1.15 Proposition.(page 33) Let $F:\mathbb{R} \to \mathbb{R}$ be increasing and right continuous. If $(a_j,b_j]$ $(j=1,\dots,n)$ are disjoint $h-$intervals, let
$$\mu_0\bigg(\bigcup_{j=1}^n (a_j,b_j] \bigg)=\sum_{j=1}^n [ F(b_j)-F(a_j)]$$
and let $\mu_0(\emptyset)=0$. Then $\mu_0$ is a premeasure on the algebra $\mathcal{A}$.
Note: the algebra $\mathcal{A}$ is the the collection of finite disjoint unions of $h-$intervals, defined to be any set of the form $(a,b]$ or $(a,\infty)$ or $\emptyset$, where $-\infty\leq a<b<\infty$.
At the end of the proof of Proposition 1.15 on page 34, Folland uses a compactness argument to show that $\mu_0$ is countably additive in the case of an interval $I=(a,b]$ where both $a$ and $b$ are finite. He then mentions that the same reasoning can be applied when $a=-\infty$ or to $b=\infty$ to get
(i) $F(b)-F(-M)\leq\sum_{j=1}^{\infty}\mu_0(I_j)+2\epsilon$
(ii) $F(M)-F(a)\leq\sum_{j=1}^{\infty}\mu_0(I_j)+2\epsilon$
respectively, where $0<M<\infty$ is arbitrary. I have three questions:
In the case $a=-\infty$ I think that we actually have $F(b)-F(-M)\leq\sum_{j=1}^{\infty}\mu_0(I_j)+\epsilon$ since in that case we don't have to find a $\delta<0$ such that $F(a+\delta)-F(a)<\epsilon$. Am I correct?
In the case $b=\infty$ we might have $b_j=\infty$ for some $j$. In that case do we define $\delta_j=0$ and proceed with the same argument?
What about the case where both $a=-\infty$ and $b=\infty$ ? Is there a reason why Folland doesn't mention this case?
Thanks a lot for your help. My goal is to redo the proof as in the book in each of the 3 cases.