In Royden's book, he gives the following book for the proof that Lebesgue measure is countable additive. I will just give a sketch of the proof.
Let $\{E_k \}_{k=1}^{\infty}$ be a collection of disjoint measurable sets. We only need to show that $$ m(\cup_{k=1}^{\infty} E_k) \geq \sum_{k=1}^{\infty} m(E_k) $$ since it follows by countable subadditivtiy that $$ m(\cup_{k=1}^{\infty} E_k) \leq \sum_{k=1}^{\infty} m(E_k) $$ For any finite subcollection of $\{E_k \}_{k=1}^{\infty}$, it follows by monotonicty that $$ m(\cup_{k=1}^{\infty} E_k) \geq \sum_{k=1}^{n} m(E_k) $$ for each n. What confuses me is the next statement: "Since the left hand side of the inequality is independent of n it follows that"
$$ m(\cup_{k=1}^{\infty} E_k) \geq \sum_{k=1}^{\infty} m(E_k) $$
as desired. Why is it true that the left hand side of this inequality is independent of n?
Here is an example which may help elucidate the proof: consider the partial sums
$$
\sum_{k=1}^n \frac{1}{2^k} < 1
$$
If we let $n\to \infty$ then we only have
$$
\sum_{k=1}^n \frac{1}{2^k} \leq 1
$$
Showing that the limit may have equality.
Another useful observation is to note that if there is a finite $n$ where the finite sum is equal to the measure of the union of these measurable sets, then the rest of measurable sets must have measure 0, so the equality holds in the limit.
Best Answer
On the left hand side, you're taking the measure of the union of all $E_k$ ($k = 1,\ldots,\infty$), which is not defined in terms of $n$.
What he's saying is that since $$ m(\cup_{k=1}^\infty E_k) \geq \sum_{i=1}^n E_i $$ for all values $n$ you can let $n \to \infty$ to get the final inequality.