Let $(A_n)_{n\in\mathbb{N}}\subseteq\mathcal{A}$ be a sequence of mutually disjoint sets. Show that
$$ u1_{\cup_nA_n} \in\mathcal{L}^1(\mu) \iff \left( u1_{A_n} \in\mathcal{L}^1(\mu)\quad\text{ and } \quad \sum_{n=1}^{\infty} \int_{A_{n}}|u| d \mu<\infty \right) $$
It does not say in the problem, but $u:X\to\mathbb{R}$
I know that $1_{\cup_nA_n} = \sum_{n=1}^\infty 1_{A_n}$. So I know that $u\sum_{n=1}^\infty 1_{A_n}\in\mathcal{L}^1(\mu)$. Since $u\sum_{n=1}^\infty 1_{A_n}$ is measurable, then $\sum_{n=1}^\infty u1_{A_n}$ is also measurable, right? In from that it follows that $u1_{A_n}$ is measurable, right? Then I just have to show that $\sum_{n=1}^{\infty} \int_{A_{n}}|u| d \mu<\infty$, right?
I think I need the below lemma:
Lemma 10.8 On the measure space $(X, \mathscr{A}, \mu)$ let $u \in \mathcal{M}^{+}(\mathscr{A}) .$ The set function $$ \nu: A \mapsto \int_{A} u d \mu=\int \mathbb{1}_{A} u d \mu, \quad A \in \mathscr{A} $$ is a measure on $(X, \mathscr{A}) .$ It is called the measure with density (function) $u$ with respect to $\mu .$ We write $\nu=u \mu$ or $d \nu=u d \mu$
Just apply the Lemma to the measure $\nu_1$ and $\nu_2$ defined by $\nu_1 (A)=\int_A u^{+} d\mu$ and $\nu_2 (A)=\int_A u^{-} d\mu$ and recall that $f \in L^{1}(\mu)$ if and only if $\int |f| d\mu <\infty$ if and only if $\int f^{+}d\mu <\infty$ and $\int f^{-}d\mu <\infty$ . We actually have the identity $\int u1_{\cup A_n} d\mu=\sum \int u1_{\cup A_n}$ obtained by applying the lemma to $u^{+}$ and $u^{-}$.