Consider functions $f_j\in\mathcal{L}^1, j\in\mathbb{N}$. Show: If $\sum_{j=1}^{\infty}\int\lvert f_j\rvert <\infty$, then $\sum_{j=1}^{\infty}f_j$ converges a.e. against an integrable function and it is $$ \int\sum_{j=1}^{\infty}f_j = \sum_{j=1}^{\infty}\int f_j. $$
Hm, it's rather difficult to me to show that.
Let $\sum_{j=1}^{\infty}\int\lvert f_j\rvert <\infty$. Then $$ \int\lvert f_j\rvert\to 0\mbox{ a.e.} $$ and therefore $\lvert f_j\rvert\to 0$ a.e.?
But I don't know how to continue.
Call the space $X$ and the measure $\mu$.
Define $h_n(x) = \sum_{j=1}^n \left|f_j(x)\right|$. By the monotone convergence theorem: $$ \int \sum_{j=1}^\infty \left|f_j\right| = \int \lim_{n\to\infty} h_n = \lim_{n\to\infty} \int h_n = \sum_{j=1}^\infty \int \left|f_j\right| < \infty $$
Define:
$$ E = \{x \in X: \sum_{j=1}^\infty \left|f_j(x)\right| = \infty \} $$
If $\mu(E) > 0$, then $\int \sum_{j=1}^\infty \left|f_j\right|$ would be $\infty$, a contradiction. It follows that $\mu(E) = 0$, and $\sum_{j=1}^\infty \left|f_j\right|$ converges a.e.
Define $\varphi(x) = \sum_{j=1}^\infty \left|f_j(x)\right|$ and $g_n(x) = \sum_{j=1}^n f_j(x)$. We have $\varphi \in \mathcal{L}^1$ and $\left|g_n\right| \le \varphi \ \forall n \in \mathbb N$. By the dominated convergence theorem: $$ \sum_{j=1}^\infty \int f_j = \lim_{n\to\infty} \int g_n = \int \lim_{n\to\infty} g_n = \int \sum_{j=1}^\infty f_j $$