Bourbaki

Lemma 1

Tag 09D0, permanent. Cite it as [Bourbaki, Tag 09D0].

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

— *Let X,Y,ZX, Y, Z be three topological spaces, with YY locally compact. Let (x,y)↦xy(x, y) \mapsto xy be a continuous mapping of X×YX \times Y into ZZ, which defines a mapping x↦uxx \mapsto u_x of XX into F(Y;Z)\mathscr{F}(Y; Z) by the relation ux(y)=xyu_x(y) = xy. Let ff be a continuous function on ZZ with values in R\mathbf{R} or in a Banach space, S the support of ff, and μ\mu a measure on YY. Assume that for every x0∈Xx_0 \in X, there exists a neighborhood VV of x0x_0 in XX such that ⋃x∈Vux−1(S)\bigcup_{x \in V} u_x^{-1}(S) is relatively compact in YY. Then: a) for every x∈Xx \in X, f∘uxf \circ u_x is continuous on YY, with compact support; b) the mapping x↦∫Yf(xy) dμ(y)x \mapsto \int_Y f(xy)\, d\mu(y), which is defined by a), is continuous on XX.

The assertion a) is obvious. Let us prove b). Since continuity is a local property, we may reduce to the case that ⋃x∈Xux−1(S)\bigcup_{x \in X} u_x^{-1}(S) is contained in a compact subset Y′Y' of YY. Since the function (x,y)↦f(xy)(x, y) \mapsto f(xy) is continuous on X×YX \times Y, f∘uxf \circ u_x tends to f∘ux0f \circ u_{x_0} uniformly on Y′Y' as xx tends to x0x_0

(GT, X, §3, No. 4, Th. 3), therefore μ(f∘ux)\mu(f \circ u_x) tends to μ(f∘ux0)\mu(f \circ u_{x_0}). Whence the lemma.

Let us now return to the previous notations.

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

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Cites

  • X, §3, No. 4, Th. 3 not yet on this site
Transcribed from the printed volume. Not translated.