Using continuity of the measure in a proof... I'm trying to understand the following proof:

I don't understand how the conclusion came from the equation in the green box, did they use continuity of the measure?
 A: The existing answers are confused about what the relevant part of the proof is trying to conclude. Let's recap: The green box and the conclusion that followed at the bottom are all part of the $\Rightarrow$ direction, which is

If equality $\mu^*(E) = \mu^+(E)$ holds, then there exists an $A \in \mathcal M^*$ such that $E \subset A$ and $\mu^*(E) = \mu^*(A)$.

Since the asker appears to accept that $B \in \mathcal M^*$ and $E \subset B$, what we need to prove is that $\mu^*(E) = \mu^*(B)$ (not $\mu^*(E) = \mu^*(E)$ as was written in the proof – that would be a complete triviality).
We use the monotonicity of outer measures. Note that $E \subset B$ implies
$$
\mu^*(E)
\leq \mu^*(B). \tag{1}
$$
On the other hand, by the definition of $B$, we have $B \subset A_{1/n}$ for all $n$. So monotonicity and the green box tells us that
$$
\mu^*(B)
\leq \mu^*(A_{1/n})
\leq \mu^*(E) + \frac{1}{n}
\;\text{for all $n \in \mathbb Z_{>0}$}. \tag{2}
$$
The statements $(1)$ and $(2)$ combine to force $\mu^*(E) = \mu^*(B)$, as desired.
A: It looks like there is a typo in the green box. The far right should probably be $$\mu^+(E) + \frac{1}{n}.$$ And so we then have that for all $n$, $$\mu^*(E) \le \mu^+(E) + \frac{1}{n}.$$ This implies that the same inequality holds without the $\frac{1}{n}$. But from the middle of the large box, we have the reverse inequality. Therefore, equality holds as desired. (Of course, I'm explaining only the last sentence of the proof.)
