Lemma 4

Let $T$ be a locally compact space, $a$ a point of $T$, $M$ a subset of $\mathscr{M}(T)$, and $\mathfrak{F}$ a filter on $M$. Assume that: (i) for every compact subset $K$ of $T$, the numbers $|\mu|(K)$, for $\mu \in M$, are bounded above; (ii) $\lim_{\mu,\mathfrak{F}} |\mu|(K) = 0$ for every compact subset $K$ of $T - \{a\}$. (iii) there exists a compact neighborhood $V$ of $a$ in $T$ such that $\lim_{\mu,\mathfrak{F}} \mu(V) = 1$.

Then the filter $\mathfrak{F}$ converges to $\varepsilon_a$ in $\mathscr{M}(T)$ equipped with the topology of compact convergence in $\mathscr{H}(T)$.

By the hypothesis (i), $M$ is an equicontinuous subset of $\mathscr{M}(T)$ since it is vaguely bounded and $\mathscr{H}(T)$ is barreled (TVS, III, §4, No. 2, Th. 1). It therefore suffices (GT, X, §2, No. 4, Th. 1) to prove that if $f \in \mathscr{H}(T)$, then $\lim_{\mu,\mathfrak{F}} \mu(f) = f(a)$. Let $K$ be the union of $V$ and the support of $f$; if $K'$ is the closure of $K - V$, one has

since $K'$ is compact and does not contain $a$, one concludes from this that $\lim_{\mu,\mathfrak{F}} \mu(K) = 1$. Let $\varepsilon > 0$, and let $W$ be an open neighborhood of $a$ in $K$ such that $|f(t) - f(a)| \leq \varepsilon$ for $t \in W$; one can write

the integral over $K$ may be written as the sum of the analogous integrals over $W$ and $K - W$; if $C = \sup |f|$, one therefore has

Since the first and third terms on the right side tend to 0 with respect to $\mathfrak{F}$, one sees that indeed $\lim_{\mu,\mathfrak{F}} \mu(f) = f(a)$.