Exercise 17, § 2

Let $\mu$ be Lebesgue measure on $T = [0, 1]$; the mapping $f \mapsto f \cdot \mu$ of $C(T)$ into $M(T)$ is continuous for the strong topology on $M(T)$, hence is a vectorial measure $m$ with values in that Banach space. Show that $m$ is scalarly of base $\mu$ and absolutely continuous with respect to $\mu$ for the strong topology, but that $m$ does not have base $\mu$ (for the strong topology on $M(T)$). (Observe that $m$ has base $\mu$ for the vague topology $\sigma(M(T), C(T))$, and deduce from this that if one had $m = g \cdot \mu$, with $g$ scalarly $\mu$-integrable (for the strong topology on $M(T)$), one would necessarily have $g(t) = \varepsilon_t$ almost everywhere. Show that this leads to a contradiction, by observing that for every bounded numerical function $\theta$ on $T$, the linear form $z' : \lambda \mapsto \sum_{t \in T} \theta(t) \lambda(\{t\})$ on $M(T)$ is continuous but that $\langle g, z' \rangle$ is not necessarily $\mu$-measurable.)