How to prove $\mu^\ast(A)=\mu(A)$ for every elementary set A? How to prove $\mu^\ast(A)=\mu(A)$ for every elementary set A? ( $\mu^\ast(A)$ is the outer measure of $A$)
I'm doing Lebesgue Theory chapter of Baby Rudin and I don't understand the underlined part in the picture.
Do we have $\mu^\ast(A)\le\mu^\ast(G)\le\mu(G)$?

Definitions:



 A: The definition is:
$$\mu^*(A) = \inf \left\{\sum_{n=1}^\infty \mu(A_n): (A_n)_{n=1}^\infty \text{ are open elementary sets such that } A \subseteq \bigcup_{n=1}^\infty A_n\right\}$$
We assumed that $G$ is an open elementary set such that $A \subseteq G$. Define a sequence of open elementary sets $(G_n)_{n=1}^\infty = (G, \emptyset, \emptyset, \emptyset, \ldots)$.
We have $A \subseteq G = \bigcup_{n=1}^\infty G_n$ so by definition of $\mu^*(A)$, we have $$\mu^*(A) \le \sum_{n=1}^\infty \mu(G_n) = \mu(G) + \mu(\emptyset) + \mu(\emptyset) + \ldots = \mu(G)$$
Notice that this really means that the infimum goes over all finite or countable coverings of $A$ by elementary sets.
$$\mu^*(A) = \inf \left\{\sum_{i\in I} \mu(A_i): \{A_i\}_{i\in I} \text{ are open elementary sets such that } A \subseteq \bigcup_{i\in I} A_i \text{ and } I \subseteq \mathbb{N}\right\}$$

This implies that $\mu^*(A) \le \mu(G) \le \mu(A) + \varepsilon$. So, for arbitrary $\varepsilon > 0$ we have $\mu^*(A) \le \mu(A) + \varepsilon$ therefore we can conclude $\mu^*(A) \le \mu(A)$.
