Convolution preserves compact support and is associative

For fixed $f,g\in C_c(G)$ the kernel $F(x,y):=f(y)g(y^{-1}x)$ is continuous on $G\times G$, and $F(x,y)\mathrm{\ne }0$ forces $y\in \mathrm{supp}f$ and $y^{-1}x\in \mathrm{supp}g$, that is $x\in \mathrm{supp}f\cdot \mathrm{supp}g$; by [F4] the set $(\mathrm{supp}f\cdot \mathrm{supp}g)\times \mathrm{supp}f$ is compact by [F5] and closed, so $\mathrm{supp}F\subseteq (\mathrm{supp}f\cdot \mathrm{supp}g)\times \mathrm{supp}f$ and $F\in C_c(G\times G;\mathbb{C})$.

The inner integral of the first expression equals the inner integral of the second: substituting $z=yw$ and using left invariance of $\mu$ from [F3] gives ${\int }_{G}g(y^{-1}z)h(z^{-1}x)d\mu (z)={\int }_{G}g(w)h(w^{-1}{y}^{-1}x)d\mu (w)$, since $(yw{)}^{-1}=w^{-1}{y}^{-1}$ and ${\int }_{G}H(z)d\mu (z)={\int }_{G}H(yw)d\mu (w)$ for the compactly supported continuous $H$ occurring here.

Applying [F2] to $F$ (with [A1] supplying its choice hypothesis) shows that the partial integral $x\mapsto {\int }_{G}F(x,y)d\mu (y)=(f\ast g)(x)$ is continuous with compact support; hence $f\ast g\in C_c(G)$ and $C_c(G)$ is closed under convolution.

For $f,g,h\in C_c(G)$ and $x\in G$, expanding the definitions gives $((f\ast g)\ast h)(x)={\int }_G{\int }_{G}f(y)g(y^{-1}z)h(z^{-1}x)d\mu (y)d\mu (z)$ and $(f\ast (g\ast h))(x)={\int }_G{\int }_{G}f(y)g(w)h(w^{-1}{y}^{-1}x)d\mu (w)d\mu (y)$: in both expressions the iterated integrals exist by two applications of [F2] to the continuous compactly supported kernels obtained as in step 1.1.

Combining steps 3.1 and 1.2 gives $((f\ast g)\ast h)(x)=(f\ast (g\ast h))(x)$ for every $x\in G$, hence $(f\ast g)\ast h=f\ast (g\ast h)$, which is the asserted associativity. ∎