Question on Radon Nikodym Theorem Let $\mu,\lambda$ and $\nu$ be $\sigma$-finite measures on $(X,M)$ such the $\nu\ll \mu$. Let $\mu= \nu + \lambda$. Then if $f$ is the Radon-Nikodym derivate of $\nu$ wrt $\mu$, we have $0\leq f\lt 1~\mu$-a.e. where $f = \frac{d\nu}{d\mu}$.  
Approach.   
Suppose to the contrary that $f\geq 1$. Then since I can infer $\nu \ll \mu$, $$\nu(E) = \int_E f~d\mu \geq \int_E 1 d\mu = \mu(E).$$ But then $\nu(E) \leq \nu(E) + \lambda(E) = \mu(E)$. So a contradiction. Thus $f\lt 1~\mu$-a.e.  
Please, does this look okay?

Why don't you just use the fact that since $\nu(E) + \lambda(E) = \mu(E)$ for all measurable $E$, then $\nu(E) \leq \mu(E)$. Writing this in integral form, we have $\int_E \frac{d \nu}{d \mu} d \mu \leq \int_E 1 d \mu$. From this you should be able to conclude that $\frac{d \nu}{d \mu} \leq 1$ $\mu$-a.e.
