Example

— Let us take for $m$ the identity mapping of $\mathscr{H}(T)$. To say that $m$ is scalarly of base $\mu$ means that every real measure on $T$ has base $\mu$. In particular, the point measure $\varepsilon_t$ ($t \in T$) must have base $\mu$, which requires that $\mu(\{t\}) > 0$ for every $t \in T$, and implies in particular that every compact subset of $T$ is countable.

In this No. we are going to prove a result that generalizes one of the consequences of the Lebesgue–Nikodym theorem, namely, that the dual of $L^1(\mu)$ is $L^\infty(\mu)$ (Ch. V, §5, No. 8, Th. 4), and that gives a sufficient condition for a vectorial measure that is scalarly of base $\mu$ to have base $\mu$.

Let $\pi$ be the canonical mapping of $\mathscr{L}^\infty(\mu)$ onto $L^\infty(\mu)$. We shall say that a linear subspace $G$ of $L^\infty$ has the lifting property if there exists a linear mapping $\rho$ of $G$ into $\mathscr{L}^\infty(\mu)$ (called a lifting of $G$) such that $\pi \circ \rho$ is the identity on $G$ and $|\rho(f)(t)| \leq N_\infty(f)$ for all $t \in T$ and $f \in G$.

One proves that if $\mu$ is Lebesgue measure on $\mathbf{R}^n$, the entire space $L^\infty(\mathbf{R}^n, \mu)$ has the lifting property (Exer. 18).