Theorem 2 (Egoroff)

— *Let X be a locally compact space, μ a measure on X, A a countable set, $\mathfrak{F}$ a filter on A having a countable base, and $(f_\alpha)_{\alpha \in A}$ a family of measurable mappings of X into a metrizable space F. Assume that $\lim_{\mathfrak{F}} f_\alpha(x) = f(x)$ exists in the complement of a locally negligible subset N of X. Under these conditions, 1° the function f (extended to N in any manner whatsoever) is measurable; 2° for every compact subset K of X and every $\varepsilon > 0$, there exists a compact set $K_1 \subset K$ such that $|\mu|(K - K_1) \leq \varepsilon$ and such that the restrictions of the $f_\alpha$ to $K_1$ are continuous and converge uniformly to f on $K_1$.

The first assertion obviously follows from the second, which we are going to prove. There exists a compact set $K_0 \subset K$ such that $|\mu|(K - K_0) \leq \varepsilon/2$ and such that the restrictions to $K_0$ of all the functions $f_\alpha$ are continuous (No. 1, Prop. 2). Let $(A_n)$ be a decreasing countable base for the filter $\mathfrak{F}$; let d be a metric on F compatible with the topology. For every pair of integers $n > 0,\ r > 0$, let $B_{n,r}$ be the set of points $x \in K_0$ such that, for at least one pair of indices $\alpha, \beta$ belonging to $A_n$, $d(f_\alpha(x), f_\beta(x)) \geq 1/r$; for fixed $\alpha$ and $\beta$, the set of $x \in K_0$ such that $d(f_\alpha(x), f_\beta(x)) \geq 1/r$ is closed in $K_0$, hence is compact; consequently, $B_{n,r}$ is a countable union of compact sets contained in $K_0$, hence is integrable (§ 4, No. 5, Props. 6 and 8). If $r$ is fixed, the intersection of the decreasing sequence of sets $B_{n,r}$ ($n = 1, 2, \ldots$) has measure zero, since $f_\alpha(x)$ tends to $f(x)$ almost everywhere in $K_0$ with respect to the filter $\mathfrak{F}$; thus $\lim_{n \to \infty} |\mu|(B_{n,r}) = 0$ (§ 4, No. 5, Cor. of Prop. 7), consequently there exists an integer $n_r$ such that $|\mu|(B_{n_r,r}) \leq \varepsilon/2^{r+2}$. Let $B$ be the union (for $r = 1, 2, \ldots$) of the sets $B_{n_r,r}$; $B$ is integrable and

(§ 4, No. 5, Cor. of Prop. 8). Let $C$ be the complement of $B$ in $K_0$; by construction, $f_\alpha(x)$ converges uniformly to $f(x)$ in $C$ with respect to the filter $\mathfrak{F}$, and since the restrictions of the $f_\alpha$ to $C$ are continuous, so is the restriction of $f$ to $C$. It then suffices to take a compact set $K_1 \subset C$ such that $|\mu|(C - K_1) \leq \varepsilon/4$ in order to satisfy the conditions of the statement, since $|\mu|(K - K_1) = |\mu|(K - K_0) + |\mu|(B) + |\mu|(C - K_1) \leq \varepsilon$.

The conclusions of Th. 2 do not necessarily hold if $F$ is not metrizable (Exer. 1). If $F$ is metrizable, and the set $A$ is not countable but the filter $\mathfrak{F}$ has a countable base, then the first conclusion of Th. 2 is again valid; for, if $(A_n)$ is a countable base for $\mathfrak{F}$ and $\alpha_n$ is an element of $A_n$, then $f$ is the limit of the sequence $(f_{\alpha_n})$ locally almost everywhere, hence is measurable; however, the second conclusion of Th. 2 is no longer necessarily valid (cf. Exer. 4).