Proposition 5

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

Integration, VIII, p. 20, § 3. Convolution of measures on groups

Let $\lambda$ be a left (resp. right) Haar measure on G, and $\mu \in \mathscr{M}^1(G)$ . Then $\mu$ and $\lambda$ (resp. $\lambda$ and $\mu$ ) are convolvable, and $\mu * \lambda = \mu(1)\lambda$ (resp. $\lambda * \mu = \mu(1)\lambda$ ).

We can suppose that $\mu \geqslant 0$ . Let $f \in \mathscr{K}_+(G)$ . When $\lambda$ is a left Haar measure,

\int^* d\mu(x) \int^* f(xy)\, d\lambda(y) = \int^* d\mu(x) \int f(y)\, d\lambda(y) = \lambda(f)\|\mu\| ,

therefore the function $(x, y) \mapsto f(xy)$ is $(\mu \otimes \lambda)$ -integrable, and its integral for $\mu \otimes \lambda$ is $\lambda(f)\|\mu\|$ . One argues similarly when $\lambda$ is a right Haar measure.

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

Cited by

Exercise 12, § 3

Cites

Nothing.