I am learning measure theory by myself, and I encounter a puzzling proof in the textbook, Measure and Integral by Wheeden and Zygmund.
The theorem (theorem 3.14 in the textbook pg. 37) states that 'every closed set F is measurable'.
In the proof they use two lemmas:
Lemma 3.15: Suppose that $\{I_k\}^N_{k=1}$ is a finite collection of non-overlapping intervals, then $ \bigcup I_k$ is measurable and $|\bigcup I_k|=\sum |I_k|$.
Lemma 3.16: If $d(E_1,E_2)>0$, then $|E_1\cup E_2|_e=|E_1|_e+|E_2|_e$.
Then, the proof goes like this: Choose an open set $G$ s.t. $F\subset G$ and $|G|_e<|F|_e+\epsilon$. $G\backslash F$ is open, thus it can be written as a countable union of non-overlapping intervals. Thus, $G\backslash F=\bigcup_{k=1} ^{\infty} I_k$. Then, $G=F\cup \bigcup_{k=1} ^\infty I_k$. For any $N<\infty$, we must have $F\cup \bigcup_{k=1} ^N I_k\subset (F\cup \bigcup_{k=1} ^\infty I_k)$. Note that by Heine-Borel Theorem, the finite collection of closed and bounded interval, $\bigcup_{k=1} ^N I_k$ is compact. Furthermore, if $E_1$ and $E_2$ are compact and disjoint, $d(E_1,E_2)>0$. Now, note that $F$ and $\bigcup_{k=1} ^N I_k$ are compact and disjoint. Thus, $d(F, \bigcup_{k=1} ^N I_k)>0$. Then, by Lemma 3.16, we must have
$$|F\cup \bigcup_{k=1} ^N I_k|_e= |F|_e+|\bigcup_{k=1} ^N I_k|_e, $$ then by Lemma 3.15, $|\bigcup_{k=1} ^N I_k|_e=|\bigcup_{k=1} ^N I_k|=\sum _{k=1} ^N |I_k|$. Furthermore, by the property of $|\cdot|_e$ and the fact that $(F\cup \bigcup_{k=1} ^N I_k) \subset G$,
$$|F\cup \bigcup_{k=1} ^N I_k|_e= |F|_e+|\bigcup_{k=1} ^N I_k|_e=|F|_e+\sum_{k=1} ^N |I_k|_e\leq |G|_e~~\text{for any $N$}.$$
And, then it proceeds to say that, as for any $N$, the inequality is true, the following must be true too: $$|F|_e+\sum_{k=1} ^{\infty} |I_k|_e\leq |G|_e.$$
This is the part where I got lost. I understand that $|F|_e+\sum_{k=1} ^N |I_k|_e\leq |G|_e$ holds for any $N$, but here $N$ must be finite I believe as we want to have $\bigcup_{k=1} ^N I_k $ be compact (i.e., a collection of closed and bounded intervals must be finite to have it compact). Then, the proof says as $|F|_e+\sum_{k=1} ^N |I_k|_e\leq |G|_e$ is true for any $N$, it must be true for $N$ countably infinite. I am not sure what I am missing here.