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 $\int fd\alpha$ 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 $fd\alpha$ 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] $\alpha ({\eta }_i)(f(t_{i+1})-f(t_i))+f(t_i)(\alpha ({\eta }_i)-\alpha (t_i))+f(t_{i+1})(\alpha (t_{i+1})-\alpha ({\eta }_i))=f(t_{i+1})\alpha (t_{i+1})-f(t_i)\alpha (t_i)\mathrm{.}$ The refined mesh does not exceed $\parallel P\parallel$, so the complementary sums converge to $\int fd\alpha$. 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.