Bourbaki

Corollary 2

Tag 09E0, permanent. Cite it as [Bourbaki, Tag 09E0].

Integration, chapter VII, § 1. Construction of a Haar measure

— Let E be a real vector space of finite dimension n, (e1,e2,…,en)(e_1, e_2, \ldots, e_n) a basis of E, P the set of x=∑i=1nξiei∈Ex = \sum_{i=1}^n \xi_i e_i \in E such that 0≤ξi≤10 \leq \xi_i \leq 1 for all i, and μ\mu the unique Haar measure on the additive group E such that μ(P)=1\mu(P) = 1. Let x1,…,xnx_1, \ldots, x_n be points of E, S the closed convex envelope in E of the set {0,x1,…,xn}\{0, x_1, \ldots, x_n\}. Writing xi=∑j=1nαijejx_i = \sum_{j=1}^n \alpha_{ij} e_j, one has

μ(S)=μ(S˙)=1n!∣det⁡(αij)∣. \mu(S) = \mu(\dot{S}) = \frac{1}{n!} |\det(\alpha_{ij})|.

We shall identify E with Rn\mathbf{R}^n by means of the isomorphism that transforms (ei)(e_i) into the canonical basis of Rn\mathbf{R}^n. Then μ\mu is identified with the Lebesgue measure μn\mu_n on Rn\mathbf{R}^n.

Suppose first that xi=eix_i = e_i for all i. Then S is the set SnS_n of x=(ξi)x = (\xi_i) in Rn\mathbf{R}^n such that

ξi≥0 for all iandξ1+⋯+ξn≤1. \xi_i \geq 0 \text{ for all } i \quad \text{and} \quad \xi_1 + \cdots + \xi_n \leq 1.

Set μn(Sn)=an\mu_n(S_n) = a_n. Let λ∈R\lambda \in \mathbf{R}. Identifying Rn\mathbf{R}^n with Rn−1×R\mathbf{R}^{n-1} \times \mathbf{R}, we may consider the section CλC_\lambda of SnS_n at λ\lambda. This section is empty if λ<0\lambda < 0 or λ>1\lambda > 1; if 0≤λ≤10 \leq \lambda \leq 1, CλC_\lambda is the set of (ξ1,…,ξn−1)∈Rn−1(\xi_1, \ldots, \xi_{n-1}) \in \mathbf{R}^{n-1} such that

ξ1≥0,…,ξn−1≥0,ξ1+⋯+ξn−1≤1−λ, \xi_1 \geq 0, \ldots, \xi_{n-1} \geq 0, \quad \xi_1 + \cdots + \xi_{n-1} \leq 1 - \lambda,

hence may be deduced from Sn−1S_{n-1} by a homothety of ratio 1−λ1 - \lambda, so that μn−1(Cλ)=(1−λ)n−1an−1\mu_{n-1}(C_\lambda) = (1 - \lambda)^{n-1} a_{n-1}. By the Lebesgue–Fubini theorem,

an=∫01(1−λ)n−1an−1dλ=1nan−1. a_n = \int_0^1 (1 - \lambda)^{n-1} a_{n-1} d\lambda = \frac{1}{n} a_{n-1}.

Since a1=1a_1 = 1, one sees that an=1n!a_n = \frac{1}{n!}.

Let us return to the general case of the corollary. Let uu be the endomorphism of Rn\mathbf{R}^n such that u(ei)=xiu(e_i) = x_i for all ii. One has u(Sn)=Su(S_n) = S. If uu is invertible, Prop. 15 proves that

μn(S)=1n!∣det⁡u∣=1n!∣det⁡(αij)∣. \mu_n(S) = \frac{1}{n!} |\det u| = \frac{1}{n!} |\det(\alpha_{ij})|.

Since S−S∘S - \overset{\circ}{S} is contained in a finite number of hyperplanes, μ(S∘)=μ(S)\mu(\overset{\circ}{S}) = \mu(S). Finally, if uu is not invertible, then SS is contained in a hyperplane, so that μ(S)=0=det⁡(αij)\mu(S) = 0 = \det(\alpha_{ij}).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed volume. Not translated.