Theorem 5

Let $\Phi$ be a set of subsets of a locally compact space X, satisfying (PCI) and (PC_{II}), and let $\alpha$ be a real-valued function, defined on $\Phi$, satisfying the conditions (PM_I), (PM_{II}) and (PM_{III}). In order that there exist a positive measure $\mu$ on X such that the sets of $\Phi$ are $\mu$-integrable and $\mu(M) = \alpha(M)$ for all $M \in \Phi$, it is necessary and sufficient that $\alpha$ satisfy in addition the following condition:*

(PM_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$, there exist a compact set $K \subset M$ and an open set $U \supset M$ such that, for every $N \in \Phi$ satisfying the relation $K \subset N \subset U$, one has $|\alpha(N) - \alpha(M)| \leq \varepsilon$.

Moreover, if the condition (PM{IV}) is satisfied, then the measure $\mu$ is unique; for every compact set $K$, $\mu(K) = \inf_{M \in \Phi, M \supset K} \alpha(M)$; for every open set $U$, $\mu^*(U) = \sup_{M \in \Phi, M \subset U} \alpha(M)$.*

Note that the condition (PM_{IV}) is equivalent to the conjunction of the following two:

(PM'_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$, there exists an open set $U \supset M$ such that, for every $N \in \Phi$ contained in $U$, $\alpha(N) \leq \alpha(M) + \varepsilon$. (PM''_{IV}) For every $\varepsilon > 0$ and every $M \in \Phi$, there exists a compact set $K \subset M$ such that, for every $N \in \Phi$ containing $K$, $\alpha(N) \geq \alpha(M) - \varepsilon$.

For, it is obvious that (PM'_{IV}) and (PM''_{IV}) imply (PM_{IV}). Conversely, let us show for example that (PM_{IV}) implies (PM'_{IV}): let K be a compact set and U an open set such that $K \subset M \subset U$ and $|\alpha(P) - \alpha(M)| \leq \varepsilon$ for every $P \in \Phi$ satisfying $K \subset P \subset U$. Then, if $N \in \Phi$ and $N \subset U$, $M \cup N$ belongs to $\Phi$ and $K \subset M \cup N \subset U$, whence $\alpha(M \cup N) \leq \alpha(M) + \varepsilon$ and a fortiori $\alpha(N) \leq \alpha(M) + \varepsilon$.

When the set $\Phi$, satisfying (PC_I) and (PC_{II}), consists of compact sets, then the condition (PM''_{IV}) is trivially verified, and (PM_{IV}) is then equivalent to (PM'_{IV}).

The condition (PM_{IV}) is necessary: this follows at once from Th. 4 of No. 6 on the 'approximation' of an integrable set by a compact set and an open set. To prove the other assertions of the theorem, we proceed in several steps.

1° Definition of a topology on $\mathfrak{P}(X)$.

For every pair $(K, U)$ consisting of a compact set $K$ and an open set $U$ in $X$, we denote by $I(K, U)$ the set of subsets $M \subset X$ such that $K \subset M \subset U$; in order that $I(K, U)$ be nonempty, it is necessary and sufficient that $K \subset U$. If $(K', U')$ is a second pair, formed by a compact set $K'$ and an open set $U'$, we have

Let $T$ be the topology on $\mathfrak{P}(X)$ generated by the set of subsets $I(K, U)$ as $K$ runs over the set of compact subsets of $X$, and $U$ over the set of open subsets of $X$; by the foregoing, the $I(K, U)$ form a base for the topology $T$ (GT, I, §1, No. 3).

We observe that the definition of $T$ implies that, in $\mathfrak{P}(X)$, the set of compact subsets of $X$ is dense. The condition (PC_{II}) expresses that $\Phi$ is dense in $\mathfrak{P}(X)$, and condition (PM_{IV}) expresses that the function $\alpha$ is continuous on $\Phi$ for the topology induced by $T$. Finally, Th. 4 of No. 6 expresses that the function $M \mapsto \mu(M)$ is continuous on the clan of $\mu$-integrable sets, for the topology induced by $T$.

2° Uniqueness of $\mu$.

We denote by $\overline{\Phi}$ the set of subsets $M \subset X$ such that $\alpha(N)$ tends to a finite limit as $N$ tends to $M$ (for the topology $T$) while remaining in $\Phi$; we may then extend $\alpha$ in only one way to a continuous mapping $\overline{\alpha}$ of $\overline{\Phi}$ into $\mathbf{R}$ (GT, I, §8, No. 5, Th. 1). If there exists a measure $\mu$ meeting the requirements, the above remarks prove that the clan $\Psi$ of $\mu$-integrable sets is contained in $\overline{\Phi}$ and that $\mu(M) = \overline{\alpha}(M)$ for every $M \in \Psi$; this relation holds in particular for every compact subset $M$ of $X$, which proves the uniqueness of $\mu$ (No. 10, Cor. 3 of Prop. 19).

3° Extension of $\alpha$ to the compact sets.

Without assuming the existence of $\mu$, we are now going to study the set $\overline{\Phi}$ and the extension $\overline{\alpha}$ of $\alpha$ to $\overline{\Phi}$. We first show that every compact set $K$ belongs to $\overline{\Phi}$ and that $\overline{\alpha}(K) = \inf_{P \in \Phi, P \supset K} \alpha(P)$. Set $a = \inf_{P \in \Phi, P \supset K} \alpha(P)$; for every $\varepsilon > 0$, there exists an $M \in \Phi$ such that $K \subset M$ and $\alpha(M) \leq a + \varepsilon$. By (PM'_{IV}), there exists an open set $U \supset M$ such that, for every $N \in \Phi$ contained in $U$, we have $\alpha(N) \leq \alpha(M) + \varepsilon \leq a + 2\varepsilon$; for every $N \in \Phi$ such that $K \subset N \subset U$, we therefore have $a \leq \alpha(N) \leq a + 2\varepsilon$, which, by the definitions, shows that $K \in \overline{\Phi}$ and $\overline{\alpha}(K) = a$.

This result proves at once that if $K_1$ and $K_2$ are two compact sets such that $K_1 \subset K_2$, then $\overline{\alpha}(K_1) \leq \overline{\alpha}(K_2)$. If $K_1$ and $K_2$ are any two compact sets, we have $\overline{\alpha}(K_1 \cup K_2) \leq \overline{\alpha}(K_1) + \overline{\alpha}(K_2)$ by (PM_{II}). We shall see, moreover, that if $K_1$ and $K_2$ are disjoint then $\overline{\alpha}(K_1 \cup K_2) = \overline{\alpha}(K_1) + \overline{\alpha}(K_2)$. For, there then exist two disjoint open sets $U_1, U_2$ such that $K_1 \subset U_1$, $K_2 \subset U_2$ (GT, II, §4, Prop. 4). Therefore, by (PC_{II}), there also exist two sets $M_1 \in \Phi$, $M_2 \in \Phi$ such that $K_1 \subset M_1 \subset U_1$ and $K_2 \subset M_2 \subset U_2$. Now let $P$ be any set of $\Phi$ containing $K_1 \cup K_2$; the union of the two sets $P \cap M_1$ and $P \cap M_2$ belongs to $\Phi$ by (PC_{I}), and since these two sets are disjoint, application of (PM_I) and (PM_{III}) yields

which establishes our assertion.

4° Extension of $\alpha$ to the open sets.

We shall now see that, for an open set $U$ to belong to $\overline{\Phi}$, it is necessary and sufficient that, as $K$ runs over the set of compact subsets of $U$, the supremum of the numbers $\overline{\alpha}(K)$ is finite; moreover, $\overline{\alpha}(U)$ is then equal to this supremum.

For, let $U$ be an open set belonging to $\overline{\Phi}$; for every $\varepsilon > 0$ there exists a compact set $K \subset U$ such that, for every set $M \in \Phi$ satisfying $K \subset M \subset U$, one has $|\overline{\alpha}(U) - \alpha(M)| \leq \varepsilon$, whence $|\overline{\alpha}(U) - \overline{\alpha}(K)| \leq \varepsilon$; on the other hand, if $K'$ is any compact set contained in $U$, then $K \subset K \cup K' \subset U$, whence $|\overline{\alpha}(U) - \overline{\alpha}(K \cup K')| \leq \varepsilon$ and so $\overline{\alpha}(U) \geq \overline{\alpha}(K \cup K') - \varepsilon \geq \overline{\alpha}(K') - \varepsilon$; $\overline{\alpha}(U)$ is therefore indeed equal to the supremum of the numbers $\overline{\alpha}(K)$ as $K$ runs over the set of compact subsets of $U$.

Conversely, let $U$ be an open set such that $b = \sup_{K \subset U} \overline{\alpha}(K) < +\infty$ ($K$ running over the set of compact subsets of $U$), and let us show that $U \in \overline{\Phi}$. For every $\varepsilon > 0$, there exists a compact set $K \subset U$ such that $b - \varepsilon \leq \overline{\alpha}(K) \leq b$; by (PM''_{IV}), for every set $M \in \Phi$ such that $K \subset M \subset U$, there exists a compact set $K' \subset M$ such that

therefore $b - \varepsilon \leq \alpha(M) \leq b + \varepsilon$, which proves that $U \in \overline{\Phi}$.

From this characterization of the open sets $U \in \overline{\Phi}$, and of $\overline{\alpha}(U)$, it follows first of all that if $U_1$ and $U_2$ are two open sets such that $U_1 \subset U_2$ and $U_2 \in \overline{\Phi}$, then $U_1 \in \overline{\Phi}$ and $\overline{\alpha}(U_1) \leq \overline{\alpha}(U_2)$. On the other hand, if $U_1$ and $U_2$ are two open sets belonging to $\overline{\Phi}$, then the same is true of $U_1 \cup U_2$, and $\overline{\alpha}(U_1 \cup U_2) \leq \overline{\alpha}(U_1) + \overline{\alpha}(U_2)$. For, let $K$ be any compact set contained in $U_1 \cup U_2$; for every point $x \in K$, there exists a compact neighborhood of $x$ contained in either $U_1$ or $U_2$; one can therefore cover $K$ by a finite number of these neighborhoods; if $K_1$ (resp. $K_2$) is the union of those that are contained in $U_1$ (resp. $U_2$), then $K \subset K_1 \cup K_2$, whence

which establishes the asserted property.

5° Properties of $\overline{\Phi}$ and $\overline{\alpha}$.

The definition of $\overline{\Phi}$ and $\overline{\alpha}$ can now be transformed as follows (taking into account (PC_{II})): in order that $M \in \overline{\Phi}$, it is necessary and sufficient that, for every $\varepsilon > 0$, there exist a compact set $K$ and an open set $U \in \overline{\Phi}$ such that $K \subset M \subset U$ and $\overline{\alpha}(U) - \overline{\alpha}(K) \leq \varepsilon$; $\overline{\alpha}(M)$ is, moreover, the infimum of the $\overline{\alpha}(U)$ for the open sets $U \in \overline{\Phi}$ containing $M$, and the supremum of the $\overline{\alpha}(K)$ for the compact sets $K \subset M$.

From this, we shall first deduce that if $M_1, M_2$ and $M_1 \cup M_2$ belong to $\overline{\Phi}$, then $\overline{\alpha}(M_1 \cup M_2) \leq \overline{\alpha}(M_1) + \overline{\alpha}(M_2)$. Indeed, if $U_1$ and $U_2$ are two open sets of $\overline{\Phi}$ containing $M_1$ and $M_2$, respectively, and such that $\overline{\alpha}(U_1) \leq \overline{\alpha}(M_1) + \varepsilon$ and $\overline{\alpha}(U_2) \leq \overline{\alpha}(M_2) + \varepsilon$, then $U_1 \cup U_2$ belongs to $\overline{\Phi}$, contains $M_1 \cup M_2$, and consequently

whence our assertion.

Next, let us show that if $K$ is a compact set and $U$ is an open set of $\overline{\Phi}$ such that $K \subset U$, then $\overline{\alpha}(U - K) = \overline{\alpha}(U) - \overline{\alpha}(K)$. By the foregoing, we have $\overline{\alpha}(U) \leq \overline{\alpha}(K) + \overline{\alpha}(U - K)$. On the other hand, for every compact set $K' \subset U - K$,

since $U - K$ is open and belongs to $\overline{\Phi}$, $\overline{\alpha}(U - K)$ is the supremum of the $\overline{\alpha}(K')$, which shows that $\overline{\alpha}(K) + \overline{\alpha}(U - K) \leq \overline{\alpha}(U)$.

The definition of $\overline{\Phi}$ may therefore now be expressed in the following way: in order that $M \in \overline{\Phi}$, it is necessary and sufficient that, for every $\varepsilon > 0$, there exist a compact set $K$ and an open set $U \in \overline{\Phi}$ such that $K \subset M \subset U$ and $\overline{\alpha}(U - K) \leq \varepsilon$.

We are now in a position to prove that $\overline{\Phi}$ is a clan and $\overline{\alpha}$ an additive set function on $\overline{\Phi}$. We first show that if $M$ and $N$ belong to $\overline{\Phi}$ then so do $M \cap \mathbf{C}N$ and $M \cup N$. By hypothesis, for every $\varepsilon > 0$ there exist two compact sets $K, K'$ and two open sets $U, U'$ of $\overline{\Phi}$ such that

The set $K'' = K \cap \mathbf{C}U'$ is compact, the set $U'' = U \cap \mathbf{C}K'$ is open and belongs to $\overline{\Phi}$, and $K'' \subset M \cap \mathbf{C}N \subset U''$; on the other hand, $U'' - K''$ is contained in the union of $U \cap \mathbf{C}K$ and $U' \cap \mathbf{C}K'$, whence $\overline{\alpha}(U'' - K'') \leq 2\varepsilon$, which proves that $M \cap \mathbf{C}N \in \overline{\Phi}$. Similarly, $U_1 = U \cup U'$ is open and belongs to $\overline{\Phi}$, $K_1 = K \cup K'$ is compact, and $K_1 \subset M \cup N \subset U_1$; on the other hand, $U_1 - K_1$ is contained in the union of $U - K$ and $U' - K'$, whence again $\overline{\alpha}(U_1 - K_1) \leq 2\varepsilon$, and $M \cup N$ belongs to $\overline{\Phi}$. Finally, if $M$ and $N$ are disjoint, then

consequently $\overline{\alpha}(M \cup N) \geq \overline{\alpha}(M) + \overline{\alpha}(N) - 2\varepsilon$; since $\varepsilon$ is arbitrary, we have $\overline{\alpha}(M \cup N) = \overline{\alpha}(M) + \overline{\alpha}(N)$.

6° Existence of the measure $\mu$.

By Prop. 18 of No. 9, there exists one and only one positive linear form $\beta$ on the vector space $\mathscr{E}(\overline{\Phi})$ of $\overline{\Phi}$-step functions, such that $\beta(\varphi_M) = \overline{\alpha}(M)$ for all $M \in \overline{\Phi}$. For every compact subset $K$ of $X$, let us denote by $\mathscr{G}(K)$ the space of uniform limits of functions of $\mathscr{E}(\overline{\Phi})$ whose support is contained in $K$. Since $\beta$ is positive, $|\beta(f)| \leq \overline{\alpha}(K) \cdot \|f\|$ for every function $f \in \mathscr{E}(\overline{\Phi})$ whose support is contained in $K$; the restriction of $\beta$ to the space of these functions is a continuous linear form for the topology of uniform convergence; it may therefore be extended to a positive continuous linear form $\overline{\beta}_K$ on $\mathscr{G}(K)$. Moreover, if $K \subset K_1$, then the restriction of $\overline{\beta}_{K_1}$ to $\mathscr{G}(K)$ is identical to $\overline{\beta}_K$, therefore there exists a positive linear form $\overline{\beta}$ on the union $\mathscr{G}$ of the $\mathscr{G}(K)$, that extends each of the forms $\overline{\beta}_K$.

Now, since every compact set belongs to $\overline{\Phi}$, the space $\mathscr{H}$ of continuous real-valued functions with compact support is a subspace of $\mathscr{G}$ (No. 10, Prop. 19); the restriction to $\mathscr{H}$ of the positive linear form $\overline{\beta}$ is therefore a positive measure $\mu$. Let us show that for every compact set $K$, $\mu(K) = \overline{\alpha}(K)$. For every $\varepsilon > 0$, there exists an open set $U \in \overline{\Phi}$ such that $K \subset U$, $\mu(U) \leq \mu(K) + \varepsilon$ and $\overline{\alpha}(U) \leq \overline{\alpha}(K) + \varepsilon$. Let $f$ be a continuous mapping of $X$ into $[0, 1]$ whose support is contained in $U$ and such that $f(x) = 1$ on $K$ (Ch. III, §1, No. 2, Lemma 1). Then $\mu(K) \leq \mu(f) \leq \mu(U) \leq \mu(K) + \varepsilon$, and, on the other hand,

since $\mu(f) = \overline{\beta}(f)$, we see that $|\mu(K) - \overline{\alpha}(K)| \leq \varepsilon$, and since $\varepsilon$ is arbitrary, $\mu(K) = \overline{\alpha}(K)$.

The characterization of the open sets belonging to $\overline{\Phi}$, combined with Cor. 4 of Th. 4 of No. 6, then shows that the open sets belonging to $\overline{\Phi}$ are none other than the $\mu$-integrable open sets, and that, for such a set $U$, we have $\mu(U) = \overline{\alpha}(U)$. Th. 4 of No. 6 and the characterization of the sets of $\overline{\Phi}$ given in 5° then show that the $\mu$-integrable sets are the sets of $\overline{\Phi}$ and that, for such a set $M$, $\mu(M) = \overline{\alpha}(M)$. Finally, the fact that $\mu^*(U) = \sup_{M \in \Phi, M \subset U} \alpha(M)$ for every open set $U$ follows at once from (PC$_{II}$) and Cor. 4 of Th. 4 of No. 6.

Theorem 5 is thus completely proved.