On an identity about integrals Suppose you have two finite Borel measures $\mu$ and $\nu$ on $(0,\infty)$. I would like to show that there exists a finite Borel measure $\omega$ such that 
$$\int_0^{\infty} f(z) d\omega(z) = \int_0^{\infty}\int_0^{\infty} f(st) d\mu(s)d\nu(t).$$ 
I could try to use a change of variable formula, but the two integration domains are not diffeomorphic. So I really don't know how to start. Any help would be appreciated! This is not an homework, I am currently practising for an exam. 
Thanks!
 A: When we have no idea about the problem, the question we have to ask ourselves is: "if a measure $\omega$ works, what should it have to satisfy?". 
We know that for a Borel measure that it's important to know them on intervals of the form $(0,a]$, $a>0$ (because we can deduce their value on $(a,b]$ for $a<b$, and on finite unions of these intervals). So we are tempted to define 
$$\omega((0,a]):=\int_{(0,+\infty)^2}\chi_{(0,a]}(st)d\mu(s)d\nu(t)=\int_{(0,+\infty)}\mu(0,At^{-1}])d\nu(t).$$
Note that if the collection $\{(a_i,b_i]\}_{i=1}^N\}$ consists of pairwise disjoint elements, so are for each $t$ the $(a_it^{-1},b_it^{-1}]$, which allows us to define $\omega$ over the ring which consists of finite disjoint unions of elements of the form $(a,b]$, $0<a<b<+\infty$. Then we extend it to Borel sets by Caratheodory's extension theorem.
As the involved measures are finite, $\omega$ is actually uniquely determined.  
A: This operation on measures is called convolution for measures on the locally compact abelian group $(0,+\infty)$ under multiplication.   
Using exp and log we may convert this to the usual convolution on the LCA group $(-\infty,+\infty)$ under addition.  In that case, the convolution $\omega = \mu \ast \nu$ satisfies
$$
\int_{-\infty}^{+\infty} f(z)\; \omega(dz) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(x+y) \;\mu(dx)\;\nu(dy)
$$
