— Let F be a metrizable uniform space, and let B⊂A be a countable union of μ-integrable sets. Then, for every Cauchy sequence (fn) in S(A,μ;F), there exists a subsequence (fnk) of (fn) such that (fnk(x)) is a Cauchy sequence in F for almost every x∈B.
Suppose first that B is integrable, and denote by d a metric compatible with the uniform structure of F. We shall define recursively a double sequence (fmn) of functions in S(A,μ;F) such that f0n=fn for every n, (fmn)n≥0 is a subsequence of (fm−1,n)n≥0 for every m>0 and, finally, such that for every m>0 the set Mmn of x∈B for which d(fmn(x),fm,n+1(x))>1/2m+n+1 has measure ∣μ∣(Mmn)≤1/2m+n+1;
the possibility of such a definition follows from the fact that (fn) is a Cauchy sequence in S(A,μ;F). Set Mm=⋃n≥0Mmn; then
∣μ∣(Mm)≤n=0∑∞∣μ∣(Mmn)≤1/2m
and, for every x∈B−Mm, we have d(fmn(x),fm,n+p(x))≤1/2m+n for all n≥0 and all p>0; the sequence (fmn(x))n≥0 is therefore a Cauchy sequence in F. Now let N=⋂m=0∞Mm; N is negligible. Set gn=fnn for every n≥0; for every m, the sequence (gn)n≥m is a subsequence of the sequence (fmn)n≥0; if x∈B−N, there exists an index m such that x∈/Mm, which proves that the sequence (gn(x)) is a Cauchy sequence in F.
If now B is the union of a sequence (Bm) of integrable sets, one can define recursively a double sequence (gmn) such that g0n=fn, (gmn)n≥0 is a subsequence of (gm−1,n)n≥0 for every m>0, and such that the sequence (gmn(x))n≥0 is a Cauchy sequence in Bm−Pm, where Pm is negligible. Set hn=gnn for every n≥0, so that, for every m, the sequence (hn)n≥m is a subsequence of (gmn)n≥0; the sequence (hn(x))n≥0 is then a Cauchy sequence in F for every x∈B−P, where P=⋃m=0∞Pm is negligible.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.