How can I show $f*g = g*f$ for Lebesgue integral? I want to show that
$(f*g)(x) = (g*f)(x)$ for almost every $x\in\mathbb R^n$ in  Lebesgue integration case, where  * is convolution on $\mathbb R^n$, that is $$
(f*g) (x)=\int_{\mathbb R^n} f(x-y)g(y) dy
$$
Try to substitute $x-y=u$. This makes a trick.
If $G$ is locally compact group then $L^1(G)$ is commutative if and only if $G$ is abelian group.[ Hewitt-Ross, Abstract harmonic analysis, 20.24 theorem]. Now since $R$ is locally compact abelian group so $L^1(R)$ is cummotative.