— *Let (fn) be a sequence of functions ≥0 (finite or not) defined on X. For 1≤p<+∞,
(5)
Np(n=1∑∞fn)≤n=1∑∞Np(fn).
Set f=∑n=1∞fn; f is the upper envelope of the increasing sequence of functions gn=∑k=1nfk; the definition of Np(f), and Th. 3 of §1, No. 3, show that Np(f)=supnNp(gn). But Np(gn)≤∑k=1nNp(fk) by Prop. 1, whence the inequality (5).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.