# Proof relating to sum of Lebesgue integrable functions

I've been asked to find the proof for the following proposition:

Let $$(f_{n})_{n \geq1}$$ be a sequence of Lebesgue measure functions defined on a Lebesgue measurable set $$A$$ with values in either $$R$$ or $$C$$ such that $$\sum_{n = 1}^{\infty}\int_{A}|f_{n}(x)|dx < \infty$$. Then let the series defined by $$f(x) = \sum_{n = 1}^{\infty}f_{n}(x)$$ be absolutely convergent almost everywhere for $$x \in A, f \in L^{1}(A)$$. Prove that: $$\int_{A}f(x)dx = \sum_{n = 1}^{\infty}\int_{A}f_{n}(x)dx$$

I'm not really sure how to go about solving this problem and appreciate the help in solving it. Many thanks in advance!

• This is false. You need to remove the absolute values in the conclusion. As a hint: what very well-known measure theoretic result allows you to interchange integrals and limits (series are limits of sums)? Nov 6, 2019 at 14:37
• Yep that was my mistake and I've changed that now. I know using the monotone and dominated convergence theorem will help here but I'm just not sure how to make my argument combining these two theorems. Nov 6, 2019 at 14:47

This is for $$\int f_1 + \int f_2 = \int (f_1 + f_2)$$ case, the extension follows naturally.
For any function $$f_1$$ there exists simple function $$\phi_n$$ with $$0 \leq \phi_n \leq f_1$$ such that $$\phi_1 \leq \phi_2 \leq \cdots$$ and $$\phi_x(x) \underset{n \to \infty}{\to} f_1(x)$$ for all x.
Similarly, there exists $$\psi_n$$ such that $$0 \leq \psi_n \leq f_2$$ with $$\psi_1 \leq \psi_2 \leq \cdots$$ and $$\psi_n(x) \to f_2(x)$$.
Now, $$\phi_n + \psi_n$$ is simple, and increasing in n so $$\phi_n(x) + \psi_n(x) \to f_1(x) + f_2(x)$$.
\begin{aligned} \int(f_1 + f_2) &\underset{MCT}{=} \lim_{n \to \infty} \int (\phi_n + \psi_n))\\ &= \lim_{n \to \infty} \int \phi_n + \lim_{n \to \infty} \int \psi_n\\ &\underset{MCT}{\longrightarrow} \int f_1 + \int f_2 \end{aligned}