Existence of Riemann–Stieltjes integral How to prove that for existence of Riemann–Stieltjes integral $\int\limits_a^b f(x) \, \mathrm{d}g$, where  $g \in V([a,b])$, it is required that function $f$ is continuous in the point of discontinuity of function $g$?
It sounds obvious, but how to prove that carefully?
 A: If $\int_{a}^{b}fdg$ exists, then, for every $\varepsilon > 0$, there exists $\delta > 0$ such that
$$
            \left|\sum_{\mathcal{P}}f(x_k^{\star})\{g(x_k)-g(x_{k-1})\}-
       \sum_{\mathcal{P'}}f(x_k'^{\star})\{g(x_k')-g(x_{k-1}')\}\right| < \varepsilon
$$
whenever $\|\mathcal{P}\| < \delta$ and $\|\mathcal{P'}\| < \delta$. If you use the same partition points for $\mathcal{P}$ and $\mathcal{P'}$ and allow one of the evaluation points to vary, one consequence of the above is that
$$
                |f(x_k^{\star})-f(x_{k}'^{\star})||g(x_{k})-g(x_{k-1})| < \varepsilon \;\;\;\; (\dagger)
$$
whenever $x_k \le x_k^{\star},x_k'^{\star} \le x_{k+1} <x_k+\delta$. Suppose $g$ is discontinuous at some $x \in (a,b)$; $g$ has limits from the left and the right of $x$, which means that either $g(x+0)\ne g(x)$ or $g(x-0)\ne g(x)$, or both. There are two cases to consider


*

*Case 1: $g(x-0)=g(x+0)$.Then $g(x+0)\ne g(x)$ and $g(x-0)\ne g(x)$, which means there exists $\sigma > 0$ such that $|g(x_l)-g(x)| > \sigma$ and $|g(x)-g(x_r)| > \sigma$ for all $x-\delta_0 < x_l < x < x_r < x+\delta_0$, provided $\delta_0$ is chosen sufficient small. Let $\varepsilon > 0$ be given, and choose $\delta$ so that $(\dagger)$ holds with $\varepsilon$ replaced by $\sigma\varepsilon$. Let $\delta_1$ be the minimum of $\delta_0$ and $\delta$.
Then
$$
         |f(x')-f(x)|\sigma < \sigma\varepsilon,\;\;\; x-\delta_1 < x' < x \\
         |f(x)-f(x'')|\sigma < \sigma\varepsilon,\;\;\; x < x'' < x+\delta_1.
$$
Hence, $f(x-0)=f(x)=f(x+0)$, making $f$ continuous at $x$.

*Case 2: $g(x-0)\ne g(x+0)$. In this case there exists $\sigma > 0$ such that $|g(x_l)-g(x_r)| > \sigma$ for all $x-\delta_0 < x_l < x < x_r < x+\delta_0$, provided $\delta_0$ is chosen sufficiently small. Let $\varepsilon > 0$ be given. Choose $\delta > 0$ so that $(\dagger)$ holds with $\varepsilon$ replaced by $\sigma\varepsilon$. Let $\delta_1$ be the minimum of $\delta_0$ and $\delta$. Then
$$
          |f(x')-f(x'')|\sigma < \sigma\varepsilon,\;\;\; x-\delta_1 < x' , x'' < x+\delta_1
$$
Set $x''=x$ in order to concluded that $\lim_{x'\rightarrow x}f(x')=f(x)$.
I'll leave the endpoint cases at $x=a$, $x=b$ to you.
