I am preparing for an exam and came across this question that I thought would be beneficial to go through:
Let $\mu$ and $\nu$ be finite non-zero positive measures on a measure space $(X,M)$ such that $\nu \ll \mu \ll \nu.$ Show the strict inequality $$ 0 < \frac{d \nu}{d(\mu + \nu)} < 1.$$
I am not super confident in the material and feel like my argument may lack some rigor or may be missing something. Here is my solution:
As $\nu \ll \mu$ and $\nu \ll \nu + \mu$, we have $$\int_{E} \frac{d \nu}{d(\nu + \mu)} \, d (\nu + \mu) = \nu(E) = \int_{E} \frac{d \nu}{d \mu} \, d \mu.$$ Then by the chain rule we have $$\nu(E) = \int_{E} \frac{d \nu}{d(\nu + \mu)} \, d (\nu + \mu) = \int_{E} \frac{d \nu}{d(\nu + \mu)} \frac{d(\nu + \mu)}{d \mu} \, d \mu, $$ and so it follows that $$\frac{d \nu}{d \mu} = \frac{d \nu}{d(\nu + \mu)} \frac{d (\nu + \mu)}{d \mu} = \frac{d \nu}{d(\nu + \mu)} \left(\frac{d \nu}{d \mu} + \frac{d \mu}{d \mu} \right) = \frac{d \nu}{d(\nu + \mu)} \left( \frac{d \nu}{d \mu} + 1 \right) \quad \mu\text{-a.e.}.$$
Since $\nu$ is positive, $d\nu/d \mu \geq 0$ $\mu$-a.e. and so $1 + d \nu/d \mu > 0$ $\mu$-a.e. and so we have that $$ \frac{d \nu/d \mu}{1 + d \nu/d \mu} = \frac{d \nu}{d (\nu + \mu)}\quad \mu\text{-a.e.}$$ However this implies that $$0 \leq \frac{d \nu}{d(\nu + \mu)} < 1 $$ as $0 \leq d \nu/d\mu < 1 + d \nu/d \mu$.
Does everything seem correct up to this point? Moreover, the problem was pretty explicit that the inequality is strict, and I am unsure exactly why it is strictly larger than 0.
Let $A$ be the set on which $$\frac{d\nu}{d(\mu+\nu)}=1.$$ Then, $$\mu(A)+\nu(A)=\int_Ad(\mu+\nu)=\int_A\frac{d\nu}{d(\mu+\nu)}\,d(\mu+\nu)=\int_Ad\nu=\nu(A),$$ which implies that $\mu(A)=0$. Therefore $\nu(A)=0$ as well, and $A$ has $\mu+\nu$ measure $0$.
If, now, $B$ is the set where $$\frac{d\nu}{d(\mu+\nu)}=0,$$ then $$\nu(B)=\int_Bd\nu=\int_B\frac{d\nu}{d(\mu+\nu)}\,d(\mu+\nu)=0.$$