set of Borel sets that contain $A_n$ being stable for countable intersections, there exists for each $n$ a Borel set $B_n$ such that $A_n \subset B_n$ and $I(A_n) = I(B_n)$ (cf. the proof of Prop. 2). Set $C_n = \bigcap_{p \geq n} B_p$; $C_n$ is Borel, and $A_n \subset C_n \subset B_n$, therefore $I(A_n) = I(C_n)$. On the other hand, the sequence $(C_n)$ is increasing. Let $C = \bigcup_n C_n$: the relation $A \subset C$ implies that

(2) Every Souslin space has a countable base for open sets (GT, IX, §6, No. 2, Prop. 4), hence is Lindelöf (GT, I, §9, Exer. 14).

whence the equality $I(A) = \lim_n I(A_n)$ is immediate. Consequently, I is a capacity.

If $(H_n)$ is a decreasing sequence of closed sets in T, obviously $I\left( \bigcap_n H_n \right) = \inf_n I(H_n)$. It follows that every Souslin subset F of T is capacitable for I (TG, IX, §6, No. 10, Prop. 15). In particular, every Borel set A of T is capacitable (loc. cit., §6, No. 3, Prop. 10).(3) In other words,

where K runs over the set of compact sets contained in A; we have proved that I is inner regular.