Even with the Riemann Integral, we do not usually use the definition (as a limit of Riemann sums, or by verifying that the limit of the upper sums and the lower sums both exist and are equal) to compute integrals. Instead, we use the Fundamental Theorem of Calculus, or theorems about convergence. The following are taken from Frank E. Burk's A Garden of Integrals, which I recommend. One can use these theorems to compute integrals without having to go down all the way to the definition (when they are applicable).
Theorem (Theorem 3.8.1 in AGoI; Convergence for Riemann Integrable Functions) If $\{f_k\}$ is a sequence of Riemann integrable functions converging uniformly to the function $f$ on $[a,b]$, then $f$ is Riemann integrable on $[a,b]$ and
$$R\int_a^b f(x)\,dx = \lim_{k\to\infty}R\int_a^b f_k(x)\,dx$$
(where "$R\int_a^b f(x)\,dx$" means "the Riemann integral of $f(x)$").
Theorem (Theorem 3.7.1 in AGoI; Fundamental Theorem of Calculus for the Riemann Integral) If $F$ is a differentiable function on $[a,b]$, and $F'$ is bounded and continuous almost everywhere on $[a,b]$, then:
- $F'$ is Riemann-integrable on $[a,b]$, and
- $\displaystyle R\int_a^x F'(t)\,dt = F(x) - F(a)$ for each $x\in [a,b]$.
Likewise, for Riemann-Stieltjes, we don't usually go by the definition; instead we try, as far as possible, to use theorems that tell us how to evaluate them. For example:
Theorem (Theorem 4.3.1 in AGoI) Suppose $f$ is continuous and $\phi$ is differentiable, with $\phi'$ being Riemann integrable on $[a,b]$. Then the Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists, and
$$\text{R-S}\int_a^b f(x)d\phi(x) = R\int_a^b f(x)\phi'(x)\,dx$$
where $\text{R-S}\int_a^bf(x)d\phi(x)$ is the Riemann-Stieltjes integral of $f$ with respect to $d\phi(x)$.
Theorem (Theorem 4.3.2 in AGoI) Suppose $f$ and $\phi$ are bounded functions with no common discontinuities on the interval $[a,b]$, and that the Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists. Then the Riemann-Stieltjes integral of $\phi$ with respect to $f$ exists, and
$$\text{R-S}\int_a^b \phi(x)df(x) = f(b)\phi(b) - f(a)\phi(a) - \text{R-S}\int_a^bf(x)d\phi(x).$$
Theorem. (Theorem 4.4.1 in AGoI; FTC for Riemann-Stieltjes Integrals) If $f$ is continuous on $[a,b]$ and $\phi$ is monotone increasing on $[a,b]$, then $$\displaystyle \text{R-S}\int_a^b f(x)d\phi(x)$$
exists. Defining a function $F$ on $[a,b]$ by
$$F(x) =\text{R-S}\int_a^x f(t)d\phi(t),$$
then
- $F$ is continuous at any point where $\phi$ is continuous; and
- $F$ is differentiable at each point where $\phi$ is differentiable (almost everywhere), and at such points $F'=f\phi'$.
Theorem. (Theorem 4.6.1 in AGoI; Convergence Theorem for the Riemann-Stieltjes integral.) Suppose $\{f_k\}$ is a sequence of continuous functions converging uniformly to $f$ on $[a,b]$ and that $\phi$ is monotone increasing on $[a,b]$. Then
The Riemann-Stieltjes integral of $f_k$ with respect to $\phi$ exists for all $k$; and
The Riemann-Stieltjes integral of $f$ with respect to $\phi$ exists; and
$\displaystyle \text{R-S}\int_a^b f(x)d\phi(x) = \lim_{k\to\infty} \text{R-S}\int_a^b f_k(x)d\phi(x)$.
One reason why one often restricts the Riemann-Stieltjes integral to $\phi$ of bounded variation is that every function of bounded variation is the difference of two monotone increasing functions, so we can apply theorems like the above when $\phi$ is of bounded variation.
For the Lebesgue integral, there are a lot of "convergence" theorems: theorems that relate the integral of a limit of functions with the limit of the integrals; these are very useful to compute integrals. Among them:
Theorem (Theorem 6.3.2 in AGoI) If $\{f_k\}$ is a monotone increasing sequence of nonnegative measurable functions converging pointwise to the function $f$ on $[a,b]$, then the Lebesgue integral of $f$ exists and
$$L\int_a^b fd\mu = \lim_{k\to\infty} L\int_a^b f_kd\mu.$$
Theorem (Lebesgue's Dominated Convergence Theorem; Theorem 6.3.3 in AGoI) Suppose $\{f_k\}$ is a sequence of Lebesgue integrable functions ($f_k$ measurable and $L\int_a^b|f_k|d\mu\lt\infty$ for all $k$) converging pointwise almost everywhere to $f$ on $[a,b]$. Let $g$ be a Lebesgue integrable function such that $|f_k|\leq g$ on $[a,b]$ for all $k$. Then $f$ is Lebesgue integrable on $[a,b]$ and
$$L\int_a^b fd\mu = \lim_{k\to\infty} L\int_a^b f_kd\mu.$$
Theorem (Theorem 6.4.2 in AGoI) If $F$ is a differentiable function, and the derivative $F'$ is bounded on the interval $[a,b]$, then $F'$ is Lebesgue integrable on $[a,b]$ and
$$L\int_a^x F'd\mu = F(x) - F(a)$$
for all $x$ in $[a,b]$.
Theorem (Theorem 6.4.3 in AGoI) If $F$ is absolutely continuous on $[a,b]$, then $F'$ is Lebesgue integrable and
$$L\int_a^x F'd\mu = F(x) - F(a),\qquad\text{for }x\text{ in }[a,b].$$
Theorem (Theorem 6.4.4 in AGoI) If $f$ is continuous and $\phi$ is absolutely continuous on an interval $[a,b]$, then the Riemann-Stieltjes integral of $f$ with respect to $\phi$ is the Lebesgue integral of $f\phi'$ on $[a,b]$:
$$\text{R-S}\int_a^b f(x)d\phi(x) = L\int_a^b f\phi'd\mu.$$
For Lebesgue-Stieltjes Integrals, you also have an FTC:
Theorem. (Theorem 7.7.1 in AGoI; FTC for Lebesgue-Stieltjes Integrals) If $g$ is a Lebesgue measurable function on $R$, $f$ is a nonnegative Lebesgue integrable function on $\mathbb{R}$, and $F(x) = L\int_{-\infty}^xd\mu$, then
- $F$ is bounded, monotone increasing, absolutely continuous, and differentiable almost everywhere with $F' = f$ almost everywhere;
- There is a Lebesgue-Stieltjes measure $\mu_f$ so that, for any Lebesgue measurable set $E$, $\mu_f(E) = L\int_E fd\mu$, and $\mu_f$ is absolutely continuous with respect to Lebesgue measure.
- $\displaystyle \text{L-S}\int_{\mathbb{R}} gd\mu_f = L\int_{\mathbb{R}}gfd\mu = L\int_{\mathbb{R}} gF'd\mu$.
The Henstock-Kurzweil integral likewise has monotone convergence theorems (if $\{f_k\}$ is a monotone sequence of H-K integrable functions that converge pointwise to $f$, then $f$ is H-K integrable if and only if the integrals of the $f_k$ are bounded, and in that case the integral of the limit equals the limit of the integrals); a dominated convergence theorem (very similar to Lebesgue's dominated convergence); an FTC that says that if $F$ is differentiable on $[a,b]$, then $F'$ is H-K integrable and
$$\text{H-K}\int_a^x F'(t)dt = F(x) - F(a);$$
(this holds if $F$ is continuous on $[a,b]$ and has at most countably many exceptional points on $[a,b]$ as well); and a "2nd FTC" theorem.
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
Firstly, as Berci mentions in the comments, Riemann-Stieltjes integrals are defined on closed intervals. Often, Riemann integrals are defined on closed intervals too, but are extended to open or half-open intervals with limits.
I'm going to assume that when you write about $f$ being integrable wrt $g$ on $(a,b]$ and it being the same as $f$ wrt $g$ on $[a,b]$, you mean that $\displaystyle \lim_{\delta \to 0} \int_{a+ \delta}^b fdg$ exists and is equal to $\displaystyle \int_a^b fdg$.
But as you mention, this does not need to be true, even if $\displaystyle \int_a^b fdg$ exists. For example, consider $g(x) = 1$ if $x = 0$ and $0$ otherwise, and let $f$ be a function which is $1$ in a neighborhood $0$, say $[-\alpha,\alpha]$ for some small $\alpha$, of and $0$ everywhere else. Then $\displaystyle \int_0^1 fdg = 1$. But for any $\epsilon > 0$, $\displaystyle \int_\epsilon^1 fdg = 0$ for the exact reason you stated - we don't capture the first endpoint. And if $g$ is discontinuous there, that endpoint really matters. Similarly, if $g$ is terribly discontinuous near $0$, then the integral on $(a,b]$ will have many problems.
But if you have limitation on $g$, like saying that $g$ is continuous (or weaker, of bounded variation), then the limit will exist. But still, the limit on $(a,b]$ does not need to equal the integral on $[a,b]$ even with bounded variation, because of the missing endpoint. (I note that bounded-variation also usually is defined on closed intervals, so there is a little bit of fuzzyness to my argument).