Proposition 11
Tag 09B1, permanent. Cite it as [Bourbaki, Tag 09B1].
Integration, chapter VI, § 2. Vectorial measures . "
— Let be three Banach spaces, a continuous bilinear mapping of into . Let be a majorizable vectorial measure on , with values in . Then there exists one and only one continuous linear mapping of into such that, for every and every numerical function integrable for , one has . *Moreover,
for every function .
If there exists such a mapping, its value for a step function over the -integrable sets is uniquely determined: for, it is known that one can then write , where the are -integrable and disjoint, and the (Ch. IV, §4, No. 9, Lemma). The value of must therefore be equal to . Now, we have (No. 3, Prop. 5)
which shows first of all that the element of does not depend on the particular expression of in the form , hence that we may denote it by . One verifies immediately that the mapping so defined is linear on the space of step functions over the -integrable sets: for, it suffices to write two functions of in the form and with the same finite family of pairwise disjoint -integrable sets (which is possible by virtue of the Lemma of Ch. IV, §4, No. 9). The inequality (5) then shows that is continuous on , and since this subspace is dense in (Ch. IV, §4, No. 10, Cor. 1 of Prop. 19 and Ch. V, §1, No. 3), one deduces from this the existence and uniqueness of , as well as the inequality (4).
One says that is the integral of with respect to (relative to the bilinear mapping ); when the value of the bilinear mapping at the point is denoted , we shall write instead of .
With the notations of No. 6, the integral is none other than with and .
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.