I'm reading a proof on proving that countable unions of measurable sets are again measurable. The proof is as follows
Let $T$ be a test set. $E_1\cup E_2$ is measurable if Caratheodory's criterion holds i.e $$\mu((E_1\cup E_2)\cap T)+\mu((E_1\cup E_2)^c\cap T)=\mu(T)\longleftrightarrow \mu(T_1\cup T_2\cup T_3)+\mu(T_4) = \mu(T)$$ Since $E_2$ is measurble $$\mu(E_2 \cap(T_1\cup T_2))+\mu(E_2^c \cap(T_1\cup T_2))=\mu(T_1\cup T_2)$$ By Caratheodory's criterion with test set $T_1\cup T_2$ we have $\mu(T_1)+\mu(T_2)= \mu(T_1\cup T_2)$
This is just for the case of two measurable sets but it goes on to say by iterating it would work for countably many sets. I'm having trouble understanding what's happening here to me it seems like all they are doing is juggling terms. Does anyone have an alternate proof not necessarily involving Caratheodory's criterion as I want to understand the intuition behind it.
This follows directly for the definition of a measurable set. Here's an alternative proof using the outer measure only.
Let $\{E_n\}_{n \in \mathbb{N}}$ be a countable collection of measurable sets and let $E = \bigcup_{n \in \mathbb{N}} E_n$. Let $\epsilon > 0$ be given. For each $n \in \mathbb{N}$, we can find an open set $U_n \supset E_n$ such that $$\mu^*(U_n \backslash E_n) < \frac{\epsilon}{2^n}.$$ Now, let $U = \bigcup_{n \in \mathbb{N}} U_n$. Obviously, $U$ is open and covers $E$. Since $$U \backslash E \subset \bigcup_{n \in \mathbb{N}} U_n \backslash E_n,$$ by countable subadditivity of $\mu^*$ we have $$\mu^*(U \backslash E) \leq \mu^*(\cup_{n \in \mathbb{N}} U_n \backslash E_n) \leq \sum_{n \in \mathbb{N}} \mu^*(U_n \backslash E_n) < \sum_{n \in \mathbb{N}} \frac{\epsilon}{2^n} = \epsilon.$$
Thus, $E$ is measurable.