why does the integral of convolution equal to the product of their integral separately?

Question

$(f*g)(x)$ is called convolution and is the integral of $f(x-y)g(y)$ with respect to $y$ on $\mathbb{R}^n$. But why the integral of $f*g$ is equal to product of integral of $f$ and $g$. Wiki says it follows from Fubini's Theorem but I don't see why.

Answer 1

The standard proof uses the definition of convolution and Fubini: \begin{align*} \int_{\Bbb R^n} (f\ast g)(x)\,dx &= \int_{\Bbb R^n} \bigg( \int_{\Bbb R^n} f(x-y)g(y)\,dy \bigg) \,dx \\ &= \int_{\Bbb R^n} g(y) \bigg( \int_{\Bbb R^n} f(x-y) \,dx \bigg) \,dy \\ &= \int_{\Bbb R^n} g(y) \bigg( \int_{\Bbb R^n} f(z) \,dz \bigg) \,dy, \end{align*} using the change of variables $z=x-y$ for each fixed $y$.