Remark

— 1) The mapping $t \mapsto \varepsilon_t$ is a homeomorphism of T into the space $\mathscr{C}'(T)$; for, if L is a compact subset of $\mathscr{C}(T)$, and $t_0 \in T$, there exists (GT, X, §2, No. 5, Cor. 3 of Th. 2) a neighborhood W of $t_0$ such that $|g(t) - g(t_0)| \leq 1$ for every $t \in W$ and every function $g \in L$, therefore $\varepsilon_t - \varepsilon_{t_0} \in L^\circ$ for $t \in W$, which proves the continuity of $t \mapsto \varepsilon_t$ (cf. Ch. IV, §4, No. 8, Prop. 15); one knows in addition that the inverse mapping is already continuous for the vague topology (Ch. III, §1, No. 9, Prop. 13), hence a fortiori for the topology of uniform convergence in the compact subset of $\mathscr{C}(T)$. If one then identifies T with its image in $\mathscr{C}'(T)$ via $t \mapsto \varepsilon_t$, one can say that $\lambda \mapsto \int f d\lambda$ is the unique continuous extension of f to a linear mapping.

2) Note that in the proof of the continuity of $\lambda \mapsto \int f d\lambda$, we have not used the fact that F is quasi-complete. The conclusion of Prop. 14 is therefore still valid without this hypothesis, provided one knows in addition that $\int f d\mu \in F$ for every positive measure $\mu$ with compact support.

Suppose now that $f(T)$ is a bounded subset of F. Then, for every bounded positive measure $\mu$ on T, f is scalarly $\mu$-integrable and $\int f d\mu \in F$

(No. 2, Prop. 8). If $\lambda$ is any bounded real measure on $T$, then $\lambda^+$ and $\lambda^-$ are bounded, and one sees immediately that $\lambda \mapsto \int f\, d\lambda$, defined as above, is a linear mapping of the space $\mathscr{M}^1(T)$ of bounded measures on $T$, into the locally convex space $F$, that obviously extends the mapping $\lambda \mapsto \int f\, d\lambda$ of $\mathscr{C}'(T)$ into $F$.