Applying Fubini's Theorem to Show the Identitiy

Question

$F(x)=\mu((-\infty,x])$ and $G(x)=\nu((-\infty,x])$ are two distribution functions with discontinuities. How could we show that $$\int_{(r,s]}F(y)dG(y)+\int_{(r,s]}G(y)dF(y)=F(s)G(s)-F(r)G(r)+\sum_{x\in(r,s]}\mu(\{x\})\nu(\{x\}) $$

Thanks!

Answer 1

\begin{align*} &\int_{(r,s]}F(y)\,dG(y)+\int_{(r,s]}G(y)\,dF(y)\\ &\quad =\int_{(r,s]}[F(y)-F(r)]\,dG(y)+F(r)[G(s)-G(r)]\\ &\qquad +\int_{(r,s]}[G(y)-G(r)]\,dF(y)+G(r)[F(s)-F(r)]\\ &\quad=\iint_{r<u\le v\le s}dF(u)dG(v)+\iint_{r<v\le u\le s}dG(v)dF(u)\\ &\qquad+F(r)G(s)+F(s)G(r)-2F(r)G(r)\\ &\quad=\iint_{r<u\le s,r<v\le s}dF(u)dG(v)+\iint_{r<u=v\le s}dF(u)dG(v)\\ &\qquad+F(r)G(s)+F(s)G(r)-2F(r)G(r)\\ &\quad=[F(s)-F(r)][G(s)-G(r)]+\sum_{x\in(r,s]}\mu(\{x\})\nu(\{x\})\\ &\qquad +F(r)G(s)+F(s)G(r)-2F(r)G(r)\\ &\quad=F(s)G(s)-F(r)G(r)+\sum_{x\in(r,s]}\mu(\{x\})\nu(\{x\}) \end{align*}