Proposition 2

The space $\mathscr{L}_F^\infty$ is complete; the space $L_F^\infty$ is a Banach space.

For, let $(f_n)$ be a Cauchy sequence in $\mathscr{L}_F^\infty$; for every integer $n$, there exists an integer $k_n$ such that $N_\infty(f_r - f_s) \leq 1/n$ for $r \geq k_n$ and $s \geq k_n$; thus, there exists a locally negligible set $A_{rs}$ such that $|f_r(x) - f_s(x)| \leq 1/n$ for all $x \notin A_{rs}$. If $A_n$ is the union of the sets $A_{rs}$ (for $r \geq k_n$ and $s \geq k_n$), then $A_n$ is locally negligible and, for every $x \notin A_n$, $|f_r(x) - f_s(x)| \leq 1/n$ for all indices $r \geq k_n, s \geq k_n$. Let $A$ be the locally negligible set formed by the union of the $A_n$, and set $g_n(x) = f_n(x)$ for $x \notin A$, $g_n(x) = 0$ for $x \in A$; then $g_n$ belongs to $\mathscr{L}_F^\infty$ and, by the definition of $A$, the sequence $(g_n)$ converges uniformly on $X$ to a function $g$. It follows that the function $g$ is measurable (§ 5, No. 4, Th. 2); moreover, $g$ is bounded on the set of $x \in X$ where $|g_{k_1}(x)| \leq N_\infty(g_{k_1})$ and, since the complement of this set is locally negligible, $g$ belongs to $\mathscr{L}_F^\infty$. It is clear that in $\mathscr{L}_F^\infty$, the sequence $(g_n)$ has limit $g$, and the same is therefore true of the sequence $(f_n)$, since $N_\infty(f_n - g_n) = 0$ for all $n$. The second part of the proposition may be deduced immediately from this.