I am re-reading an old textbook "Introduction to Hilbert spaces and applications" by Lokenath Debnath and Piotr Mikusinski, and there is a proof of a lemma in a chapter about the Lebesgue integral that I just can't rap my head around. It's this one:
Lemma 2.2.2. Let $[a_1,b_1),[a_2,b_2),...$ be disjoint subintervals of an interval $[a,b)$ such that $$\bigcup_{n=1}^\infty[a_n,b_n)=[a,b). \tag{2.2.2}$$ Then $$\sum_{n=1}^\infty(b_n-a_n)=b-a. \tag{2.2.3}$$
Proof. Let $S\subset[a,b)$ consist of all points $c$ such that the lemma holds for the interval $[a,c)$ and the sequence of subintervals $[a_n,b_n)\cap[a,c)$. Therefore, if $c\in S$, then $$c-a=\sum_n(b_{c,n}-a_n),$$ where $b_{c,n}=\min\{b_n,c\}$ and the summation is over all those $n$ for which $a_n \lt b_{c,n}$. It suffices to prove that $b \in S$. To this end we first prove that $\mathbf{LUB}\, S \in S$. Indeed, if $s=\mathbf{LUB}\, S$ and $\{s_n\}$ is an non-decreasing sequence of elelments of $S$ convergent to $s$, then $$s_n-a=\sum_m(b_{s_n,m}-a_m) \le \sum_m{(b_{s,m}-a_m)} \le s-a. \tag{2.2.4}$$ Since $s_n-a \to s-a$, (2.2.4) implies $$\sum_m{(b_{s,m}-a_m)}=s-a,$$ and consequently $s \in S$. Next we show that $s=b$. Suppose $s \lt b$. Then $$s \in [a_k,b_k) \, \text{ for some } k \in \mathbf{N},$$ and thus $b_k \in S$. Since this contradicts the definition of $s$, we conclude $s=b$. The proof is now complete.
My problem with the proof is the idea that the end point of an open interval can belong to the interval. How can $b \in S$ if $S \subset [a,b)$?
In a newer edition of the book, the proof has been rewritten. In effect, $S \subset (a,b]$ or $S \subseteq (a,b]$ has replaced the old definition. Sorry for wasting people's time.