Assume that R is irreducible. Let B={α1,…,αl} be a basis of R, f the connection index of R (§ 1, no. 9) and α~=n1α1+⋯+nlαl the highest root of R (for the order defined by B). Then the order of W is equal to
(l!)n1n2…nlf.
Let (ω1,…,ωl) be the basis of P(R−) dual to B. By the Cor. of Prop. 5, the open simplex C with vertices 0,n1−1ω1,…,nl−1ωl is an alcove of E. Choose a Haar measure μ on the additive group V∗. Let A be the set of elements of V∗ of the form ξ1ω1+⋯+ξlωl, with 0<ξi<1 for i=1,…,l. By Cor. 2 of Prop. 15 of Integration, Chap. VII, § 1, no. 10,
μ(A)/μ(C)=(l!)n1n2…nl.(5) On the other hand, let
A′ be the set of elements of
V∗ of the form
ξ1α1~+⋯+ξlαl~, with
0<ξi<1 for
i=1,…,l. Since
(α1~,…,αl~) is a basis of the
Z-module
Q(R−), we can apply Lemma 1 with
X=V∗,
G=Wa,
G′=Q,
U=C and
U′=A′. This gives
μ(A′)/μ(C)=(Wa:Q)=Card W.(6)
Finally, Lemma 1 can be applied again, taking X = V*, G = P, G' = Q, U = A and U' = A'. This gives
μ(A′)/μ(A)=(P:Q)=(P(R−):Q(R−))=f.
The proposition now follows by comparing formulas (5), (6) and (7).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.