A question on the definition of the Riemann–Stieltjes integral

I have been reading Wikipedia's article on the Riemann–Stieltjes integral (https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral) and I don't understand why the Riemann–Stieltjes integral isn't equaivalent to the Generalized Riemann–Stieltjes integral. I mean, if instead of $$g$$ we take the identity function (i.e. we consider the case of the Riemann integral) then they are the same. Why does this change for other $$g$$'s? I am quite confused because my textbook uses the definition Generalized Riemann–Stieltjes integral as the definition for the Riemann–Stieltjes integral and this is why I want to understand the difference between them.
EDIT: I would also like to know which properties of the generalized integral transfer to the ungeneralized one and which do not to fully grasp them.

• ${}\,+1.$ One who edits Wikipedia articles should understand the way I edited the punctuation in this question. Sep 6 '20 at 13:31
• Pollard, Henry (1920). "The Stieltjes integral and its generalizations". The Quarterly Journal of Pure and Applied Mathematics. 49. babel.hathitrust.org/cgi/… Sep 6 '20 at 13:53
• Pollard writes: "The integral so obtained exists and is identical with the integral previously defined whenever this exists, and exists in certain cases where this does not." I don't yet know what those "certain cases" are. $\qquad$ Sep 6 '20 at 13:55

On page 80, Pollard defines two functions: $$f(x) = \begin{cases} 0 & x < 1, \\ k & x\ge 1, \end{cases}$$ $$\varphi(x) = \begin{cases} 0 & x\le 1, \\ 1 & x>1. \end{cases}$$ He seems to claim that as a generalized Riemann–Stieltjes integral, apparently defined as a Moore–Smith limit of a net indexed by partitions of the interval $$[0,2],$$ the integral $$\int_0^2 f(x)\,d\varphi(x)$$ exists and is equal to $$k(\varphi(2) - \varphi(1)),$$ but that as a Riemann–Stieltjes integral, defined as a limit as the mesh of the partition approaches $$0,$$ that integral does not exist. Pollard calls the the generalized Riemann–Stieltjes integral "the modified Stieltjes integral."
• If the partition does not contain the point $1$, then by choosing the corresponding $c$ either $< 1$ or $\geqslant 1$ you can make the sum $0$ or $k$. But if the partition contains the point $1$, then for the $[x_{k-1},1]$ interval you always have $f(c)\cdot (0 - 0)$ and for the $[1, x_{k+1}]$ interval you always have $k\cdot(1-0)$. Sep 6 '20 at 14:30
• I seem to recall Baby Rudin saying the Riemann–Stieltjes integral is undefined in cases where $f$ and $\varphi$ have a point of discontinuity in common. So that's where the "generalized" version goes beyond the ungeneralized one. Sep 6 '20 at 14:50
• @JustAnAmateur : I am inclined to doubt that any useful properties of the generalized version fail to apply to the version defined by Stieltjes except that the one defined by Stieltjes fails to be defined when $f$ and $\varphi$ have a common discontinuity. The paper by Pollard might be the first place I'd look for an answer to that. Sep 6 '20 at 17:13