Proposition 8

— Let $\mu$ be a positive measure on $T$, $m$ a measure on $T$ with values in $F$, with base $\mu$ and density $f$ with respect to $\mu$. Let $q$ be a lower semi-continuous semi-norm on $F$.

a) If, for every compact subset $K$ of $T$, the upper integral $\int_K^*(q \circ f)\,d\mu$ is finite, then $m$ is $q$-majorizable.

b) If $m$ is $q$-majorizable, then $q(m)$ has base $\mu$; if in addition $f$ is $\mu$-measurable as a mapping of $T$ into $F_\sigma$, then $q \circ f$ is locally $\mu$-integrable and $q(m) = (q \circ f) \cdot \mu$.

a) For every finite subset $J$ of $A'_q$, let us denote by $\lambda_J$ the supremum of the measures $|z' \circ m|$, where $z'$ runs over $J$; if $g_J = \sup_{z' \in J} |\langle z', f \rangle|$ then $\lambda_J = g_J \cdot \mu$ (Ch. V, §5, No. 2, Prop. 2). For every relatively compact open subset $U$ of $T$, let $\lambda_{J,U}$ be the restriction of $\lambda_J$ to $U$; let us first show that as $J$ runs over the directed set $\mathfrak{F}$ of finite subsets of $A'_q$, the family $(\lambda_{J,U})$ is bounded above in $\mathscr{M}(U)$. Indeed, for every function $h \geq 0$ in $K(U)$,

whence our assertion (Ch. II, §2, No. 2). Let $\nu_U$ be the supremum of this family of measures in $\mathscr{M}(U)$. If $U'$ is a second relatively compact open subset of $T$ such that $U \subset U'$, then $\nu_U$ is the restriction of $\nu_{U'}$ to $U$, as follows immediately from the expression of the supremum of an increasing directed set of measures (Ch. II, §2, No. 2) and the fact that $\lambda_{J,U}$ is the restriction to $U$ of $\lambda_{J,U'}$. Thus there is one and only one positive measure $\nu$ whose restriction to each $U$ is $\nu_U$ (Ch. III, §2, No. 1, Prop. 1), and it is clear that $\nu = q(m)$.

b) Since the measures $\lambda_J$ have base $\mu$, so does their supremum $q(m)$ (Ch. V, §5, No. 5, Th. 2). If $f$ is $\mu$-measurable for the topology $\sigma(F, F')$ on $F$, it follows at once from the definitions that the mapping $g : t \mapsto (g_J(t))_{J \in \mathfrak{F}}$ of $T$ into the product space $\mathbf{R}^\mathfrak{F}$ is $\mu$-measurable. The restriction of $q \circ f = \sup_{J \in \mathfrak{F}} g_J$ to every compact subset of $T$ on which $g$ is continuous is lower semi-continuous; consequently $q \circ f$ is $\mu$-measurable (Ch. IV, §5, No. 5, Cor. of Prop. 8 and No. 10, Prop. 15). Let $K$ be a compact subset of T; it admits a partition consisting of a $\mu$-negligible set and a sequence $(K_n)$ of compact sets on which $g$ is continuous. Then $\int_{K_n}^*(q \circ f)\, d\mu = \sup_J \int_{K_n} g_J\, d\mu \leq \int_{K_n} dq(m)$ for all $n$ (Ch. IV, §1, No. 1, Th. 1 and Ch. V, §7, No. 1, Prop. 1), whence $\int_K^*(q \circ f)\, d\mu = \sum_n \int_{K_n}^*(q \circ f)\, d\mu \leq \int_K dq(m)$. But this proves that $q \circ f$ is locally $\mu$-integrable and that $\lambda_J \leq (q \circ f) \cdot \mu \leq q(m)$ for all $J \in \mathscr{F}$; whence, by definition, $q(m) = (q \circ f) \cdot \mu$.