integrating limits of a sequence of integrable functions

Question

Let $f_1, f_2, \dots$ be measurable functions on some measure space $(X, \mathcal{A}, \mu)$. Consider the following theorem: If $\sum_{n=1}^\infty f_n$ converges almost everywhere, and for all $n$,$|\sum_{k=1}^n f_k| \leq g$ almost everywhere, where $g$ is integrable, then $\sum_{n=1}^\infty f_n$ is integrable and $\int_X \sum_{n=1}^\infty f_n d\mu = \sum_{n=1}^\infty \int_X f_n d\mu.$ Prove this by using Fubini’s theorem on $X × \{1, 2, \dots\}$.

I want to use Fubini's theorem in the following way: if $\int_X \sum_{n=1}^\infty f_n d\mu < \infty$ then also $\sum_{n=1}^\infty \int_X f_n d\mu < \infty$ and $\int_X \sum_{n=1}^\infty f_n d\mu = \sum_{n=1}^\infty \int_X f_n d\mu.$ I am stuck proving $\int_X \sum_{n=1}^\infty f_n d\mu < \infty$.

My try here: define $u_k := \sum_{k=1}^n f_k $. Then $\forall k \in \mathbb{N}$ we have $u_k \leq |u_k| \leq g$. This implies that $u_k$ is integrable for all $k \in \mathbb{N}$. We do not know if $u_k$ is an increasing sequence or not, but we do know that $u := \lim_{k\to\infty}u_k = \sum_{n=1}^\infty f_n$ exists. I know $\int u_n d\mu \leq \int g d\mu.$ I suspect $u \leq g$. Is my suspicion right? And how do I prove it? I am trying to work towards $\int u d\mu \leq \int g d\mu.$

Answer 1

Use Fatou's lemma on the series $|u_k|$ as you have defined it to show $|u| = |\sum_1^\infty f_n |$ and so $\sum_1^\infty f_n $ is integrable.

Specifically, since $|u_k| \leq g$ for all $k$ $$\int |u_k| \, d\mu \leq \int g \, d\mu $$ for all $k$.

So using Fatou's lemma, $$\int |\sum_1^\infty f_n | \, d\mu = \int \lim \inf_{k \to \infty} |u_k| \, d\mu \leq \lim \inf_{k \to \infty} \int |u_k| \, d\mu \leq \int g \, d\mu < \infty$$

since you assumed the partial sums $|u_k|$ converge almost everywhere.

Then $$\int_X \sum_1^\infty f_n \, d\mu \leq \int_X |\sum_1^\infty f_n | \, d\mu < \infty$$

and you get past your sticking point in the answer you started.