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!

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
$$
\int_{-\infty}^{+\infty} f(z)\; \omega(dz) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} f(x+y) \;\mu(dx)\;\nu(dy)
$$
