Radon-Nikodym derivative of sum of two measures Problem Statement: Suppose that $\mu$ and $\nu$ are two finite measures such that $\nu \ll \mu$, let $\rho = \mu + \nu$, and note that since $\mu(A) \le \rho(A)$, and $\nu(A) \le \rho(A)$, we have $\mu \ll \rho$ and $\nu \ll \rho$. Prove that if $f = d\mu/d\rho$ and $g = d\nu/d\rho$, then $f +g = 1$ and $d\nu = (g/f) \, d\mu$.
My attempt at a solution: For the first part, I said
$$\mu(A) = \int_A f \, d\rho \ \ \ \text{and} \ \ \ \nu(A) = \int_A g \, d\rho,$$
therefore, $\mu(A) + \nu(A) = \int_A (f+g) \, d\rho$. Now, $\mu(A) + \nu(A) = (\mu+\nu)(A)$, and $\mu + \nu$ is another finite measure, so we have that $f+g$ is the Radon-Nikodym derivative of $\mu + \nu$ w.r.t $\rho$, written as $f + g = \frac{d(\mu+\nu)}{d\rho}$. Now, we consider $\int_A 1 \, d\rho = \int \chi_A \, d\rho = \rho(A) = \mu(A) + \nu(A) = (\mu+\nu)(A)$, and by the uniqueness of the Radon-Nikodym derivative, $f + g = 1$ a.e. However, I am not sure how to prove that $f+ g = 1$ everywhere - am I going about this proof completely wrong, or is there something else I need to do from here to show this?
For this next part, I am not entirely sure what the question is asking. We know that there is some non-negative, integrable w.r.t. $\mu$ function $h$ such that $\nu(A) = \int_A h \, d\mu$. Are we being asked to prove that $h = g/f$? If so, it would be easy to write something like $\int_A (g/f) \, d\mu = \int_A\frac{d\nu \, d\rho}{d\rho \, d\mu} \, d\mu = \int_A \, d\nu = \nu(A)$, thus showing that $g/f = h$ a.e.; however, I don't know that we can manipulate Radon-Nikodym derivatives in the way, yet, so there must be something else going on here, and I don't know how to go about proving it!
 A: You cannot prove $f+g=1$ everywhere, because counterexamples can readily be found.  Where the exercise says $f+g=1$ there must be a convention in force that says that is to be construed as almost everywhere.
Since $\mu \ll \rho$, there is a Radon--Nikodym derivative $f=d\mu/d\rho$.
$$
\mu(A) = \int_A f\,d\rho = \int_A f \,\Big( \frac 1 g\, d\nu \Big) \tag 1
$$
The second equality in $(1)$ above should work if $\rho \ll \nu$. See if you can prove that by using the hypothesis that $\nu\ll\mu$ and the definition of $\rho$.
A: You have that
$$\rho(A) = \int_{A} 1 d\rho$$
And
$$\rho(A) = \mu(A)+\nu(A) =  \int_{A} f d\rho+ \int_{A} g d\rho= \int_{A} f+g d\rho$$
And as this is true for all $A$ measurable, it's a classical result that this imply that $f+g=1$ a.e (for $\rho$)
For the second one, just remark that
$$\int_A 1 d\nu = \int_A g d\rho = \int_A \frac{g}{f} f d\rho = \int_A  \frac{g}{f} d\mu$$
A: *

*Your argument is OK. You can only prove that $f+g=1$ a.e.$[\rho]$, because if you change $f$ in a set of $\rho$-measure 0, then the new function, let us call it $\tilde f$, will also satisfy $\tilde f = d\mu/d\rho$  and  $\mu(A) = \int_A \tilde f \, d\rho$.

*Note that $\nu \ll \mu \ll \rho$.  It is easy to show that 
$$\frac{d\nu}{d\rho} =  \frac{d\nu}{d\mu}\frac{d\mu}{d\rho}$$  
Let $\frac{d\nu}{d\mu}=h$. We have then $\nu(A) = \int_A h \, d\mu=\int_A hf \, d\rho$, then by the uniqueness of the Radon-Nikodym derivative, $g=hf$ a.e.
Now (and this is an important part) because $\rho=\mu+\nu$ and $\nu \ll \mu$, we also have that $\rho \ll \mu$ which allows us to conclude that $f=\frac{d\mu}{d\rho}>0$ a.e.. So from  $g=hf$ a.e. we get $h=g/f$ a.e., that is $\frac{d\nu}{d\mu}=g/f$.
