During the detailed reading of below lemma I ran into the following questions. I have spent almost 2 days in order to understand it completely and some of questions I have got but some not.
1) The author trying to show that $\bar{\mu}$ is $\sigma$-additive on $\mathcal{A}$. Since $A\in \mathcal{A}$ then $$\bar{\mu}(A)=\lim \limits_{n\to \infty} \mu_n(A\cap S_n)<\infty$$ but why is it increasing? I know that $A\cap S_n\subset A\cap S_{n+1}$ but we cannot say that $\mu_n(A\cap S_n)\leq \mu_{n+1}(A\cap S_n)$ because $\mu_n, \mu_{n+1}$ are different measures. So we cannot use here monotonicity of measure.
2) I was trying to show that if $A:=\sqcup_{i}A_i$ then $\bar{\mu}(A)\leq \sum \limits_{i=1}^{\infty}\bar{\mu}(A_i)$. Here is my approach: Since $A\in \mathcal{A}$ then $$\bar{\mu}(\sqcup_{i}A_i)=\lim \limits_{n\to \infty}\mu_n((\sqcup_{i}A_i)\cap S)=\lim \limits_{n\to \infty}\sum \limits_{i=1}^{\infty}\mu_n(A\cap S_n) \underbrace{\leq}_{\text{is it true?}} $$ $$\sum \limits_{i=1}^{\infty}\lim \limits_{n\to \infty}\mu_n(A_i\cap S_n)=\sum \limits_{i=1}^{\infty}\lim \limits_{n\to \infty}\bar{\mu}(A_i),$$ the last equation holds because $A_i\in \mathcal{A}$. But the problem is the following: is this inequality true or not? I was not able to prove it. Maube there is some alternative way of proving it.
3) The third question is more philosophical. What is the essence of this lemma? Bogachev applies it in the constructing of Lebesgue measure. But can anyone explain it in simple way?
Would be very grateful for your help! Please explain my questions in detail!
