No, the Lebesgue integral is not more general than the improper Riemann one, it just has some very nice properties that make it convenient to work with. Remember that, once you define the concept of Lebesgue integrability, an important theorem says that $f$ is Lebesgue integrable if and only if $|f|$ is so. Consider now the function $\Bbb e^{\Bbb i x^2}$: its modulus is $1$, which is clearly not integrable on $\Bbb R$; nevertheless, its improper Riemann integral exists as $\lim \limits _{R \to \infty} \int \limits _{-R} ^R \Bbb e^{\Bbb i x^2} \Bbb d x = \sqrt {\pi \Bbb i}$, so you may still assign a value to it.
As you can see, there are moments when the "humbler" improper Riemann integral is capable of producing better results than the Lebesgue one. Let us see why and when. When mathematicians use the Lebesgue integral, they usually do so in order to use the already established (and very powerful) theory of Lebesgue spaces, which are Banach spaces. Being Banach spaces, we usually use various inequalities regarding their norms; nevertheless, most of our approaches rely on the following starting point: $| \int f | \le \int |f|$ (or something similar). If you think of this, and of the example in the above paragraph, you will see that there is a class of functions (those that are not absolutely integrable) for which these techniques will not produce useful results.
(If you think further about this issue, Lebesgue integrability is like absolute convergence for series: it works well and produces powerful results for many series, but what do you do with the following example: $\sum \limits _n (-1)^n \dfrac {\Bbb e ^{\Bbb i n x}} n$? Absolute integrability will betray you in this case, you'll have to resort to the Abel-Dirichlet test that will confirm the convergence of the above.)
To summarize, for absolutely $p$-integrable functions the Lebesgue integral works best, no need to use anything else; for non-absolutely $p$-integrable functions, you'll have to try alternative approaches, such as improper Riemann integrability.
Of course, the above makes sense on spaces of infinite measure. If your space is a compact subset of $\Bbb R^n$ with the usual measure and $f$ is a Riemann integrable function, then it will also be Lebesgue integrable and its two integrals will coincide, which is very nice if you think about it (the advantage of Riemann integrability on compact spaces being that we know how to explicitly compute things, essentially repeatedly simplifying our problem until we may use the Leibniz-Newton theorem).
Best Answer
Pollard's paper appears to be where the generalization was introduced. The paper begins on page 73 and that page has only the title, the author's name and affiliation, and the references. The link above is to page 74.
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."
But I haven't carefully sifted through all the details.