A continuous integrand is Riemann–Stieltjes integrable against every bounded-variation integrator

Put $V=\operatorname{Var}_{[a,b]}(\alpha)$. If $V=0$, every integrator increment is $0$, so every Stieltjes sum is $0$ and the integral exists. Assume $V>0$. Given $\varepsilon>0$, uniform continuity gives $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon/(4V)$. Any two tagged partitions of mesh below $\delta$ have all local oscillations below $\varepsilon/(4V)$; comparing both with their common refinement through [L1] makes their sums differ by less than $\varepsilon$.

Choose one tagged uniform partition with mesh tending to $0$ for each natural index and call its sum $s_n$. Step 1.1 makes $(s_n)$ Cauchy, so [L5] gives a real limit $I$.

Given $\varepsilon>0$, choose the threshold in step 1.1 for error $\varepsilon/2$ and then a uniform sum $s_n$ beyond that threshold with $|s_n-I|<\varepsilon/2$. Every arbitrary tagged sum with sufficiently small mesh differs from $s_n$ by less than $\varepsilon/2$, hence is within $\varepsilon$ of $I$. This is the mesh-limit definition, and [L2] identifies the unique value.