Riemann–Stieltjes integration by parts

For a partition $P=(n,t)$, finite summation by parts gives the exact identity $\sum_{i<n}f(t_i)(\alpha(t_{i+1})-\alpha(t_i))+\sum_{i<n}\alpha(t_{i+1})(f(t_{i+1})-f(t_i))=f(b)\alpha(b)-f(a)\alpha(a)$.

Suppose $\int f\,d\alpha$ exists and consider an arbitrary tagged sum $S_f(\alpha;P,\eta)$. Refine each $[t_i,t_{i+1}]$ by inserting its tag $\eta_i$. On $[t_i,\eta_i]$ tag the complementary $f\,d\alpha$ sum at $t_i$, and on $[\eta_i,t_{i+1}]$ tag it at $t_{i+1}$. Direct expansion on the $i$th interval gives [step 1.1, L1, L2, L3, L4] $\alpha(\eta_i)(f(t_{i+1})-f(t_i))+f(t_i)(\alpha(\eta_i)-\alpha(t_i))+f(t_{i+1})(\alpha(t_{i+1})-\alpha(\eta_i))=f(t_{i+1})\alpha(t_{i+1})-f(t_i)\alpha(t_i).$ The refined mesh does not exceed $\lVert P\rVert$, so the complementary sums converge to $\int f\,d\alpha$. Telescoping the displayed identities forces every fine tagged sum for $\int\alpha\,df$ to converge to the endpoint product minus that integral.

Exchanging $f$ and $\alpha$ proves the converse. Adding the two values yields the displayed formula, including the singleton and reversed-orientation cases.