The function $g$ is continuous in discontinuity point

Let $g \in V[a, b]$, where $V[a, b]$ is the set of functions of bounded variation on $[a,b]$. Show that existence of the Riemann-Stieltjes integral $\int_a^b f dg$ implies that $g$ is continuous in any discontinuity point of $f$.

May be I can show that the integral exist when $g$ is monotone increasing and continuous by substracting the upper and lower Riemann-Stieltjes sums and showing this quantity is smaller than any $\epsilon$?

First prove (1) this is true if $g$ is a monotone increasing function and then (2) use the fact that any function of bounded variation is the difference of two increasing functions.

For (1) we show that if $g$ is not continuous at a point where $f$ is discontinuous, then the Riemann-Stieltjes integral cannot exist.

Suppose $g$ is monotone increasing and that WLOG $f$ and $g$ are discontinuous from the right at $\xi \in (a,b).$ Consider any partition $P = (x_0,x_1, \ldots, x_{i-1},\xi, x_i, \ldots, x_n)$ with $\xi$ as a partition point and $x_i - \xi = \delta_i$

There exists $\epsilon > 0$ such that for every $\delta > 0$ (including $\delta_i$), there are points $y_1, y_2 \in (\xi, \xi + \delta)$ such that $|f(y_1) - f(\xi)| \geqslant \epsilon$ and $|g(y_2) - g(\xi)| \geqslant \epsilon$.

Then we have

$$U(P,f,g) - L(P,f,g) \geqslant \epsilon^2,$$

since $g(x_i) - g(\xi) \geqslant g(y_2) - g(\xi) \geqslant \epsilon$ and $\sup_{x \in [\xi,x_i]} f(x) - \inf_{x \in [\xi,x_i]} f(x) \geqslant \epsilon$

Therefore, the Riemann criterion is not satisfied and $f$ is not RS integrable with respect to $g$ on $[a,b]$.

I'll let you consider part (2).