Riemann–Stieltjes sums, upper and lower sums, and the Riemann–Stieltjes integral

Let $a<b$, let $f,\alpha :[a,b]\to \mathbb{R}$, and let $P=(n,t)$ be a partition (Partition of $[a,b]$ as a finite strictly increasing list $a=t_0<t_1<\cdots <t_n=b$, its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions). A choice of tags ${\xi }_i\in [t_i,t_{i+1}]$ for $i<n$ makes $(P,\xi )$ a tagged partition as in Tagged partitions of $[a,b]$, with a tag ${\xi }_i$ in each subinterval, and the Riemann sum $S(f,P,\xi )={\sum }_{i}f({\xi }_i){\mathrm{\Delta }}_i$. Its Riemann-Stieltjes sum is

$S(f,\alpha ;P,\xi ):={\sum }_{i<n}f({\xi }_i)(\alpha (t_{i+1})-\alpha (t_i))\mathrm{.}$

The function $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $[a,b]$ if there is $I\in \mathbb{R}$ such that for every $\epsilon >0$ there is $\delta >0$ for which every tagged partition with $\mathrm{\parallel }P\mathrm{\parallel }<\delta$ satisfies $\mathrm{\mid }S(f,\alpha ;P,\xi )-I\mathrm{\mid }<\epsilon$. Then $I={\int }_a^{b}fd\alpha$.

If $f$ is bounded (Lower bound, bounded below, bounded set) and $\alpha$ is nondecreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $\mathbb{R}$, with the dictionary to monotone sequences), put

$m_i:={\inf }_{x\in [t_i,t_{i+1}]}f(x),M_i:={\sup }_{x\in [t_i,t_{i+1}]}f(x),$ $L_{\alpha }(f,P):={\sum }_{i<n}{m}_i{\mathrm{\Delta }}_i\alpha ,U_{\alpha }(f,P):={\sum }_{i<n}{M}_i{\mathrm{\Delta }}_i\alpha ,$

where ${\mathrm{\Delta }}_i\alpha =\alpha (t_{i+1})-\alpha (t_i)\ge 0$. These are the lower and upper Stieltjes sums. Each subinterval is nonempty and its image under bounded $f$ is bounded above and below, so the suprema exist by Complete ordered field (least-upper-bound property) and the infima by Greatest lower bound (infimum) and Every nonempty set bounded below has an infimum. Finite sums use Finite sums and finite products, by recursion and Laws of finite sums and finite products. On $[a,a]$ the integral is $0$; for $b<a$ set ${\int }_a^{b}fd\alpha =-{\int }_b^{a}fd\alpha$, matching The integral with oriented limits: ${\int }_a^{a}f:=0$ and ${\int }_b^{a}f:=-{\int }_a^{b}f$.