# 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$. Any suggestions is greatly appreciated. • By MCT,$\int \sum |f_j| = \sum \int |f_j|$, hence$\sum |f_j| \in L^1$, hence$\sum |f_j|$converges a.e.. Since$\sum f_j$converges absolutely a.e., it converges a.e..$\left|\sum f_j\right| \leq \sum |f_j|$shows that$\sum f_j \in L^1$. – Qiyu Wen Jun 28 '16 at 6:26 ## 1 Answer @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 $$\int \sum_{1}^{\infty} f_n = \lim_{k \to \infty} \int \sum_{1}^{k}f_n = \lim_{k \to \infty}\sum_{1}^{k} \int f_n = \sum_{1}^{\infty} \int f_n$$

• In the last step of equalities, $$\int \sum_{1}^{\infty} f_n = \lim_{k \to \infty} \int \sum_{1}^{k}f_n = \lim_{k \to \infty}\sum_{1}^{k} \int f_n = \sum_{1}^{\infty} \int f_n$$ how are you using the Dominated Convergence Theorem? What is the f? – Axion004 Oct 2 '18 at 1:36