Baby Rudin Theorem 6.16 Explanation I am trying to work through rudin and I am very confused with 6.16.  An explanation of what it is saying and working through the steps would be immensely helpful. For reference the actual proof is stated on page 139 at    this link:  proof.    Thanks!
It is stated as such:
6.16 Theorem: 
Suppose $c_n \geq $ 0 for n = 1, 2, 3, ...., $\sum c_n $ converges, $\{s_n\}$ is a sequence of distinct points in (a,b), and $\alpha(x) = \sum_{n=1}^{\infty} {c_n} I(x - s_n)$.  Let f be continuous on [a, b]. Then
$$\int_{a}^{b} f\,d\alpha =  \sum_{n=1}^{\infty} {c_n} f(s_n).$$
 A: I think the motivation is given in Remark 6.18, which is that the Riemann Stieltjes integral can be used to treat series as well.
If $\alpha$ is a step function (that is, the sum of multiples of shifted unit functions), then instead of writing $\sum_n c_n f(x_n)$, we can
write $\int f d \alpha$, where $\alpha = \sum_n c_n 1_{[x_n,\infty)}$, the point being to unify the treatment.
A further motivation might be to show that integration with respect to a monotonic, right continuous $\alpha$ can be split into a continuous $\alpha_1$ and a 'step' function $\alpha_2$ (the latter being the content of Theorem 6.16). In this case we let $\alpha_2(x) = \lim_{y \downarrow x} \alpha(x) - \lim_{y \uparrow x} \alpha(x)$, and $\alpha_1(x) = \alpha(x)-\alpha_2(x)$. A little work shows that $\alpha_2$ is zero except at a possibly
countable set of points, and the sum of the discontinuities is bounded. Furthermore, $\alpha_1$ is continuous.
If $\alpha = \sum_n c_n 1_{[x_n,\infty)}$, where the sum is finite, then Theorem 6.12(e) shows that $\int f d \alpha = \sum_n c_n f(x_n)$, Theorem 6.16 extends this to an infinite sum using Theorem 6.12(d), which shows that the 'tail' of the integration becomes arbitrarily small.
The 'meat' of the argument is Theorem 6.12(d,e).
