Proposition 6

Let $X, Y, Z$ be locally compact spaces, $\varphi$ a continuous mapping of $X \times Y$ into $Z$, and $\lambda, \mu$ measures on $X, Y$. For $\lambda$ and $\mu$ to be $\varphi$-convolvable, it is necessary and sufficient that the mapping $(x, y) \mapsto \varepsilon_{\varphi(x,y)} = \varepsilon_x * \varepsilon_y$ of $X \times Y$ into $\mathscr{M}(Z)$ be scalarly $(\lambda \otimes \mu)$-integrable for the topology $\sigma(\mathscr{M}(Z), \mathscr{H}(Z))$, in which case

To say that $\lambda$ and $\mu$ are $\varphi$-convolvable signifies that, for every $f \in \mathscr{H}(Z)$, $f \circ \varphi$ is $(\lambda \otimes \mu)$-integrable, that is, for every $f \in \mathscr{H}(Z)$ the function $(x, y) \mapsto \langle f, \varepsilon_{\varphi(x,y)} \rangle$ is $(\lambda \otimes \mu)$-integrable, that is, again, that the mapping $(x, y) \mapsto \varepsilon_{\varphi(x,y)}$ of $X \times Y$ into $\mathscr{M}(Z)$ is scalarly $(\lambda \otimes \mu)$-integrable for $\sigma(\mathscr{M}(Z), \mathscr{K}(Z))$. If this is the case, then

whence $\lambda * \mu = \int_{X \times Y} \varepsilon_{\varphi(x,y)} d\lambda(x) d\mu(y)$.