double integral wrt minimum How can one calculate the following integral?
$$
\int_0^1{\int_0^1 xy \quad d[min(x,y)]}
$$
I have no idea how to hande the $d(min(x,y))$.
Does anyone have an idea for this problem?

Certainly this is a (Lebesgue-)Stieltjes integral. Nevertheless, fortunately $C_1=xy$ and $C_2=\min(x,y)$ are copulas and for copulas the following holds (partial dderivatives exist a.e. of course):
$$\iint_{[0,1]^2}C_1(x,y)\,\mathrm{d}C_2(x,y)=\frac{1}{2}-\iint_{[0,1]^2}\frac{\partial}{\partial x}C_1(x,y)\frac{\partial}{\partial y}C_2(x,y)\,\mathrm{d}(x,y).$$
Therefore $\iint_{[0,1]^2}xy\,\mathrm{d}\min(x,y)=\frac{1}{2}-\iint_{T}y\,\mathrm{d}(x,y),$ where $T=\{(x,y)\in[0,1]^2: ~0\leq x\leq 1, ~0\leq y<x\}$.
Finally $\iint_{[0,1]^2}xy\,\mathrm{d}\min(x,y)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}$.
