For every function f∈O(U;A), the integral ∫Ufω is an element of A depending only on a and on the germ of f in a neighbourhood of Spn(a). It satisfies the inequality
(4) ∥∫fω∥⩽(∫∥ω∥) sup ∥f(z)∥,
U U z∈Supp(h)
and
∫U(af)ω=a∫Ufω
for every a∈A.
We have seen above that the integral is defined for f∈C(U;A) and satisfies
∥∫fω∥⩽∫∥f∥∥ω∥⩽(∫∥ω∥) sup ∥f(z)∥.
U U U z∈Supp(h)
Moreover, for every a∈A and every f∈C(U;A), we have
∫U(af)ω=a∫Ufω
(INT, VI, §2, no2, Prop. 2 applied to multiplication by a).
Let (h′,u1′,...,un′) be a sequence adapted to a such that Supp(h′)⊂U. Let ω′ be the associated differential form. By Lemma 6 of I, p. 57, there exists a differential form ψ on Cn of degree n−1, with coefficients in A and with support contained in Supp(h)∪ Supp(h′)⊂U, such that
ω−ω′=d(ψ∧dz1∧⋯∧dzn)
Let f∈O(U;A). Since the mapping f is holomorphic, we have
df=∑n∂zi∂fdzi
i=1
(VAR, R2, p. 24, 8.8.9) and hence
f(ω−ω′)=fd(ψ∧dz1∧⋯∧dzn)=d(fψ∧dz1∧⋯∧dzn)
By Stokes' formula (VAR, R2, p. 48, 11.2.3), we then have
∫Uf(ω−ω′)=0
Thus the element ∫Ufω does not depend on the choice of the sequence (h,u1,...,un).
To conclude, let us prove that ∫Ufω depends only on the germ of f in a neighbourhood of Spn(a). Let U and U′ be open neighbourhoods of Spn(a). Let f∈O(U;A) and f′∈O(U′;A) be such that f and f′ coincide on an open neighbourhood U′′ of Spn(a). There exists a mapping h of Cn into C, indefinitely differentiable, equal to 1 in a neighbourhood of Spn(a), and with compact support contained in U′′ (Lemma 1 of I, p. 52), and there exists (u1,...,un) such that the sequence (h,u1,...,un) is adapted to a (Lemma 3 of I, p. 54). Let ω be the associated differential form. Since Supp(ω )⊂ Supp(h)⊂U′′ (Lemma 4, b) of I, p. 54), we have
∫Ufω=∫U′′fω=∫U′′f′ω=∫U′f′ω
which completes the proof.
This lemma proves that there exists a unique a-linear mapping Θa of O(Spn(a);A) into A such that
(5) Θa(f)=n! n∫fω
(2iπ)U
for every open set U and every representative f∈O(U;A) of a germ f∈O(Spn(a);A). The linear mapping Θa is continuous by inequality (4) and EVT, II, p. 29, Proposition 5.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.