# 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 $$\int \sum_{1}^{\infty}|f_n| \leq \sum_{1}^{\infty}\int |f_n|$$