Let $f:[a,b]\to\mathbb{R}$ and let $\alpha,\beta:[a,b]\to\mathbb{R}$ be functions of bounded variation . Suppose the Riemann–Stieltjes integrals $\int_a^b f\,d\alpha$ and $\int_a^b f\,d\beta$ exist. Let $c,d\in\mathbb{R}$ and define a new integrator $\gamma=c\alpha+d\beta.$

Proposition: The integral $\int_a^b f\,d\gamma$ exists and $\int_a^b f\,d(c\alpha+d\beta)=c\int_a^b f\,d\alpha+d\int_a^b f\,d\beta.$

This is the “integrator-side” linearity of the Riemann–Stieltjes integral. Together with integrand-side linearity, it makes $\int f\,d\alpha$ bilinear in $(f,\alpha)$ (within the class where the integral exists).

Proof sketch: For a tagged partition , the Riemann–Stieltjes sums satisfy $ \sum f(t_i)\bigl(\gamma(x_i)-\gamma(x_{i-1})\bigr)

c\sum f(t_i)\bigl(\alpha(x_i)-\alpha(x_{i-1})\bigr) +d\sum f(t_i)\bigl(\beta(x_i)-\beta(x_{i-1})\bigr). $If the right-hand sums converge as the mesh tends to$0$, then so does the left-hand sum, with the stated limit. A careful argument uses the definition of Riemann–Stieltjes integrability via control of upper/lower sums or via the Cauchy criterion for these sums.