Ash's construction of the Lebesgue-Stieltjes Measure from a distribution function

Solution 1:

There is an equivalence between Stieltjes measures on the real line, and Stieltjes measures on the compactified real line that assign measure 0 to the added boundary points at infinity.

For any measure of one type there is a unique measure of the other type that assigns the same values to finite intervals in $R$. This is nothing more than the notation of "improper integrals" from calculus. As in calculus it is a notational convenience used to avoid constantly writing about limits of integrals. The proof could be written or read in terms that avoid any extension of the space or the measure, just as any calculation with improper integrals can be presented as a limit of calculations on finite intervals.

Solution 2:

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 $\mu(\cup A_n) = \sum \mu(A_n)$