— For every pair of numerical functions f≥0, f′≥0 defined on T and T′, *respectively,
∬∙f(t)f′(t′)dμ(t)dμ′(t′)=(∫∙f(t)dμ(t))(∫∙f′(t′)dμ′(t′)).
We begin by treating the case that μ and μ′ are measures with compact support; the same is then true of μ⊗μ′, and all of the symbols ∫∙, ∬∙ may be replaced by upper integrals. By Prop. 6,
∬∗f(t)f′(t′)dμ(t)dμ′(t′)≥∫∗dμ(t)∫∗f(t)f′(t′)dμ′(t′)
=(∫∗f(t)dμ(t))(∫∗f′(t′)dμ′(t′)).
To establish the reverse inequality, let us choose a function h≥f (resp. h′≥f′), the lower envelope of a sequence (hn) (resp. (hn′)) of lower semi-continuous functions, such that
∫∗h(t)dμ(t)=∫∗f(t)dμ(t)
(resp. ∫∗h′(t′)dμ′(t′)=∫∗f′(t′)dμ′(t′)); the existence of such functions follows immediately from the definition of upper integral (Ch. IV, §1, No. 3, Def. 3) and Lebesgue’s theorem. Applying Prop. 7 to the measurable function h⊗h′, we have
∬∗f(t)f′(t′)dμ(t)dμ′(t′)≤∬∗h(t)h′(t′)dμ(t)dμ′(t′)
=(∫∗h(t)dμ(t))(∫∗h′(t′)dμ′(t′))
=(∫∗f(t)dμ(t))(∫∗f′(t′)dμ′(t′)),
which is the sought-for inequality. Thus the proposition is established when μ and μ′ are measures with compact support. To treat the general case, it suffices to represent μ (resp. μ′) as the sum of a family (μα)α∈A (resp. (μβ′)β∈B) of measures with compact support (§ 2, No. 3, Prop. 4), write the formula (10) for each measure μα⊗μβ′, and sum on (α,β), taking into account Prop. 1 (§ 2, No. 2, Prop. 1).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.