Theorem 4

— Let F be a metrizable space and let d be a metric on F compatible with the topology. For a mapping f of X into F to be measurable, it is necessary and sufficient that it satisfy the following two conditions: a) the inverse image under f of every closed ball of F is measurable; b) for every compact set K ⊂ X, there exists a countable subset H of F such that $f(x) \in \overline{H}$ for almost every $x \in K$.

The condition a) is necessary by Prop. 7; on the other hand, in the notations of Th. 3, condition b) is satisfied by taking H to be the countable set formed by the values of all of the functions $g_n$.

Let us now show that the conditions a) and b) are sufficient. Let K be any compact subset of X; there exists a negligible subset N of K such that $f(K - N)$ is contained in the closure of a countable set of points of F, which we arrange in a sequence $(a_n)$. Let $A_{n,p}$ be the set of $x \in K - N$ such that $d(f(x), a_n) \leq 1/p$. It follows from condition a) that $A_{n,p}$ is measurable. For fixed $p$, define a sequence of sets $B_{n,p} \subset K - N$ recursively by setting

the sets $B_{n,p}$ are measurable, and those that are nonempty form a partition of $K - N$. Let $g_{m,p}$ be the function equal to $a_i$ on the set $B_{i,p}$ for $1 \leq i \leq m$ and equal to a constant $b \in F$ on the complement of the union of these sets; $g_{m,p}$ is a measurable step function; as $m$ tends to infinity, $g_{m,p}$ converges pointwise to the function $f_p$ equal to $a_n$ on $B_{n,p}$ ($n \geq 1$) and to $b$ on $N \cup C K$, therefore (Th. 2) $f_p$ is measurable. As $p$ tends to infinity, $f_p(x)$ tends to $f(x)$ for every $x \in K - N$, and to $b$ for $x \in N \cup C K$; the limit of the $f_p$ is therefore measurable, and the principle of localization proves that $f$ itself is measurable.