If $f\in\mathcal R(\alpha)$, then $\alpha\in\mathcal R(f)$ and $\int_a^b f\,d\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\int_a^b\alpha\,df$.Suggest a slogan
Let \(a<b\) and let \(f,\alpha\colon[a,b]\to\RR\). Write \(f\in\mathcal R(\alpha)\) if there is \(A\in\RR\) such that for every \(\varepsilon>0\) there is a partition \(P_\varepsilon\) of \([a,b]\) with \[ \abs{\sum_{i=1}^k f(t_i)\qty{\alpha(x_i)-\alpha(x_{i-1})}-A}<\varepsilon \] for every partition \(P=\theset{a=x_0<\cdots<x_k=b}\) refining \(P_\varepsilon\) and every choice of \(t_i\in[x_{i-1},x_i]\); then \(\int_a^b f\,d\alpha\coloneqq A\). If \(f\in\mathcal R(\alpha)\) on \([a,b]\), then \(\alpha\in\mathcal R(f)\) on \([a,b]\) and \[ \int_a^b f\,d\alpha=f(b)\alpha(b)-f(a)\alpha(a)-\int_a^b\alpha\,df. \] [@Apo74].