If f1 and f2 are two functions in I+ then
μ∗(f1+f2)=μ∗(f1)+μ∗(f2).
As φ1 (resp. φ2) runs over the set of functions in K+ such that φ1≤f1 (resp. φ2≤f2), the functions φ1+φ2 form a directed set (for ≤) whose upper envelope is f1+f2. Therefore, by Th. 1,
μ∗(f1+f2)=supμ(φ1+φ2)=sup(μ(φ1)+μ(φ2)),
where (φ1,φ2) runs over the set of pairs of functions in K+ such that φ1≤f1 and φ2≤f2; since
sup(μ(φ1)+μ(φ2))=supμ(φ1)+supμ(φ2)
(GT, IV, §5, No. 7, Cor. 2 of Prop. 12), the theorem is proved.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.