Throughout this post I refer to these as "Darboux integrals/integration" and to these as "Riemann integrals/intnegration". It is a standard result of Analysis that these notions of integrability coincide.
I've tried (and failed) looking for a book that introduces integration in terms of Riemann integrals: Rudin's POMA, Tao's Analysis I, Spivak's Calculus and Abbott's Understanding Analysis all use Darboux integration. Wikipedia explains the reason in its article on Darboux integration:
"The definition of the Darboux integral has the advantage of being easier to apply in computations or proofs than that of the Riemann integral. Consequently, introductory textbooks on calculus and real analysis often develop Riemann integration using the Darboux integral, rather than the true Riemann integral."
Itching my curiosity I've shown that some elementary theorems on integration can be proven directly using Riemann integrability (such as that $\int(f+g)=\int f + \int g$, as well as $c\int f = \int cf$). Currently I'm stuck trying to prove Rudin's 6.17 and hence I'm asking for a hand.
Theorem 6.17: Assume $\alpha$ increases monotonically on $[a,b]$ and $\alpha' \in \mathcal{R}$ (i.e. $\alpha'$ is Darboux integrable). Let $f$ be a bounded real function on $[a,b]$. Then $f\in \mathcal{R}(\alpha)$ if and only if $f\alpha'\in\mathcal{R}$. In that case
$$\int_a^bf\ d\alpha = \int_a^bf(x)\alpha'(x)\ dx.$$
Can 6.17 be shown by using Riemann integration?
To be clearer, showing the result by directly employing the criterion in the definition of Riemann integrability, as opposed to using other (equivalent) criteria.
There would be no point in using the criterion on the definition of Darboux integrability, nor to use, as Rudin did, the fact that a function is Riemann integrable if and only if for any $\epsilon>0$, there is some partition $P$ such that
$$\left|U(f,\alpha,P)-L(f,\alpha,P)\right|<\epsilon.$$
Best Answer
Here is a proof (without refering to Darboux integrals) of the reverse implication of 6.17 that if $f \alpha'$ is Riemann integrable on $[a,b]$ then $f$ is Riemann-Stieltjes integrable with respect to the integrator $\alpha$ on $[a,b]$ and
$$\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x)\, dx$$
The objective is to show that for any $\epsilon >0$, there exists a partition $P_\epsilon$ such that for any refinement
$$P = \{x_0,x_1,\ldots,x_n:a = x_0 < x_1 < \ldots <x_n = b\}$$
and any choice of intermediate points $\xi_j \in [x_{j-1},x_j]$, we have
$$\left|S(P,f,\alpha,\{\xi_j\})- \int_a^bf(x) \alpha'(x) \, dx\right|= \left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|< \epsilon$$
By the mean value theorem, there exists $\eta_j \in (x_{j-1},x_j)$ for $j=1,2,\ldots,n$ such that
$$\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]= \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j) (x_j - x_{j-1})$$
Thus,
$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|\\= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx + \sum_{j=1}^nf(\xi_j)\alpha'(\eta_j)(x_j-x_{j-1})- \sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})\right|\\ \leqslant \underbrace{\left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|}_{A}+ \underbrace{\sum_{j=1}^n |f(\xi_j)||\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})}_{B}$$ Because $f\alpha'$ is Riemann integrable it follows that there exists a partition $P_A$ such that if $P$ is a refinement, then $$\tag{1}A= \left|\sum_{j=1}^nf(\xi_j)\alpha'(\xi_j)(x_j-x_{j-1})- \int_a^bf(x) \alpha'(x) \, dx\right|< \frac{\epsilon}{2}$$
Since $f$ is bounded, there exists $M>0$ such that $|f(x)| < M$ for all $x \in [a,b]$ and, hence
$$B = \sum_{j=1}^n |f(\xi_j)|[\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < M \sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) $$
It will be shown below that because $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then
$$\tag{2}\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})<\frac{\epsilon}{2M},$$
whence,
$$\tag{3}B < M\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1}) < \frac{\epsilon}{2}$$
Hence, if P is a refinement of the common refinement $P_\epsilon = P_A \cup P_B$, then by (1) and (3) we have
$$\left|\sum_{j=1}^nf(\xi_j)[\alpha(x_j) - \alpha(x_{j-1})]- \int_a^bf(x) \alpha'(x) \, dx\right|=A+B < \epsilon,$$
and we have proved that $\int_a^b f \, d\alpha = \int_a^b f(x) \alpha'(x) \, dx$ as was to be shown.
The forward implication can be proved in a similar way.
Proof of (2)
Since $\alpha'$ is Riemann integrable, there exists a partition $P_B$ such that if $P$ is a refinement, then for any choice of intermediate points $t_j \in [x_{j-1},x_j]$, we have
$$\tag{4}\left|\sum_{j=1}^n \alpha'(t_j)(x_j - x_{j-1}) - \int_a^b \alpha'(x) \, dx\right|< \frac{\epsilon}{8M}$$
Denoting $M_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$ and $m_j = \sup\{\alpha'(x): x \in [x_{j-1},x_j]\}$, by properties of the supremum and infimum there exist points $t_j', t_j''$ such that
$$M_j - \frac{\epsilon}{4M(b-a)} < \alpha'(t_j') \leqslant M_j, \quad m_j \leqslant \alpha'(t_j'') < m_j + \frac{\epsilon}{8(b-a)},$$
and, thus,
$$\sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) < \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1}) -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{2M}\\ = \sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx + \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1}) + \frac{\epsilon}{4M}$$
Hence, since the LHS is nonnegative,
$$\tag{5} \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1}) \\< \left|\sum_{j=1}^n \alpha'(t_j')(x_j- x_{j-1})- \int_a^b\alpha'(x) \, dx \right|+ \left| \int_a^b \alpha'(x) \, dx -\sum_{j=1}^n \alpha'(t_j'')(x_j- x_{j-1})\right|+ \frac{\epsilon}{4M}$$
Using (4), it follows from substitution into (5) that
$$ \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})< \frac{\epsilon}{8M} + \frac{\epsilon}{8M} + \frac{\epsilon}{8M} = \frac{\epsilon}{2M}$$
Noting that $\xi_j, \eta_j \in [x_{j-1},x_j]$ and $|\alpha'(\xi_j)- \alpha'(\eta_j)| \leqslant M_j - m_j$, we obtain the result
$$\sum_{j=1}^n |\alpha'(\eta_j) - \alpha'(\xi_j)|(x_j - x_{j-1})\leqslant \sum_{j=1}^n (M_j - m_j)(x_j - x_{j-1})<\frac{\epsilon}{2M}$$