theorem of regularity of Lebesgue measurable set I was reading a proof regarding the condition for Lesbesgue measurable set. Specifically it is the Theorem 2.24 and the proof of the theorem here: https://www.math.ucdavis.edu/~hunter/m206/measure_notes.pdf
In the theorem, it says a set A is lebesgue measurable if and only if there is an open set G where $A\subset G$ such that $\mu^{*}(G\setminus A) \le \epsilon$ for any $\epsilon$. 
But there is another theorem previous to that which is Theorem 2.23 and the proof of Theorem 2.23 specifically the line (2.12) where it says we always have $\mu(G) \le \mu^{*}(A) + \epsilon $ for ANY set A whether A is measurable or not where G is an open set.
So I thought the condition $\mu(G) \le \mu^{*}(A) + \epsilon $ automatically implies $\mu(G) \le \mu(A) + \epsilon $ since we have $\mu(G) \le \mu^{*}(A) + \epsilon \le \mu(A) + \epsilon $ and therefore we have $\mu(G)-\mu(A)  \le \epsilon$ and hence, $\mu(G \setminus A) \le \epsilon$ whether A is measurable or not. 
So why is it that Theorem 2.24 is saying we have $\mu(G\setminus A) \le \epsilon$ if and only if A is measurable, because from Theorem 2.23, it seems it doesn't matter whether A is measuralbe or not. 
Could someone shine some light or give some hints.
Thank you
 A: Firstly i don't know how you deduce that $\mu^*(A) + \epsilon \leqslant  \mu(A)+ \epsilon$.
Now we know that the outer measure $\mu^*$ is defined on all subsets $\mathbb{R}$ 
and $\mu$ is the ristriction of $\mu^*$ in the sigma algebra of the Lebesgue measurable sets.
Now one can prove that $\mu^*(A)=\inf\{m^*(G)|G $ is open and $A\ \subseteq G\}$
for all $A \subseteq\mathbb{R}$ and thus we have the inequality: $\mu(G)=\mu^*(G) \leqslant \mu^*(A) +\epsilon$
But the inequality $\mu(G$\ $A)\leqslant \epsilon$ is a stronger statement and is used in the case where $A$ is measurable, because $G=A \cup (G$ \ $A)$
thus $\mu(G)=\mu^*(G) \leqslant  \mu^*(A)+\mu^*(G$ \ $A)$.
But when $A$ is measurable we have equality $\mu(G) = \mu(A)+\mu(G$ \ $A)\Rightarrow \mu(G$ \ $A)=\mu(G) - \mu(A)$(Note that if $A$ is measurable then also $G$ \ $A$ is measurable.)
Thus $\mu(G$\ $A) \leqslant \epsilon \Rightarrow \mu(G) \leqslant \mu(A)+ \epsilon =\mu^*(A) + \epsilon$.
But $\mu(G) \leqslant \mu^*(A)+ \epsilon $ does not imply that $\mu^*(G$ \ $A) \leqslant \epsilon$ when $A$ is not measurable and also $\mu(G$ \ $A)$ is not defined at all if $A$ is not measurable.
