Additivity of Radon-Nikodym derivatives of signed measures In the section on the Lebesgue-Radon-Nikodym theorem, Folland's Real Analysis: Modern Techniques and Their Applications (Second Edition) pp. 91 says that "it is obvious that $d(\nu_1 + \nu_2)/d\mu = d\nu_1/d\mu + d\nu_2/d\mu$."
I can almost see this, but I don't quite see how additivity necessarily applies in the underlying integrals.
Let $\nu_1$ and $\nu_2$ be $\sigma$-finite signed measures and $\mu$ be a $\sigma$-finite measure on space $(\Omega, \mathcal{F})$.  Let $\nu_1, \nu_2 \ll \mu$, and define $\nu := \nu_1 + \nu_2$.   Then for all measurable sets $A \in \mathcal{F}$, we have
$$ \nu(A) = \nu_1(A) + \nu_2(A) \stackrel{\text{Radon-Nikodym}}{=} \int_A f_1 \, d\mu + \int_A f_2 \, d\mu  \stackrel{?}{=} \int_A (f_1 + f_2) \, d\mu$$
where $f_1, f_2$ are Radon-Nikodym derivatives of $\nu_1$ and $\nu_2$, respectively, with respect to $\mu$.
My question is this: how do we know that the additivity used in the final equality applies?  Isn't it possible that we obtain a sum of the form $\infty - \infty$ or $-\infty + \infty$?
Indeed, according to the statement of the Lebesgue-Radon-Nikodym Theorem in Folland (pp.90), we only know  that $f_1$ and $f_2$ are extended integrable functions, not integrable functions.  Based on that, either of $\int_A f_i d\mu = \infty$ and $\int_A f_i d\mu = -\infty$ would seem to be possible.
 A: Additivity of Radon-Nikodym Derivatives.  Let $\nu = \nu_1 + \nu_2$ be a $\sigma$-finite signed measure on $(\Omega,\mathcal{F})$.  If $\nu_1, \nu_2 \ll \mu$, then $\nu_1 + \nu_2 \ll \mu$, and
$$\frac{d(\nu_1+\nu_2)}{d\mu} = \frac{d\nu_1}{d\mu} + \frac{d\nu_2}{d\mu} \quad \mu-\text{a.e.} $$
Proof. That $\nu_1, \nu_2 \ll \mu \implies \nu_1 + \nu_2 \ll \mu$ is immediate from the definition of absolute continuity.  Now for all $A \in \mathcal{F}$,
\begin{align*}
\nu(A) &= \nu_1(A) + \nu_2(A)  \\
& \stackrel{\text{where $f_1, f_2$ are Radon-Nikodym Derivatives}}{=} \int_A f_1 d\mu + \int_A f_2 d\mu \\
& \stackrel{\text{Additivity applies (see below)}}{=}  \int_A (f_1 + f_2) d\mu 
\end{align*}
So $f_1+f_2$ must be a Radon-Nikodym derivative of $\nu$ w.r.t to $\mu$.
Additivity applies in the final equality by the Additivity Theorem (see Thm 1.6.3 of Ash and Doleans-Dade's Probability and Measure Theory).  In particular each $\int_A f_1 d\mu + \int_A f_2 d\mu$ is a well-defined sum (not of the form $\infty-\infty$ or $-\infty + \infty$) because if it weren't, then $\nu = \nu_1 + \nu_2$ would fail to map some $A \in \mathcal{F}$ to an extended real-valued number, which would violate the definition of signed measure.
