Given any two $\sigma-$finite measures $\mu$ and $\nu$ on some measurable space $(\Omega, \mathcal F)$ how do I compute $\dfrac{d\mu}{d(\mu+\nu)}$?
If $\nu<<\mu$ then I could have said that $\dfrac{d\mu}{d\mu+d\nu}=\dfrac{1}{1+\dfrac{d\nu}{d\mu}}$.
Thus I thought of using the Lebesgue decomposition of $\nu$ w.r.t. $\mu$: $\nu=\nu_{ac}+\nu_s$ where $\nu_{ac}<<\mu$ and $\nu_s\perp \mu$. I guess that $\dfrac{d\mu}{d(\mu+\nu)}=\dfrac{1}{1+\dfrac{d\nu_{ac}}{d\mu}}$
Is my guess correct?