Real Analysis, Radon-Nikodym derivative Problem: Suppose that $\mu, \nu$ are $\sigma$-finite (positive) measures on $(X,M)$ with $\nu << \mu$, and let $\lambda = \mu+\nu$. Let $f$ be the Radon-Nikodym derivative $f = d\nu / d\lambda $. Show that $ 0 \leq f < 1$  $\mu $-a.e and $d\nu/ d\mu=f/(1-f)$.
There is something about this that just does not hit home for me. I would greatly appreciate some help solving this problem. Here is what I have done so far:
Because $\mu, \nu$ are $\sigma$-finite positive measures, so is $\lambda$. Since $\nu << \mu$, then $\lambda << \mu$. We also have $\mu << \lambda$ and $\nu <<\lambda$ because $\nu, \mu$ are positive measures. Now, $d\lambda / d\lambda = 1 = d(\mu + \nu)/d\lambda = f+d\nu / d\lambda$ $\lambda$-a.e. So $f=1-d\mu /d\lambda$. 
This is pretty much how far I got. 
 A: $f\geq 0$ $\lambda$-almost everywhere (hence $\mu$-almost everwhere) since $\nu$ is a positive measure. 
Next, let $E=\{x:f(x)\geq 1\}$, and choose sets $E_n\subset E_{n+1}$ such that $\nu(E_n)<\infty$ for all $n$ and $X=\cup_{n}E_n$. Then
$$ \nu(E\cap E_n)=\int_{E\cap E_n}f\;d\lambda=\int_{E\cap E_n}f\;d\mu+\int_{E\cap E_n}f\;d\nu\geq \int_{E\cap E_n}\;d\mu+\int_E\;d\nu$$
$$=\mu(E\cap E_n)+\nu(E\cap E_n) $$
Since $\nu(E\cap E_n)<\infty$, this shows that $\mu(E\cap E_n)=0$ for all $n$, so $\mu(E)=0$. Thus $f<1$ $\mu$-almost everywhere.
For the last part, let $g=\frac{d\nu}{d\mu}$. Then for all $E\in\mathcal{M}$ we have
$$ \int_Eg\;d\mu=\nu(E)=\int_Ef\;d\lambda=\int_Ef\;d\mu+\int_Ef\;d\nu=\int_E(f+fg)\;d\mu $$
Therefore $g=f+fg$ $\mu$-almost everywhere, and since $0\leq f<1$ we can conclude that $g=\frac{f}{1-f}$ $\mu$-almost everywhere.
A: Notice that $$0 \le f(x) = \frac{d\nu}{d\lambda}(x) \le \frac{d(\mu + \nu)}{d\lambda}(x) = 1.$$ Let $E \in M$ be such that $f(x) = 1$ for every $x \in E$ and assume by contradiction that $\mu(E) > 0$. Then $$0 < \mu(E) = \lambda(E) - \nu(E) = \lambda(E) - \int_E\frac{d\nu}{d\lambda}d\lambda = \lambda(E) - \int_E 1\,d\lambda = 0.$$
Fianlly, since $\nu << \mu << \lambda << \mu$ we have $$\frac{d\nu}{d\mu} = \frac{d\nu}{d\lambda}\frac{d\lambda}{d\mu} = f\Big(\frac{d\mu}{d\lambda}\Big)^{-1} = f\Big(\frac{d(\lambda - \nu)}{d\lambda}\Big)^{-1} = f(1 - f)^{-1}.$$
