Theorem 3

Let S be a full submonoid of X. a) If $\mu$ and $\mu'$ are two bounded complex measures on $M$, such that $\mathscr{L}\mu$ and $\mathscr{L}\mu'$ have the same restriction to $S \cap X_0$, then $\mu = \mu'$. b) Let $\mathfrak{F}$ be a filter on $\mathscr{M}_+^b(M)$, such that $\mathscr{L}\lambda(s)$ has a limit $\Phi(s) \in \mathbf{C}$ with respect to $\mathfrak{F}$ for every $s \in S$. Then the filter $\mathfrak{F}$ converges vaguely to a bounded positive measure $\mu$, and $\Phi(s) = \mathscr{L}\mu(s)$ for all $s \in S \cap X_0$. c) Under the hypotheses of b), suppose in addition that the closure of $S \cap X_0$ contains 1, and that the function $\Phi$ on $S$ is continuous at the point 1. Then $\mathfrak{F}$ converges tightly to $\mu$, and $\Phi(s) = \mathscr{L}\mu(s)$ for all $s \in S$.

We shall denote by E the algebra of continuous complex functions tending to 0 at infinity on $M$, and by $\mathfrak{A}$ the linear subspace of E generated

(1) Recall that a submonoid of a monoid A contains by definition the neutral element of A (A, I, §2, No. 1).

by $S \cap X_0$; then $\mathfrak{A}$ is a subalgebra of $E$ stable under the mapping $f \mapsto \overline{f}$; since $S$ is a full submonoid of $X$, Cor. 2 of Prop. 7 of GT, X, §4, No. 4 implies that $\mathfrak{A}$ is dense in $E$.

Let us prove a): by hypothesis, $\mu(f) = \mu'(f)$ for every $f \in \mathfrak{A}$; since $\mu$ and $\mu'$ are continuous linear forms on $E$, this implies that $\mu(f) = \mu'(f)$ for $f \in E$, and in particular for every continuous function $f$ with compact support, whence $\mu = \mu'$.

Let us place ourselves under the hypotheses of b). The number $\Phi(1) = \lim_{\lambda, \mathfrak{F}} \lambda(1)$ is real and positive; let there be given a real number $a > \Phi(1)$; since $\| \lambda \| = \mathscr{L} \lambda(1)$ for $\lambda \in \mathscr{M}_+^b(M)$, the relation $\lim_{\lambda, \mathfrak{F}} \mathscr{L} \lambda(1) = \Phi(1)$ implies that the set $H$ of measures $\lambda \in \mathscr{M}_+^b(M)$ such that $\| \lambda \| \leq a$ belongs to $\mathfrak{F}$. Since $\mathscr{M}^b(M; \mathbf{C})$ may be identified with the dual of the normed space $E$ (Ch. III, §1, No. 8 & §1, No. 2, Prop. 3), the space $H$ is compact for the topology $\sigma(\mathscr{M}^b(M; \mathbf{C}), E)$. On the other hand (TVS, III, §3, No. 4, Prop. 5), this topology coincides on $H$ with the topology of pointwise convergence in any total subset of $E$. In particular, since $\mathfrak{A}$ is dense in $E$, and the same is true of the space of continuous functions with compact support (Ch. III, §1, No. 2, Prop. 3), the topology of pointwise convergence in $S \cap X_0$ coincides on $H$ with the vague topology, and $H$ is compact for this topology. It follows at once that $\mathfrak{F}$ converges vaguely to a measure $\mu \in H$, and that $\mathscr{L} \mu(s) = \lim_{\lambda, \mathfrak{F}} \mathscr{L} \lambda(s)$ for all $s \in S \cap X_0$.

Finally, let us pass to c). Since the functions $\Phi$ and $\mathscr{L} \mu$ are continuous at the point $1 \in S$, and equal on $S \cap X_0$, and since $1$ is in the closure of $S \cap X_0$, we have $\Phi(1) = \mathscr{L} \mu(1)$. In other words, $\lim_{\lambda, \mathfrak{F}} \lambda(1) = \mu(1)$. Prop. 9 of No. 3 then shows that $\mu$ is the tight limit of the filter $\mathfrak{F}$. Every element of $S$ being a bounded continuous function on $M$, this implies that $\Phi(s) = \lim_{\lambda, \mathfrak{F}} \lambda(s) = \mu(s) = \mathscr{L} \mu(s)$ for all $s \in S$.