Proposition 7

— Let $f$ be a mapping scalarly locally integrable with respect to a positive measure $\mu$ on $T$, such that, for every function $g \in \mathscr{K}(T)$, one has $\int gf\,d\mu \in F$. Then the mapping $g \mapsto \int gf\,d\mu$ of $\mathscr{K}(T)$ into $F$ is a vectorial measure on $T$, with base $\mu$ and density $f$ with respect to $\mu$.

For (No. 1, Remark 3), it suffices to show that, setting $m(g) = \int gf\,d\mu$, $z' \circ m$ is a scalar measure for every $z' \in F'$. But since $z'(m(g)) = \int g\langle z', f \rangle\,d\mu$, one has $z' \circ m = \langle z', f \rangle \cdot \mu$, whence our assertion.