Real Analysis, Folland Theorem 2.25 Integration of Complex Functions 
Theorem 2.25 - Suppose that $\{f_j\}$ is a sequence in $L^1$ such that $\sum_{1}^{\infty}\int |f_j| < \infty$. Then $\sum_{1}^{\infty}f_j$ converges a.e. to a function in $L^1$, and $$\int \sum_{1}^{\infty} f_j = \sum_{1}^{\infty}\int f_j$$

Attempted proof - Recall from Theorem 2.15, $$\int \sum_{1}^{\infty}|f_j| = \sum_{1}^{\infty}\int |f_j| < \infty$$
Let $g = \sum_{1}^{\infty}|f_n|\in L^1$, then $$\int g = \int \sum_{1}^{\infty}|f_n| \leq \sum_{1}^{\infty}\int |f_n| < \infty$$ and for each $n$, $\left|\sum_{1}^{n}\right| \leq \sum_{1}^{n}|f_n| < \sum_{1}^{\infty}|f_n| = g$. So by the Dominated Convergence Theorem, $\sum_{1}^{\infty}f_n = \lim_{n\rightarrow \infty}\sum_{1}^{n}f_n\in L^1$ and $$\int \sum_{1}^{\infty}f_n = \lim_{n\rightarrow \infty}\int \sum_{1}^{n}f_n = \lim_{n\rightarrow \infty}\sum_{1}^{n}\int f_n = \sum_{1}^{\infty}\int f_n$$
This was a rendition of my professors proof. I am not sure if there is a mistake but I am not sure why $$\int \sum_{1}^{\infty}|f_n| \leq \sum_{1}^{\infty}\int |f_n|$$
I also don't know how to show that $\sum_{1}^{\infty}f_j$ converges a.e. to a function in L^1$.
Any suggestions is greatly appreciated.
 A: @Wolfy , From Theorem 2.15 we know $$\int \sum_{1}^{\infty}|f_j| = \sum_{1}^{\infty}\int |f_j| $$ So, of course, we also know $$\int \sum_{1}^{\infty}|f_j| \leq \sum_{1}^{\infty}\int |f_j| $$ 
but we don't need this inequality since we know the equality holds. 
Here is the proof with some adjusments / clarifications

Theorem 2.25 - Suppose that $\{f_j\}$ is a sequence in $L^1$ such that $\sum_{1}^{\infty}\int |f_j| < \infty$. Then $\sum_{1}^{\infty}f_j$ converges a.e. to a function in $L^1$, and $$\int \sum_{1}^{\infty} f_j = \sum_{1}^{\infty}\int f_j$$

Proof - Recall from Theorem 2.15, $$\int \sum_{1}^{\infty}|f_j| = \sum_{1}^{\infty}\int |f_j| $$
Let Let $g = \sum_{1}^{\infty}|f_n|$, then $g\geq 0$ and we have 
$$\int g = \int \sum_{1}^{\infty}|f_n| = \sum_{1}^{\infty}\int |f_n| < \infty$$ 
So  $g \in L^1$. 
Since $\int g <+\infty$, we have from Proposition 2.20, that $\{x: g(x)=+\infty\}$ is null set. It means that $\sum_{1}^{\infty}|f_n|$ is finite a.e.. So  $\sum_{1}^{k}f_n \to  \sum_{1}^{\infty} f_n $ a.e. (we don't know yet that $\sum_{1}^{\infty} f_n \in L^1$) and, since for all $k$, 
$$|\sum_{1}^{k}f_n| \leq \sum_{1}^{k}|f_n| \leq \sum_{1}^{\infty}|f_n|=g$$
So, by the Dominated Convergence Theorem,  $\sum_{1}^{\infty} f_n \in L^1$ and 
$$\int \sum_{1}^{\infty} f_n = \lim_{k \to \infty} \int \sum_{1}^{k}f_n = \lim_{k \to \infty}\sum_{1}^{k} \int f_n =  \sum_{1}^{\infty} \int f_n$$
