Theorem 3

For a mapping $f$ of X into a metrizable space F to be measurable, it is necessary and sufficient that, for every compact set $K \subset X$, there exist a sequence $(g_n)$ of measurable step functions, with values in F, such that $g_n(x)$ tends to $f(x)$ for almost every $x \in K$.

The condition is sufficient by Egoroff’s theorem and the principle of localization. Let us show that it is necessary: by hypothesis, there exist a negligible set $N \subset K$ and a partition $(K_m)$ of $K - N$ formed of compact sets such that the restriction of $f$ to each $K_m$ is continuous. To define the sequence $(g_n)$, it suffices to proceed in the following manner: let $d$ be a metric compatible with the topology of F; for each $K_i$ with index $i \leq n$, there exists a finite partition of $K_i$ into integrable sets $A_{ij}$ ($1 \leq j \leq q_i$) sufficiently small that the oscillation of $f$ on each of the $A_{ij}$ is $\leq 1/n$ (§ 4, No. 9, Lemma); take $g_n$ to be constant on each $A_{ij}$, equal to one of the values of $f$ in this set (for $1 \leq i \leq n,\ 1 \leq j \leq q_i$), and equal to a fixed element $a \in \mathbf{F}$ for every point of $X$ that does not belong to any of the $A_{ij}$. It is clear that the sequence $(g_n(x))$ converges to $f(x)$ at every point of $K$ not belonging to $N$.