Countable additivity. Suppose that $\{g_n\}$ is a sequence of measurable functions on X such that $$\int_X \Big( \sum_{i=1}^{\infty} |g_n| \Big) d\mu < \infty .$$ Then $$\int \Big (\sum_{n=1}^\infty g_n \Big) d\mu = \sum_{n=1}^\infty \int g_n d\mu$$
Clearly this property of Lebesgue integrals is true for functions $|g_n|$ (because it is a consequence of the monotone convergence theorem). How to extend the property of non-negative functions to $L^1(\mu)$?
Let $f_n := \sum_{k=1}^ng_k$. Then each $f_n$ is integrable and $|f_n|\le\sum_{k=1}^\infty|g_k|$, which is integrable. Now, apply dominated convergence on $(f_n)$. It's that simple.
\begin{align*} \int\left|\sum_{k=1}^{n}g_{k}-\sum_{k=1}^{m}g_{k}\right|d\mu&=\int\left|\sum_{k=m}^{n}g_{k}\right|d\mu\\ &\leq\int\sum_{k=m}^{n}|g_{k}|\\ &\leq\int\sum_{k=m}^{\infty}|g_{k}|\\ &=\int\sum_{n=1}^{\infty}|g_{n}|-\int\sum_{k=1}^{m-1}|g_{k}|\\ &\rightarrow 0 \end{align*} by Monotone Convergence Theorem. The sequence of $(s_{n})$, $s_{n}=\displaystyle\sum_{k=1}^{n}g_{k}$ is Cauchy in $L^{1}$, by completeness, we have some integrable $g$ such that \begin{align*} \int\left|\sum_{k=1}^{n}g_{k}-g\right|d\mu\rightarrow 0. \end{align*} But we know that some subsequence of $(s_{n})$ converges to $g$ a.e. so we get that $g=\displaystyle\sum_{n=1}^{\infty}g_{n}$ a.e.
The rest is easy.
A small comment: As $\int{\sum_{k=1}^{\infty}{|g_{k}|}}d\mu < \infty$ if you define $g = \sum_{k=1}^{\infty}{|g_{k}|}$ then $g(x) < \infty$ for almost every $x \in X$. Therefore the series $\sum_{k=1}^{\infty}{g_{k}(x)}$ converges absolutely for almost every $x \in X$. Then you could conclude the result directly by the dominated convergence Theorem, by defining $F_{j} = \sum_{n=1}^{j}{g_{k}(x)}$, since in this case $|F_{j}| \leq g$, for every $j \in \mathbb{N}$.
