TITLE: Ash's construction of the Lebesgue-Stieltjes Measure from a distribution function
QUESTION [11 upvotes]: I'm reading this book Probability & Measure Theory by Ash. I think I've come across a part that is a little hand-wavy. We are trying to build a Lebesgue-Stieltjes measure from a distribution function $F$ (in that the measure of interval $(a,b]$ is $F(b) - F(a)$).
He starts by adding $+\infty$ and $-\infty$ to the real line so that we can work in compact space. He defines right-semiclosed as intervals of the form $(a, b]$ and $[-\infty, b]$ and $(-\infty, b]$. He then constructs a field  by taking all finite unions of these right-semiclosed intervals.
He defines a set function over this field defined in the intuitive way (the set function takes $(a,b]$ to $F(b) - F(a)$), and he shows that this set function is countably additive.
This is where I don't understand his argument. He seems to say, ignore these points $+\infty$ and $-\infty$ so that our field no longer uses the compact space, and our set function now becomes a proper measure over a real field. Then apply the Carathéodory Extension Theorem.  
I don't see how we can go from a compact space to a non-compact space without causing harm to the properties of our set function. I'm hoping that this construction method is widely used, and someone can explain where I am confused. This is Theorem 1.4.4 in Ash, 2nd Edition.
The complete exposition can be found at http://books.google.com/books?id=TKLl3CGqsTEC&lpg=PP1&dq=probability%20and%20measure%20theory&pg=PA22#v=onepage&q&f=false from the bottom of page 22 to page 24.

REPLY [2 votes]: zyx's answer has helped me understand the proof.
Let $F: \mathbb{R}\rightarrow\mathbb{R} $ be a distribution function. We can form the function $F_C:\overline{\mathbb{R}}\rightarrow\overline{\mathbb{R}} $ by setting 
$$F_C(x) \equiv F(x), x\in\mathbb{R}$$
$$F_C(\pm\infty) \equiv \lim_{x\rightarrow \pm\infty} F(x)$$
Then we can define set function 
$$\mu_C(a,b] \equiv F_C(b)-F_C(a), a\in\overline{\mathbb{R}}, b\in \overline{\mathbb{R}}$$
$$\mu_C[-\infty, b] \equiv \mu_C(-\infty, b], b \in \overline{\mathbb{R}}$$
We can prove $\mu_C$ is countably additive on the field of finite unions of disjoint right closed intervals in $\overline{\mathbb{R}}$.  
Now we define a set function $\mu$ on the field of finite unions of disjoint right closed intervals in $\mathbb{R}$:
$$\mu(a,b] \equiv \mu_C(a,b], a\in\mathbb{R}, b\in \mathbb{R}$$
$$\mu(-\infty, b] \equiv \mu_C(-\infty, b], b \in \mathbb{R}$$
$$\mu(a, \infty) \equiv \mu_C(a, \infty], a \in \mathbb{R}$$
$$\mu(-\infty, \infty) \equiv \mu_C[\infty, \infty], a \in \mathbb{R}$$
Now suppose we have a sequence of disjoint sets $A_n$ in the field on $\mathbb{R}$, and $\cup A_n = A$ and $A$ is in the field on $\mathbb{R}$.
$$\mu(A) = \mu(\cup A_n)$$
$$\mu(\cup A_n) = \mu(A_0 \cup A_1 \cup (\cup_{n =2} ^{\infty} A_n))$$
where $A_0$ and $A_1$ are renumbered to be the unbounded sets $(a, \infty)$ and $(-\infty, b)$ if they exist. Otherwise we can take them to be the empty set.
$$\mu(A_0 \cup A_1 \cup (\cup_{n =2} ^{\infty} A_n)) = \mu(A_0) + \mu(A_1) + \mu (\cup_{n =2} ^{\infty} A_n)$$
by finite additivity.
$$\mu(A_0) + \mu(A_1) + \mu (\cup_{n =2} ^{\infty} A_n) = \mu(A_0) + \mu(A_1) + \mu_C (\cup_{n =2} ^{\infty} A_n)$$
because the $\cup_{n =2} ^{\infty} A_n$ results in finitely many disjoint interals, all of them finite. 
$$\mu(A_0) + \mu(A_1) + \mu_C (\cup_{n =2} ^{\infty} A_n) = \mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu_C (A_n)$$
by countable additivity of $\mu_C$
$$\mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu_C (A_n) = \mu(A_0) + \mu(A_1) + \sum _{n =2} ^{\infty} \mu (A_n)$$
$$= \sum \mu(A_n)$$
So the proof is done because $\mu(\cup A_n) = \sum \mu(A_n)$