I am recently reading rudin's Principles of Mathematical Analysis, and I am wondering why the monotonicity of $\alpha$ in Riemann Stieltjes integral $\int fd\alpha$ is always emphasized.
For example, if I want to integrate $\int^{a}_{b}{dx^2}$ (here $f=1$ and $\alpha = x^2$), it equals to $\int^{a}_{b}2xdx$ where $\alpha =x^2$ is not monotone. However, in theorem 6.17, it assumes that $\alpha$ is nondecreasing.
In most theorem and examples I have encountered, the monotonacity of $\alpha$ is unnecessary. So is there any example to demonstrate the importance of that?
Here are some theorem stated in Rudin's book:
Theorem 6.17
Assume $\alpha(x)$ increases monotonically and $\alpha'$ is integrable (with respect to $x$) on $[a,b]$. Let $f$ be a bounded real function on $[a,b]$.
Then $f$ is $\alpha$-integrable (that is, $\int fd\alpha$ exists) if and only if $f\alpha$ is integrable with respect to $x$. In this case: $$\int^b_afd\alpha=\int^b_af\alpha'dx$$
Theorem 6.18
Suppose $\psi$ is a strictly increasing continuous function that maps an interval $[A,B]$ onto $[a,b]$. Suppose $\alpha$ is monotonically increasing on $[a,b]$ and $f$ is $\alpha$-integrable on $[a,b]$. Define $\beta$ and $g$ on $[A,B]$ by
$$\beta(y)=\alpha(\psi(y))$$$$g(y)=f(\psi(y))$$ Then $g$ is $\beta$-integrable and $$\int^B_Agd\beta=\int^b_afd\alpha$$
Best Answer
You're correct that just about all of these theorems, where $\alpha$ is monotone, generalize instantly to functions $\alpha$ that are the difference of two increasing functions. But it's not trivial to look at a crazy function and decide whether it can be written as the difference of two increasing functions. For example, for $j=1,2$ define $$ f_j(x) = \begin{cases}0,&\text{if } x=0, \\ x^j \sin \tfrac1x,&\text{if } x>0.\end{cases} $$ It happens that on the domain $[0,1]$, the function $f_2$ can be written as the difference of two increasing functions, but $f_1$ can't.
The key property of functions of this form is that they are "of bounded variation". This is a somewhat advanced notion, and I'm guessing that this is the reason that Rudin is restricting to a simpler situation for the time being.