I was trying to understand the proof of Caratheodory's criterion from Bogachev's book and I understood it completely except one moment one moments which I am not able to solve by myself.
I have understood that $\mu^*(E\cap A)+\mu^*(E-A)=\mu^*(E)$ holds for any $E\subset X$ and $A\in \mathcal{A}$. And the author claims that it is easy to derive this inequality for the case when $A\in \mathcal{A}_{\mu}$ just using "triangle inequality" for measure, i.e. estimate (1.5.2).
I have tried different ways but no results.
Can anyone show to me how to show this unclear step?
Let $\epsilon > 0$. Following definition 1.5.1, since $A \in \mathcal{A}_\mu$, there exists $A' \in \mathcal{A}$ with $\mu^*(A \triangle A') < \epsilon$. Suppose the inequality holds for $A'$. Using the triangle inequality we can write:
$$\begin{align*} \mu^*(E \cap A) &\le \mu^*(E \cap A') + |\mu^*(E \cap A) - \mu^*(E \cap A')| \\ &\le \mu^*(E \cap A') + \mu^*((E \cap A) \triangle (E \cap A')) \end{align*}$$ using (1.5.2). But now you may verify that $(E \cap A) \triangle (E \cap A') \subseteq A \triangle A'$ (draw a Venn diagram if you like). So by monotonicity of outer measure (see note on page 17) we have $\mu^*((E \cap A) \triangle (E \cap A')) \le \mu^*(A \triangle A') < \epsilon$. Hence we have shown $$\mu^*(E \cap A) \le \mu^*(E \cap A') + \epsilon.$$
By a similar argument you may also show that $\mu^*(E \setminus A) \le \mu(E \setminus A') + \epsilon$. So we have $$\mu^*(E \cap A) + \mu^*(E \setminus A) \le \mu^*(E \cap A') + \mu(E \setminus A') + 2 \epsilon = \mu^*(E) + 2\epsilon$$ since by assumption the inequality holds for $A'$. Now since $\epsilon$ was arbitrary we must have $\mu^*(E \cap A) + \mu^*(E \setminus A) \le \mu^*(E)$ as desired.