Proposition 19

If the uniform space $F$ is metrizable and complete, then the uniform space $S(A, \mu; F)$ is complete.

There exists a locally countable family $(K_{\lambda})_{\lambda \in L}$ of compact subsets of $A$ such that the $K_{\lambda}$ are pairwise disjoint and $A - \bigcup_{\lambda} K_{\lambda}$ is locally negligible (No. 9, Prop. 14). By Prop. 17, $S(A, \mu; F)$ is isomorphic to the product $\prod_{\lambda \in L} S(K_{\lambda}, \mu; F)$; we are thus reduced to proving the proposition when $A$ is integrable; $S(A, \mu; F)$ is then metrizable (Prop. 18) and, by Lemma 4, for every Cauchy sequence $(f_n)$ in $\mathscr{S}(A, \mu; F)$ there exists a subsequence $(f_{n_k})$ that is convergent in $A - N$, where $N$ is negligible; the limit $f$ of $(f_{n_k})$ (extended in any way whatsoever to all of $A$) is then $\mu$-measurable, and it follows from the extension of Egoroff’s theorem mentioned in No. 10 that the sequence $(f_{n_k})$ converges in measure to $f$ in $A$. This implies that $f$ is a cluster point of the sequence $(f_n)$ in $\mathscr{S}(A, \mu; F)$, and since the sequence $(f_n)$ is by hypothesis a Cauchy sequence, it converges to $f$.

Q.E.D.