Remark

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

Integration, chapter VII, § 3. Applications and examples

— Let E be a finite-dimensional real affine space, T a compact space, $\mu$ a measure on T of total mass 1, and f a continuous mapping of T into E. If an origin a in E is chosen, E becomes equipped with a vector space structure, and the integral $\int_T f(t) d\mu(t)$ therefore has meaning; it represents the point x of E such that

x - a = \int_T (f(t) - a) d\mu(t) .

This point is independent of the choice of $a$ . For, let $a' \in \mathbf{E}$ and $x' \in \mathbf{E}$ be such that $x' - a' = \int_{\mathbf{T}} (f(t) - a') d\mu(t)$ . Then

x' - a = (x' - a') + (a' - a) = \int_{\mathbf{T}} (f(t) - a') d\mu(t) + \int_{\mathbf{T}} (a' - a) d\mu(t)

= \int_{\mathbf{T}} (f(t) - a) d\mu(t) = x - a,

whence $x' = x$ . We may therefore employ the symbol $\int_{\mathbf{T}} f(t) d\mu(t)$ without specifying the choice of origin in $\mathbf{E}$ . If $u$ is an affine mapping of $\mathbf{E}$ into another finite-dimensional affine space $\mathbf{E}'$ , then

u \left( \int_{\mathbf{T}} f(t) d\mu(t) \right) = \int_{\mathbf{T}} u(f(t)) d\mu(t).

For, $\mathbf{E}$ and $\mathbf{E}'$ may be identified with vector spaces in such a way that $u$ becomes a linear mapping, in which case the formula is known (Ch. III, §3, No. 2, Prop. 2 and No. 3, Prop. 7).

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

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