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