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?
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

$$
\mu^*(B)
\leq \mu^*(A_{1/n})
\leq \mu^*(E) + \frac{1}{n}
\;\text{for all $n \in \mathbb Z_{&gt;0}$}. \tag{2}
$$
The statements $(1)$ and $(2)$ combine to force $\mu^*(E) = \mu^*(B)$, as desired.
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.)