— Let E be a real vector space of finite dimension n, (e1,e2,…,en) a basis of E, P the set of x=∑i=1nξiei∈E such that 0≤ξi≤1 for all i, and μ the unique Haar measure on the additive group E such that μ(P)=1. Let x1,…,xn be points of E, S the closed convex envelope in E of the set {0,x1,…,xn}. Writing xi=∑j=1nαijej, one has
μ(S)=μ(S˙)=n!1∣det(αij)∣.
We shall identify E with Rn by means of the isomorphism that transforms (ei) into the canonical basis of Rn. Then μ is identified with the Lebesgue measure μn on Rn.
Suppose first that xi=ei for all i. Then S is the set Sn of x=(ξi) in Rn such that
ξi≥0 for all iandξ1+⋯+ξn≤1.
Set μn(Sn)=an. Let λ∈R. Identifying Rn with Rn−1×R, we may consider the section Cλ of Sn at λ. This section is empty if λ<0 or λ>1; if 0≤λ≤1, Cλ is the set of (ξ1,…,ξn−1)∈Rn−1 such that
ξ1≥0,…,ξn−1≥0,ξ1+⋯+ξn−1≤1−λ,
hence may be deduced from Sn−1 by a homothety of ratio 1−λ, so that μn−1(Cλ)=(1−λ)n−1an−1. By the Lebesgue–Fubini theorem,
an=∫01(1−λ)n−1an−1dλ=n1an−1.
Since a1=1, one sees that an=n!1.
Let us return to the general case of the corollary. Let u be the endomorphism of Rn such that u(ei)=xi for all i. One has u(Sn)=S. If u is invertible, Prop. 15 proves that
μn(S)=n!1∣detu∣=n!1∣det(αij)∣.
Since S−S∘ is contained in a finite number of hyperplanes, μ(S∘)=μ(S). Finally, if u is not invertible, then S is contained in a hyperplane, so that μ(S)=0=det(αij).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.