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

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

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