Proposition 11

— Let $(\lambda_\alpha)_{\alpha \in A}$ be a family of positive measures on T, directed for the relation $\leq$ and having a supremum $\lambda$ in $\mathscr{M}(T)$. Then, for every numerical function $f \geq 0$,

When $f$ belongs to $\mathscr{K}(T)$, this relation reduces to the definition of the supremum of a directed set in $\mathscr{M}(T)$ (Ch. II, §2, No. 2, Lemma). Suppose next that $f \leq g$ for some function $g \in \mathscr{K}_+$ (in other words, that $f$ is bounded and is zero outside a compact set K); let $\alpha$ be an index such that $\lambda_\alpha(g) \geq \lambda(g) - \varepsilon$, where $\varepsilon$ is a number $> 0$; since the measure $\nu = \lambda - \lambda_\alpha$ is positive, we have $\nu^*(f) \leq \nu(g) \leq \varepsilon$, or $\lambda_\alpha^*(f) \geq \lambda^*(f) - \varepsilon$ (Ch. IV, §1, No. 3, Prop. 15). It follows (since $\varepsilon$ is arbitrary) that the second member of (5) is $\geq$ the first; the reverse inequality being obvious, (5) is established for the special case under consideration. Next, suppose that $f$ is zero outside K but is not necessarily bounded, and set $f_n = \inf(f, n)$ for every integer $n$. Then

Finally, with no restriction made on $f$, denoting by $\mathscr{K}$ the set of compact subsets of $T$ we have