Remark

— 1) Every bounded continuous function $f$ on $X$ with values in $F$ belongs to $\mathscr{L}_F^\infty$, and

In order that $N_\infty(f) = \|f\|$ for every bounded continuous function $f$, it is necessary and sufficient that the support of the measure $\mu$ be equal to $X$. For, if there exists a continuous function $f$ with negligible compact support and not identically zero, then $N_\infty(f) = 0$ and $\|f\| > 0$. Conversely, if the support of $\mu$ is equal to $X$ then, for every bounded continuous function $f$ and every number $\alpha < \|f\|$, the set of $x \in X$ such that $|f(x)| > \alpha$ is open and nonempty, hence has outer measure $> 0$, which shows that $N_\infty(f) = \|f\|$.

When the support of $\mu$ is equal to $X$, we may therefore identify the normed space $C^b(X; F)$, of bounded continuous functions on $X$ with values in $F$, with a subspace of the space $L_F^\infty$. Since $L_F^\infty$ is not in general Hausdorff, the subspace $C^b(X; F)$ is not in general closed in $L_F^\infty$, but its canonical image in $L_F^\infty$ is a closed subspace of $L_F^\infty$ (which can moreover be identified with $C^b(X; F)$ in the case contemplated). In general, $C^b(X; F)$ is distinct from $L_F^\infty$, that is, for an arbitrary bounded measurable function $f$, there does not in general exist a continuous function $g$ equal to $f$ locally almost everywhere (§ 5, Exer. 12). This implies that the space $\mathscr{H}(X; F)$ of mappings of $X$ into $F$, continuous with compact support, is in general not dense in $L_F^\infty$, whereas it is dense in each of the spaces $L_F^p$ for $1 \leq p < +\infty$ (§ 3, No. 4. Def. 2).

2) It is immediate that the topology defined by the semi-norm $N_\infty$ is finer than the topology induced on $L_F^\infty$ by the topology of convergence in measure (§ 5, No. 11).