Proposition 13

Let F be a locally convex space that is the direct limit of a sequence of separable, metrizable locally convex spaces $F_n$, F being the union of the $F_n$. Let $F'$ be the dual of F, equipped with the topology $\sigma(F', F)$. Then, every scalarly $\mu$-measurable mapping f of T into $F'$ is $\mu$-measurable.

Suppose first that F is metrizable and separable, and let D be a countable dense set in F. Let $(V_n)$ be a decreasing fundamental sequence of balanced, convex, open neighborhoods of 0 in F; the polar sets $V_n^\circ$ are equicontinuous and their union is all of $F'$. Let $T_n = f(V_n^\circ)$; the sequence $(T_n)$ is increasing and $T = \bigcup T_n$; let us show that each of the $T_n$ is $\mu$-*measurable*. Indeed, $D \cap V_n^n$ is dense in $V_n$; for every $y \in D \cap V_n$, let $S_y$ be the set of $t \in T$ such that $|\langle y, f(t) \rangle| \leq 1$; the hypothesis implies that each of the $S_y$ is measurable, and $T_n$ is the intersection of the countable family of the $S_y$ ($y \in D \cap V_n$). This being so, for every compact subset K of T and every $\varepsilon > 0$, there exists an integer n such that $\mu(K - (K \cap T_n)) \leq \frac{\varepsilon}{4}$, then a compact subset $K_1$ of $K \cap T_n$ such that $\mu((K \cap T_n) - K_1) \leq \frac{\varepsilon}{4}$; finally, there exists a compact subset $K_2$ of $K_1$ such that $\mu(K_1 - K_2) \leq \frac{\varepsilon}{2}$ and such that the restrictions to $K_2$ of all of the functions $\langle y, f \rangle$, where $y \in D$, are continuous (Ch. IV, §5, No. 1, Prop. 2). Since the set $f(K_2) \subset f(T_n) \subset V_n^\circ$ is equicontinuous, the topology induced by $\sigma(F', F)$ on $f(K_2)$ is identical to the topology of pointwise convergence in $D$ (GT, X, §2, No. 4, Th. 1); consequently, the restriction of $f$ to $K_2$ is continuous, whence our assertion in the first case.

Let us pass to the general case. If $z'$ is a continuous linear form on $F$, its restriction $z'_n$ to $F_n$ is continuous; since $F = \bigcup_n F_n$, the dual $F'$ of $F$ may be identified (algebraically) with a linear subspace of the product $\prod_n F'_n$, and $\operatorname{pr}_n z' = z'_n$. Moreover, since each finite subset of $F$ is contained in one of the $F_n$, the topology $\sigma(F', F)$ is none other than the topology induced by the product topology of the topologies $\sigma(F'_n, F_n)$. This being so, if $f$ is scalarly $\mu$-measurable then $\operatorname{pr}_n \circ f$ is scalarly $\mu$-measurable, since for every $t \in T$, $\operatorname{pr}_n(f(t))$ is the restriction of $f(t)$ to $F_n$. The first part of the proof shows that $\operatorname{pr}_n \circ f$ is $\mu$-measurable for every $n$, therefore so is $f$ (Ch. IV, §5, No. 3, Th. 1).