A bounded function with finitely many discontinuities is Stieltjes integrable against a continuous bounded-variation integrator

If $a=b$ the integral is $0$ by the definition of the Riemann-Stieltjes integral on a singleton interval and there is nothing to prove, so assume $a<b$, which is the standing hypothesis of [L3].

First suppose that $\alpha$ is continuous and nondecreasing. Write $|f|\le M$. Given $\varepsilon>0$, [L6] and finiteness of $D$ allow pairwise disjoint closed intervals $I_x$ about the points $x\in D$ whose total $\alpha$-increment is less than $\varepsilon/(4M+1)$.

On the compact complement of the interiors of the $I_x$, the function $f$ is continuous and hence uniformly continuous by [L4]. Choose a partition containing all endpoints of the $I_x$ and fine enough that every remaining partition interval has oscillation below $\varepsilon/(1+\alpha(b)-\alpha(a))$. The intervals meeting $D$ contribute at most $2M$ times their total $\alpha$-increment, and all other intervals contribute less than $\varepsilon$. After rescaling the two preliminary bounds, the weighted oscillation sum is arbitrarily small, so [L3] gives $f\in R(\alpha)$.

For a general continuous BV $\alpha$, [L1] writes $\alpha=\alpha(a)+P_\alpha-N_\alpha$. Both $P_\alpha$ and $N_\alpha$ are continuous by [L2]. Step 2.1 gives integrability against each, and linearity in the integrator [L5] gives integrability against $\alpha$.