It's a reply to this comment. I'm writing it as an answer since there is a bit too much $\LaTeX$ in it.
Well, for each $\varepsilon$ you need $n, G_i, F_i$ such that: $\mu\Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i.
We can equivalently require: $\mu \Big(\bigcup_{i\in \mathbb N} G_i \setminus \bigcup_{i\in \mathbb N} A_i \Big) < \varepsilon$ and $\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i. So actually the problem is the second assertion.
We have $F_i$ such that $\mu(A_i\setminus F_i)< \frac{\varepsilon}{2^{i+2}}$, so $\mu \Big(\bigcup_{i\in \mathbb N} A_i \setminus \bigcup_{i\in\mathbb N} F_i\Big) \leq \sum_{i \in \mathbb N} \frac{\varepsilon}{2^{i+2}} = \frac{\varepsilon}{2}$. Thus we know: $\mu \Big(\bigcup_{i\in\mathbb N} F_i\Big) \geq \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \frac{\varepsilon}{2}.$ Finally, we use continuity of measure to conclude that there exists $n$ such that $\mu \Big(\bigcup_{i \mu\Big(\bigcup_{i\in \mathbb N} A_i\Big) - \varepsilon.$