Let $\nu$ be a bounded signed measure and $\mu$ be a $\sigma$-finite measure such that $\nu << \mu$. I want to show that there exists an $f \in \mathcal L(\mu)$ such that $\nu = f \cdot \mu$.
And I assume the result holds if $\mu$ were finite.
So the procedure is pretty straightforward: Choose a sequence of pairwise disjoint and finite measure sets $F_n\uparrow \Omega$ and define finite measures $$\mu_n(F)=\mu(F\cap F_n)$$ $$\nu_n(F)=\nu(F\cap F_n)$$ Then $\nu_n << \mu_n$ and so there exist Radon-Nikodym derivates $f_n = \frac{d\nu_n}{d\mu_n}$. We can extend these functions to all of $\Omega$ by defining them to be $0$ on $F_n^c$.
Then we set $f=\sum_{n=1}^\infty f_n$ and quite clearly $\nu(F)=\int_F f \, d \mu$ but my question is why is $f$ integrable?
We have
$$\int |f| \, d\mu = \sum_{n=1}^\infty \int_{F_n} |f| \, d\mu = \sum_{n=1}^\infty \int |f_n| \, d\mu = \sum_{n=1}^\infty \left(\int_{(f_n>0)} f_n \, d\mu - \int_{(f_n<0)} f_n \, d\mu \right)$$
How can we proceed to get a bound from there?
