Lemma 30.4.6.label Let $G$ be a locally compact group, $f, g \in L^{1}(G; \complex)$, $\mu = fdx$, and $\nu = gdx$, then $\mu * \nu = (f * g)dx$.
Proof. Let $\phi \in C_{c}(G; \complex)$, then
\begin{align*}\int_{G} \phi(x) (f * g)(x)dx&= \iint_{G \times G}\phi(x) f(y)g(y^{-1}x)dydx \\&= \iint_{G \times G}\phi(yx)g(x)dxf(y)dy \\&= \iint_{G \times G}\phi(yx) \mu(dy)\nu(dx) = \int_{G} \phi d(\mu * \nu)\end{align*}
$\square$
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