Proposition 11

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 weak topology $\sigma(F', F)$. For a mapping f of X into $F'$ to be measurable, it is necessary and sufficient that, for every $a \in F$, the scalar function $x \mapsto \langle a, f(x) \rangle$ be measurable.

The condition is necessary by No. 3, Th. 1; let us prove that it is sufficient. 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 $X_n = f^{-1}(V_n^\circ)$; the sequence $(X_n)$ is increasing and $X = \bigcup_n X_n$; let us show that each $X_n$ is $\mu$-measurable. Indeed, $D \cap V_n$ is dense in $V_n$; for every $y \in D \cap V_n$ let $S_y$ be the set of $x \in X$ such that $|\langle y, f(x) \rangle| \leq 1$; the hypothesis implies that each of the $S_y$ is measurable, and $X_n$ is the intersection of the countable family of $S_y$ for $y \in D \cap V_n$. This being so, for every compact subset K of X and every $\varepsilon > 0$, there exists an integer n such that $|\mu|(K - (K \cap X_n)) \leq \varepsilon/4$, and then a compact subset $K_1$ of $K \cap X_n$ such that $|\mu|((K \cap X_n) - K_1) \leq \varepsilon/4$; finally, there exists a compact subset $K_2$ of $K_1$ such that $|\mu|(K_1 - K_2) \leq \varepsilon/2$ and such that the restrictions to $K_2$ of all of the functions $\langle y, f \rangle$, where $y \in D$, are continuous (No. 1, Prop. 2). Since the set $f(K_2) \subset f(X_n) \subset V_n^\circ$ 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); the restriction of f to $K_2$ is therefore continuous, whence our assertion in this 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 then $\operatorname{pr}_n(z') = z'_n$. 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 $\langle a, f \rangle$ is measurable for every $a \in F$, then so is $\langle a_n, \operatorname{pr}_n \circ f \rangle$ for every n and every $a_n \in F_n$, since $\langle a_n, \operatorname{pr}_n \circ f \rangle = \langle a_n, f \rangle$; the first part of the proof then shows that $\operatorname{pr}_n \circ f$ is measurable for every n, therefore so is f (No. 3, Th. 1).