Proving the union of a countable collection of measurable sets is measurable. Let $E$ be the union of a countable collection of measurable sets.
Then there is a countable disjoint collection of measurable sets {$E_k$}$_{k=1}^\infty$ for which $E = \cup_{k=1}^\infty E_k$.
Let $A$ be any set. Let $n$ be a natural number.
Define $F_n = \cup_{k=1}^n E_k$. 
Since $F_n$ is measurable and $F_n^{c}\supset E^c$,
$m^*(A) = m^*(A\cap F_n) + m^*(A\cap F_n^c) \ge m^*(A\cap F_n) + m^*(A\cap E^c)$ , and
$m^*(A\cap F_n) = \sum_{k=1}^\infty m^*(A\cap E_k)$. (Since {$E_k$}$_{k=1}^n$ is a finite disjoint collection of measurable sets.)
Thus $m^*(A)\ge \sum_{k=1}^n m^*(A\cap E_k) + m^*(A\cap E^c) $
The left-hand side of this inequality is "independent" of n.....(omitted)
I want to ask what "indepdendent" means? and why it is "independent" of n?
 A: Suppose  you change the line "Let $A$ be any set. Let $n$ be a natural number." to "Let $A$ be any set. Then $\forall n\in \Bbb N....$"
The inequality you have difficulty with should then say $\forall n\in \Bbb N\;( m^*(A)\geq$... (et cetera)...). The last inequality holds for $every $ $n\in \Bbb N$
It follows immediately that $m^*(A)\geq$ $ m^*(A\cap E^c)+\sup_{n\in \Bbb N}\sum_{k=1}^nm^*(A\cap E_k)=$ $=m^*(A\cap E^c)+\sum_{k=1}^{\infty}m^*(E_k).$
A: We can say that it is independent of $n$ in the following way, following this reference.
By additivity of finite unions, we have: $m^*(A \cap F_n) = m^*(A \cap E_1) + \ldots m^*(A \cap E_n)$ for arbitrary $n$. 
If we let $n \rightarrow \infty$, then the LHS tends to $m^*(A \cap E)$ and the RHS tends to $\sum_{k=1}^\infty m^*(A \cap E_k)$. Since the finite sum is composed of positive numbers, if we consider the full expression $m^*(A) \geq \sum_{k=1}^n m^*(A \cap E_k) + m^*(A \cap E^C)$, for each $n$, we see that the partial sum is bounded. This means that the infinite sum is bounded, and hence "independent" of $n$.
Finally, we can use the subadditivity of of outer measure on the term $\sum_{k=1}^\infty m^*(A \cap E_k)$ and see that it is larger than the union of the $(A \cap E_k)$ terms, i.e., $\sum_{k=1}^\infty m^*(A \cap E_k) \geq m^* \bigcup_{k=1}^\infty (A \cap E_k)$.
So my takeaway is that "independent" means that we can make this argument for any $n$, and as $n$ approaches infinity, the union becomes $E$, i.e., the leftover "stuff" in $E$ that is not yet included in the union is measure zero (thought I'm not sure if this interpretation is right).
