Proposition 15

*Let $X$ be a locally compact space, $\mu$ a measure on $X$, $A$ a $\mu$-measurable subset of $X$, and $f$ a mapping of $A$ into a topological space $F$. The following conditions are equivalent:

a) The set $\mathfrak{H}$ of compact subsets $K$ of $A$ such that the restriction of $f$ to $K$ is continuous, is $\mu$-dense in $A$.

b) There exists a set $\mathfrak{K}$ of compact subsets of $A$, $\mu$-dense in $A$, such that the restriction of $f$ to every $K \in \mathfrak{K}$ is $\mu_K$-measurable.

c) There exist a homeomorphism $j$ of $F$ onto a subspace of a topological space $G$ and a $\mu$-measurable mapping $g$ of $X$ into $G$, such that $g|A = j \circ f$.*

d) Every extension of $f$ to a mapping of $X$ into $F$, constant on $X - A$, is $\mu$-measurable.

It is clear that a) implies b) and that d) implies c). The fact that c) implies a) follows from condition c) of Prop. 12 of No. 8. On the other hand, b) implies a): for, Def. 1 shows that, for each $K \in \mathscr{K}$, the set of subsets $H \in \mathscr{H}$ contained in $K$ is $\mu_K$-dense in $K$ (No. 8, Prop. 12, c)), and Prop. 13 of No. 8 shows that $\mathscr{H}$ is $\mu$-dense in $A$. It remains to see that a) implies d). Let $g$ be an extension of $f$ to $X$, constant on $X - A$. For every compact subset $L$ of $X$, $L \cap A$ and $L \cap (X - A)$ are $\mu$-integrable; therefore, for every $\varepsilon > 0$, there exist a compact subset $P \subset L \cap A$ and a compact subset $Q \subset L \cap (X - A)$ such that

On the other hand, there exists a set $H \in \mathscr{H}$ contained in $P$ such that $|\mu|(P - H) \leq \varepsilon/2$; the restriction of $g$ to the compact set $K = H \cup Q$ is then continuous ($g$ being constant on $Q$) and $|\mu|(L - K) \leq \varepsilon$, which completes the proof.