I am currently struggling with the following:
Let $\nu,\mu$ be finite measures on a measurable space $(\Omega,\mathcal{A}) ~$with$~ \nu \ll \mu$ and $f:=\frac{d\nu}{d(\mu+\nu)}$. I would like to show that: 1.) $f$ exists 2.) $\frac{d\nu}{d\mu}=\frac{f}{1-f} ~\mu-$a.e. holds.
Help would be gratefully acknowledged. Thank you in advance!
Notice that $\nu\ll\mu+ \nu$. To see that assume $\nu(A)+\mu(A)=0$, then $\mu(A)=0$ and $\nu(A)$ since a measure is nonnegative. So by Radon-Nykodym $f$ exists.
We need to show that for all measurable $A$ one has: \begin{align} \nu (A) = \int_A \frac{f}{1-f}\,d\mu \end{align} Notice that: $$\nu(A) = \int_A f \,d\mu + \int_A f\,d\nu =\int_A f\,d\mu + \int_A f\frac{d\nu}{d\mu}\,d\mu = \int_A \left( 1+\frac{d\nu}{d\mu}\right)f \,d\mu$$ This means that: \begin{align} \frac{d\nu}{d\mu} = \left( 1+\frac{d\nu}{d\mu}\right)f \end{align} Hence: \begin{align*} \frac{d\nu}{d\mu} = \frac{f}{1-f} \end{align*} Exactly what we wanted to show.