I am reading the proof of Theorem 1.1.4 in Durrett Edition 5, states as follows:
Associated with each Stieltjes measure function $F$, there is a unique measure $\mu$ on $(\mathbb{R},\mathcal{R})$ such that $\mu((a,b])=F(b)-F(a).$
Let $\mathcal{S}$ be the semi-algebra containing the empty set and all the half-open intervals $(a,b]$ with $-\infty\leq a<b\leq\infty$.
I understand most of the proof where Durrett tried to verify the assumption of Caratheodory's Extension Theorem to extend this set function $\mu$ to a unique measure on $\mathcal{R}=\sigma(\mathcal{S})$, however I don't understand his statement at the beginning of the proof, which is following:
To define $\mu$ on $\mathcal{S}$, we begin by observing that $$F(\infty)=\lim_{x\nearrow\infty}F(x)\ \text{and}\ F(-\infty)=\lim_{x\searrow-\infty}F(x)\ \text{exist}$$ and $\mu((a,b])=F(b)-F(a)$ makes sense for all $-\infty\leq a<b\leq\infty$ since $F(\infty)>-\infty$ and $F(-\infty)<\infty$.
I have two confusions:
(1) Why does $F(\infty)$ exist? $F$ is right continuous so $F(-\infty)$ exists, but can we take $x\nearrow\infty$?
(2) Why $F(-\infty)<\infty$ and $F(\infty)>-\infty$? I understand that Durrett want this to be good for $a=-\infty$ and $b=\infty$,his proof that since $F(\infty)$ and $F(-\infty)$ exist, then we just have $$\mu((-\infty,\infty])=F(\infty)-F(-\infty)<\infty,$$ so that $\mu$ is a well-defined set function.
However, perhaps $F(-\infty)\leq\infty$? instead of $<$? because $F$ is non-decreasing, instead of strictly increasing? (similarly for $F(\infty)$)
Thank you!
Best Answer
The definition I found, which it would be helpful for you to include in your post next time, is $F:\mathbb{R}\to \mathbb{R}$ is a Stieltjes measure function if it is nondecreasing and right continuous.
For both, you are missing the nondecreasing assumption, which is crucial.
1) The function is nondecreasing, so the limit exists if we allow it to be in the set $\mathbb{R}\cup \{ +\infty\}$.
2) Being nondecreasing gives both again, if $F(-\infty)=+\infty$, then $F\equiv +\infty$, and doesn't take values in $\mathbb{R}$. Similar logic for at $F(+\infty)$.