Let $A_n \subseteq \mathbb{R} ^n$ and $A_n \subseteq A_{n+1}$ so that $\cup_{i=1}^\infty{A_i}=\mathbb{R}^n$. We also have a function $f:\mathbb{R}^n \to \mathbb{R}$. Now I want to show the following equivalence:
$$\int_{\mathbb{R}^n} |f|\,d\mu< \infty \text{ iff } \forall i \in \mathbb{N}\int_{A_i}|f|\,d\mu< \infty \text{ and } \sup_{i\in \mathbb{N}}\int_{A_i}|f|\,d\mu< \infty.$$
One direction is straight forward:
$$\infty>\int_{\mathbb{R}^n}|f|\,d\mu=\int_{\cup_{i=1}^\infty{A_i}}|f|\,d\mu\geq \int_{A_i}|f|\,d\mu\text{ } \forall i\in \mathbb{N}$$ and consequently: $$\sup_{i\in\mathbb{N}} \int_{A_i}|f|\,d\mu< \infty$$
Now I think it is obvious that $sup_{i\in\mathbb{N}} \int_{A_i}|f|\,d\mu= \int_{\cup_{i=1}^\infty A_i}|f|\,d\mu$ as $A_i \subseteq A_{i+1}$ and hence $\int_{A_i}|f|d\mu\leq \int_{A_{i+1}}|f|\,d\mu$. However, I am not sure how to strictly prove this guess. If this step is done, the rest simply follows. So could you please help me clarifying this step?
Let $f_i = |f| \chi_{A_i}$. Then $(f_i)$ is monotonically increasing to $|f|$. By the monotone convergence theorem, $\int f_i \uparrow \int |f|$.
But $\int f_i = \int_{A_i} |f|$. So $$\sup_i \int_{A_i} |f| = \lim_i \int_{A_i} |f| = \int |f|$$
So $\int |f| < \infty$.