This question arose from the proof of theorem 1.1.4 in Durrett 5th Edition.
Firstly, measure on $(\mathbb{R},\mathcal{R})$are defined by giving a Stieltjes measure function with the following properties:
(1) F is non-decreasing;
(2) F is right continuous.
Theorem 1.1.4: Associated with each Stieltjes measure function $F$, there is a unique measure $\mu$ on $(\mathbb{R},\mathcal{R})$ with $$\mu((a,b])=F(b)-F(a).$$
I am okay with most part of the proof but I am stuck in the end. Denote $\bigsqcup$ to be the disjoint union.
The proof tries to use Caratheodory's Extension Theorem. That is, he defines the set function $\mu$ as above on the semi-algebra $\mathcal{S}$ which contains empty set and all the half open intervals $(a,b]$ with $-\infty\leq a<b\leq\infty$, and he tries to verify
(1) For $S, S_{i}\in \mathcal{S}$, $S=\bigsqcup_{i=1}^{n}S_{i}$, then $\mu(S)=\sum_{i=1}^{n}\mu(S_{i});$
(2) For $S, S_{i}\in\mathcal{S}$, $S=\bigsqcup_{i=1}^{\infty}S_{i}$, then $\mu(S)\leq\sum_{i=1}^{\infty}\mu(S_{i});$
(3) The extended $\overline{\mu}$ is $\sigma$-finite.
Part (1) and part (3) are really quick to show, and for (2), it suffices to verify for the case of $(a,b]\subset\sqcup_{i=1}^{\infty}(a_{i}, b_{i}]$, because then for all $S, S_{i}\in\mathcal{S}$ with $S=\bigsqcup_{i=1}^{\infty}S_{i}$, we have $S\subset\bigsqcup_{i=1}^{\infty}S_{i}$ and then result follows.
Then Durrett firstly assume $-\infty<a<b<\infty$ such that $(a,b]\subset\sqcup_{i=1}^{\infty}(a_{i}, b_{i}]$ where $-\infty<a_{i}<b_{i}<\infty$, and he showed that $$F(b)-F(a)\leq \sum_{i=1}^{\infty}(F(b_{i})-F(a_{i})),$$ as desired.
Then to go to the general case where $-\infty\leq a<b\leq \infty$, he argues in a confusing way:
(Durrett argument) To remove the last restriction, observe that if $(a,b]\subset\bigcup_{i}(a_{i},b_{i}]$ and $(A,B]\subset(a,b]$ has $-\infty<A<B<\infty$, then we have $$F(B)-F(A)\leq\sum_{i=1}^{\infty}(F(b_{i})-F(a_{i})),$$ since this result holds for any finite $(A,B]\subset(a,b]$ then the desired result follows.
Well, I agree with $$F(B)-F(A)\leq\sum_{i=1}^{\infty}(F(b_{i})-F(a_{i})),$$ since this is just the first case, but then why does the result follow? Can I generate an infinite interval with finitely many finite interval?
Thank you!
Best Answer
For instance, if $a$ is finite and $b = +\infty$, then take $A = a$ and let $B > A$ be arbitrary. Since $F$ is nondecreasing we know that $\lim_{B \to +\infty} F(B)$ exists (it could be $+\infty$), and the value of this limit is what we mean by $F(+\infty)$. Now we have $F(B) - F(A) \le \sum_{i=1}^\infty (F(b_i) - F(a_i))$ for all $B > A$, so we can pass to the limit to see that $$F(+\infty) - F(A) = \left(\lim_{B \to +\infty} F(B)\right) - F(A) = \lim_{B \to +\infty} [F(B) - F(A)] \le \sum_{i=1}^\infty (F(b_i) - F(a_i))$$ That is the desired statement. You can argue similarly if $a = -\infty$ and $b$ is finite, and if $a=-\infty, b = +\infty$.