Let $f, g : [0,2] \to \mathbb{R}$ be the function defined as $$f(x)= \begin{cases} 0 &\text{if}\, 0\leq x\leq 1\\ 1 &\text{if}\, 1<x\leq 2 \end{cases} $$
And $g$ be the greatest integer function on $[0, 2]$.
Is $f$ Riemann-Stieltjes (R-S) integrable with respect to $g$?
$f$ is discontinuous only at $x=1$ but at same point $g$ is also discontinuous. By taking different partitions of $[0, 2]$, it seems upper and lower both R-S sum are zero. Is there any theorem that guarantees R-S integrability of $f$ with respect to $g$?
I suppose your "greatest integer function" means the floor function or Gauss' step function. If so, choose the partition $P=\{0<1<2\}$. Then the Riemann-Stieltjes sum $S(P,f,g)$ of $f$ w.r.t. $P$ and $g$ is $$S(P,f,g)=f(t_1)(1-0)+f(t_2)(2-1)$$ where $t_1\in(0,1),t_2\in(1,2).$
Here we see this value is always equal to $1$, no matter what $t_1,t_2$ are, which cannot be arbitrarily small. Thus $f$ is not integrable w.r.t. $g$ on $[0,2].$
As the comment mentioned, the main reason this happened is because both $f$ and $g$ are discontinuous at $1$.