I am reading the proof of Radon-Nikodym Theorem in Follands "Real Analysis Modern Techniques and Their Applications". Specifically, Theorem 3.8 on page 90.
Folland starts with the case that both the measures are finite $\nu$ and $\mu$ are finite. However, I am a little bit confused with case 2 in which $\mu$ and $nu$ are $\sigma$-finite. Folland writes $X = \cup_j A_j$ where $A_j$ are disjoint and of finite measure. He then defines a measure $\mu_j(E) = \mu(E \cap A_j)$ and $\nu_j(E) = \nu(E \cap A_j)$. Since $\mu_j,\nu_j$ are finite measures, he applies the previous case. He defines the Radon-Nikodym derivative as $d\nu/d\mu = \sum_j f_j$ where $d\nu_j/d\mu_j = f_j$.
Based on the statement of the claim, $f$ should be integrable. But I do not see why this is true. I see that each of the $f_j$ are integrable, but why should an infinite sum of integrable functions be integrable? What am I missing?
The $f_j$'s are living in disjoint sets. So their sum is finite at each point, but in general it could be nonintegrable.