Proof that a closed set is measurable in "Measure and Integral" by Wheeden and Zygmund (Theorem 3.14)

Question

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.

Answer 1

I remember this type of argument in lectures being passed over without comment, or maybe with a comment like "take the limit of both sides" or "take the supremum of both sides", and having to sort this out myself.

Stripping away all the clutter that exists in the present context, we arrive at the following conjecture/lemma:

Lemma: Let $L$ be a nonnegative extended real number and let $\{b_N\}$ be a nondecreasing sequence of nonnegative extended real numbers, and assume that for each positive integer $N$ we have $b_N \leq L.$ Then $\lim\limits_{N \rightarrow \infty} b_N \leq L.$

Proof: If $L = +\infty,$ then the desired inequality is automatic. Thus, for the rest of the proof we assume $L$ is a nonnegative real number. For a later contradiction, assume there exists $\epsilon > 0$ such that $\lim\limits_{N \rightarrow \infty} b_N > L + \epsilon.$ Using the epsilon-N definition of "limit of a sequence" and the fact that the sequence is nondecreasing, it follows that there exists a positive integer $N'$ such that $b_{N'} \geq L + \frac{1}{2}\epsilon.$ (In fact, each term of some tail of the sequence will be $\geq L + \frac{1}{2}\epsilon$, but to get a contradiction we only need a single such term.) Now observe that the inequality involving $b_{N'}$ contradicts the assumption "for each positive integer $N$ we have $b_N \leq L$".

To apply this to your situation, let $b_N = |F|_e+\sum_\limits{k=1} ^N |I_k|_e$ and let $L = |G|_e.$

Incidentally, your phrase "must be true for $N$ countably infinite" is conceptually flawed and probably contributed to your difficulties. Most of the later parts of the Wheeden and Zygmund argument only involve sequences of numbers and limits of sequences of numbers, where the notion "countably infinite" doesn't apply in this way to $N \rightarrow \infty.$