Here is Theorem 7.20, in Chap. 7, in the book Mathematical Analysis - A Modern Approach To Advanced Calculus by Tom M. Apostol, 2nd edition:
Assume that $\alpha \nearrow$ on $[a, b]$. If $f \in R(\alpha)$ and $g \in R(\alpha)$ on $[a, b]$ and if $f(x) \leq g(x)$ for all $x$ in $[a, b]$, then we have $$ \int_a^b f(x) d\alpha(x) \leq \int_a^b g(x) d\alpha(x). $$
And, here is Apostol's proof:
For every partition $P$, the corresponding Riemann-Stieltjes sums satisfy $$ S(P, f, \alpha) = \sum_{k=1}^n f \left( t_k \right) \Delta \alpha_k \leq \sum_{k=1}^n g \left( t_k \right) \Delta \alpha_k = S(P, g, \alpha) $$ since $\alpha \nearrow$ on $[a, b]$. From this the theorem follows easily.
How to complete this argument?
My Attempt:
Let us put $$ A \colon= \int_a^b f(x) d\alpha(x), \qquad B \colon= \int_a^b g(x) d\alpha(x). \tag{0} $$ If $\alpha(a) = \alpha(b)$, then we have $A = 0 = B$. So let's assume that $\alpha(a) < \alpha(b)$ and also that $A > B$.
Then for every real number $\varepsilon$ such that $$ 0 < \varepsilon \leq \frac{A-B}{2}, $$ there exist partitions $P_\varepsilon^\prime$ and $P_\varepsilon^{\prime\prime}$ of $[a, b]$ such that $$ \big\lvert S(P, f, \alpha) - A \big\rvert < \varepsilon \tag{1} $$ whenever $P$ is finer than $P_\varepsilon^\prime$ and $$ \big\lvert S(P, g, \alpha) - B \big\rvert < \varepsilon \tag{2} $$ whenever $P$ is finer than $P_\varepsilon^{\prime\prime}$.
Let us put $$ P_\varepsilon \colon= P_\varepsilon^\prime \cup P_\varepsilon^{\prime\prime}. $$ Then (1) and (2) above hold for any partition $P$ that is finer than $P_\varepsilon$. Let $P$ be any such partition. Then we have $$ \begin{align} & \ \ \ S(P, g, \alpha) \\ &< B + \varepsilon \qquad \mbox{[using (2) above]} \\ &\leq B + \frac{A-B}{2} \qquad \mbox{[note that $0 < \varepsilon \leq (A-B)/2$]} \\ &= \frac{A + B}{2} \\ &= A - \frac{A-B}{2} \\ &\leq A - \varepsilon \qquad \mbox{[note that $0 < \varepsilon \leq (A-B)/2$]} \\ &< S(P, f, \alpha), \qquad \mbox{[using (1) above]} \end{align} $$ and hence we have $$ S(P, g, \alpha) < S(P, f, \alpha). $$ But this contradicts the fact that $$ S(P, f, \alpha) \leq S(P, g, \alpha). $$ Hence our assumption that $A> B$ is incorrect.
Is this proof correct and clear enough in each and every detail? Or, are there still some points of confusion therein?