Consider the following fragment from Folland's book "A course in abstract harmonic analysis" (question is below the image).
Can someone explain why the boxed equality is true? Don't we need that the function $$y \mapsto \int_{G}\phi(xy)\mu(dx)$$ is at least in $C_0(G)$ to apply the convolution formula? This doesn't seem to be true (for instance, consider $G= (\mathbb{R},+)$ with the Lebesgue (=Haar) measure and the function $\phi(x) = |\sin(x)|\chi_{[-2\pi, 2\pi]}(x)$). So, how can I justify the boxed equality?
By $(2.34) $ we have $$\int \psi(y)\,d(\nu*\sigma)(y)=\iint \psi (yz)\,d\nu(y)\,d\sigma(z)$$ For fixed $x$ let $\psi(y)=\phi(xy).$ Then $$\qquad \int \phi(xy)\,d(\nu*\sigma)(y)=\iint \phi (xyz)\,d\nu(y)\,d\sigma(z)\qquad (*)$$ The left hand side of the boxed equality is equal (after changing the order of integrals) $$L:=\int \left ( \int \phi(xy)\,d(\nu*\sigma)(y)\right )\,d\mu(x) $$ Making use of $(*)$ and changing the order of integration gives $$L:= \int\left ( \iint \phi (xyz)\,d\nu(y)\,d\sigma(z)\right )\,d\mu(x)= \iiint \phi(xyz)\,d\mu(x)\,d\nu(y)\,d\sigma(z)$$ Remark We can change the order of integration every time when one of the iterated integrals of the absolute value of the integrated function is finite. The functions considered here are bounded and all measures are finite. Hence the change of order of integration is justified.