Let $\phi(t)=\int e^{itx}\mu(dx)$ where $\mu$ is probability measure and $\mu({a})=0$ and $\mu({b})=0$.If $a<b$ then, $\mu(a,b)=lim_{T\rightarrow\infty}\frac{1}{2\pi}\int_{-T}^{T} \frac{e^{-ita}-e^{-itb}}{it}\phi(t)dt.$This theorem is called The inversion formula.
Let $\nu$ be another probailty measure where $\phi^{'}(s)=\int e^{ity}\mu(dy)$ and $\nu({a^{'}})=0$ and $\nu({b^{'}})=0$.If $a^{'}<b^{'}$ then, $\nu(a^{'},b^{'})=lim_{S\rightarrow\infty}\frac{1}{2\pi}\int_{-S}^{S} \frac{e^{-isa^{'}}-e^{-isb^{'}}}{is}\phi(t)dt.$
Let $m=\mu * \nu$ and $\phi(t,s)$ be characteristic function corresponding it. Following formula for $m$ is correct? $m((a,b)\times (a^{'},b^{'}))=lim_{T\rightarrow\infty S\rightarrow\infty}(\frac{1}{2\pi})^{2}\int_{-T}^{T} \int_{-S}^{S}\frac{e^{-ita}-e^{-itb}}{it}\frac{e^{-isa^{'}}-e^{-isb^{'}}}{is}\phi(t,s)dt ds$
I think this is true but i am not sure,could someone please explain for me why?