Let $\mu$ and $\nu$ be finite positive measures on a measurable space $(X,\mathcal{A})$, which are equivalent (i.e., $\nu$ is absolutely continuous with respect to $\mu$ and $\mu$ is absolutely continuous with respect to $\nu$). Let $\lambda = \mu + \nu$. Show that the Radon-Nikodym derivative satisfies $0 < \frac{d\nu}{d\lambda}< 1$.
My attempt: Is this sketch of the proof fine?
1- Using the fact the$\mu , \nu$ are positive measures and $\lambda = \mu + \nu$, we can see that $\mu\ll \lambda$, $ \nu \ll \lambda$
2- Showing that $\frac{d\nu}{d\lambda}>0$ and $\frac{d\mu}{d\lambda}>0$ (How do I show this?)
3- It is enough to prove the property that $ \frac{d(\nu+\mu)}{d\lambda}=\frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}$
4- so by $\lambda = \mu + \nu$ , we'd have $ \frac{d\lambda}{d\lambda}=\frac{d\nu}{d\lambda}+\frac{d\mu}{d\lambda}$ $\implies$ $\frac{d\nu}{d\lambda} = 1- \frac{d\mu}{d\lambda} \implies$ $0 < \frac{d\nu}{d\lambda}< 1$
Let us show that $0<\frac {d\nu} {d\lambda} <1$ almost everywhere w.r.t. $\lambda$ [hence also w.r.t. $\mu$ and w.r.t. $\nu$].
Let $E=\{ \frac {d\nu} {d\lambda}=0\}$. Then $\nu (E)=\int_E \frac {d\nu} {d\lambda} d\lambda =0$. But $\mu << \nu$ so we also get $\mu (E)=0$. Adding these two we get $\lambda (E)=0$ which means $\frac {d\nu} {d\lambda} >0$ a.e. w.r.t. $\lambda$.
To prove that $\frac {d\nu} {d\lambda} <1$ a.e. w.r.t. $\lambda$ simply repeat the above argument with $\mu$ and $\nu$ interchanged to get $\frac {d\mu} {d\lambda} >0$ a.e. w.r.t. $\lambda$ and this is equivalent to $\frac {d\nu} {d\lambda} <1$ a.e. w.r.t. $\lambda$.