Two bounded-variation functions with no common discontinuity are Riemann–Stieltjes integrable

By the no-common-discontinuity hypothesis, $f$ is continuous at every point of $D_\alpha$. For each $n\ge1$, [L1] and [L2] provide a finite step function $s_n$ with $\lVert f-s_n\rVert_\infty<1/n$ and all interior breakpoints outside $D_\alpha$. If an endpoint belongs to $D_\alpha$, continuity of $f$ there permits the value on the adjacent open component to be changed to $f$ at that endpoint while retaining the same bound after beginning with tolerance $1/(2n)$. Thus $s_n$ is continuous at every point of $D_\alpha$.

Each $s_n$ is integrable with respect to $\alpha$. Its finitely many discontinuities are points where $\alpha$ is continuous by step 1.1. By [L5], choose disjoint neighborhoods of those points whose total local variation is small. Outside them $s_n$ is locally constant, while inside them [L5] bounds differences between fine sums by the small local variation times the finite oscillation of $s_n$. Hence the fine sums are Cauchy. Choose a sequence of uniform tagged sums with mesh tending to zero; [L4] gives its limit, and comparison with a sufficiently late member of this sequence shows that every sufficiently fine tagged sum has the same limit.

Given $\varepsilon>0$, choose $n$ so that $2\operatorname{Var}_{[a,b]}(\alpha)/n<\varepsilon/2$ (the zero-variation case is immediate), and then choose a mesh bound making any two sums of $s_n$ differ by less than $\varepsilon/2$. By [L3], replacing $s_n$ by $f$ in either sum changes it by at most $\operatorname{Var}(\alpha)/n$. Hence all sufficiently fine sums of $f$ are Cauchy. Choose uniform tagged sums with mesh tending to zero; their sums form a Cauchy sequence and converge by [L4]. Comparing an arbitrary sufficiently fine sum with a late uniform sum proves convergence of the whole mesh family to that sequential limit, which is exactly the defining Stieltjes integral.