Proposition 22

— *Let F be a Banach space, X a locally compact space, $\mathscr{E}$ the set of bounded positive measures on X, $\mu$ an element of $\mathscr{E}$, and $\mathscr{B}$ a filter base on $\mathscr{E}$. Assume that $\mathscr{B}$ converges vaguely to $\mu$ and that $\| \nu \|$ tends to $\| \mu \|$ with respect to $\mathscr{B}$. Let f be a mapping of X into F satisfying the following conditions:

(i) f is bounded, and is integrable for $\mu$ and for every measure that belongs to some element of $\mathscr{B}$; (ii) the set of points of discontinuity of f is $\mu$-negligible. Then $\int f \, d\nu$ tends to $\int f \, d\mu$ with respect to $\mathscr{B}$.

Let $\varepsilon > 0$. There exist elements $a_1, \ldots, a_n$ in F, functions $g_1, \ldots, g_n$ in $\mathscr{K}(X)$, and a bounded continuous function $h \geq 0$ on X, such that

on X and $\int h \, d\mu \leq \varepsilon$ (Lemma 5). Let $M = \sup |f|$. There exist a compact subset K of X such that $\mu^*(X - K) \leq \varepsilon$ (§ 4, No. 7, Prop. 12 and No. 6, Th. 4), a compact neighborhood $K'$ of K in X, and a continuous mapping $h'$ of X into $[0, 2M]$ such that $h' = h$ on K, $h' = 2M$ on $X - K'$; replacing $h'$ by $\sup(h, h')$, we can suppose in addition that $h' \geq h$. Then $\int (h' - h) \, d\mu \leq 2M \mu^*(X - K) \leq 2M \varepsilon$. On the other hand, $h' = h_1 + 2M$, where $h_1 \in \mathscr{K}(X)$. Taking into account § 4, No. 7, Prop. 12, the number

These inequalities imply which proves the proposition.