Proposition 3

If the image under $m$ of the set of $f \in \mathscr{K}(T)$ such that $\sup_{t \in T} |f(t)| \leq 1$ is weakly relatively compact in $F$, then $\int f dm \in F$ for every bounded numerical function $f$ essentially integrable for $m$.

Let $A$ be the set of $f \in \mathscr{L}(m)$ such that $\sup_{t \in T} |f(t)| \leq 1$, and let $B = A \cap \mathscr{K}(T)$; by hypothesis, $m(B)$ is weakly relatively compact in $F$, therefore it suffices to show that $m(A)$ is contained in the closure (in ${F'}^*$) of $m(B)$ for the topology $\sigma({F'}^*, F')$; since $m(B)$ is convex and balanced, it suffices to prove that the polar of $m(B)$ in $F'$ is contained in that of $m(A)$ (TVS, II, §6, No. 3, Th. 1). Now, for a linear form $z' \in F'$ to belong to $(m(B))^\circ$, it is necessary and sufficient that $|\langle z', m(g)\rangle| = |\int g d(z'\circ m)| \leq 1$ for every function $g \in B$, which signifies that the scalar measure $|z'\circ m|$ is bounded and of norm $\leq 1$ (Ch. III, §1, No. 8); but by (2) the latter condition implies that $|\langle z', m(f)\rangle| \leq 1$ for every function $f \in A$, whence $z' \in (m(A))^\circ$.