Question is as stated above. Can someone explain if (and how) $$\iint f(x)g(y)\,dx,dy=\int f(x)\,dx*\int g(y)\,dy$$
I found this website, but I don't get why they use $x\leq y\leq b$. It makes a bit more sense (in my mind), to have $0\leq y\leq x$, but I still don't undersand how to finish/solve this logic.
If $f\in L^{1}({\bf{R}}^{n})$ and $g\in L^{1}({\bf{R}}^{m})$, then by Fubini Theorem, $fg:(x,y)\rightarrow f(x)g(y)$ is such that $fg\in L^{1}({\bf{R}}^{n}\times{\bf{R}}^{m})\cong L^{1}({\bf{R}}^{n+m})$ and that \begin{align*} \iint_{{\bf{R}}^{n+m}}f(x)g(y)dxdy&=\int_{{\bf{R}}^{n}}\left(\int_{{\bf{R}}^{m}}f(x)g(y)dy\right)dx\\ &=\int_{{\bf{R}}^{m}}\left(\int_{{\bf{R}}^{n}}f(x)g(y)dx\right)dy, \end{align*} and we have \begin{align*} \int_{{\bf{R}}^{n}}\left(\int_{{\bf{R}}^{m}}f(x)g(y)dy\right)dx&=\int_{{\bf{R}}^{n}}f(x)\left(\int_{{\bf{R}}^{m}}g(y)dy\right)dx\\ &=\left(\int_{{\bf{R}}^{n}}f(x)dx\right)\left(\int_{{\bf{R}}^{m}}g(y)dy\right). \end{align*}
