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?
 A: 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.
A: Since $\lambda=\mu + \nu$, we have $$\frac{d \lambda}{d \mu}= \frac{ d \mu }{d \mu} + \frac{ d \nu}{ d \mu}= 1 + \frac{ d \nu}{ d \mu}$$
Now, $ \lambda << \mu$, and we show $\mu << \lambda$  Let $\lambda(E)=0$, then $\mu(E)= -\nu(E)$. since $\mu, \nu$ are positive we have that $\mu(E)=\nu(E)=0$, thus $\nu << \lambda$, and  $\mu << \lambda$. Then, $$ \frac{d \mu}{d \lambda}\frac{d \lambda}{d \mu}=1$$ Thus $$\frac{d \mu}{d \lambda} = \frac{1}{1 + \frac{d \nu}{d \mu}}$$
Then we multiply by $\frac{d \nu}{d \mu}$, to give us $$ \frac{d \nu}{d \mu}\frac{ d \mu}{d \lambda}= \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}}$$
Then  since $\mu$, $\nu$, and $\lambda$ are $\sigma$-finite measure and $\nu << \mu$ and $\mu << \lambda$, we have, $$ \frac{d \nu}{d \lambda} = \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}}$$
Since $\nu$, and $\mu$ are positive and $f = \frac{d \nu}{d \lambda}$, we have $0 \le f < 1$, that is $$0 \le f = \frac{ \frac{d \nu}{d \mu}}{ 1 + \frac{ d \nu}{d \mu}} < 1$$ 
