Suppose that $f$ is $g$-integrable (in the sense of Riemann-Stieltjes). Let $g_1:[a,b]\longrightarrow\mathbb{R}$ be such that $g_1(x)=g(x)$ for a finite number of points in $(a,b)$ at which $f$ is continuous. I'm trying to prove that $f$ is $g_1$-integrable on $[a,b]$ and that $\int_a^ b fdg_1=\int_a^ b fdg$.
Suppose $g_1\neq g$ only in $c\in(a,b)$ (the case where there are $N$ discontinuities follows by induction). Lets define $h=g-g_1$. If we prove that $f$ is $h$-integrable, then by a well-know result $\int_a^ b f(g-h)=\int_a^ b fdh+\int_a^ b fdg$.
It is clear that for any partition that doesn't include the point $c$ the Riemann-Stieltjes sum $S(f,P,h)=0$ since $h(x_k)-h(x_{k-1})=0$ by definition. So lets take a partition $P$ that includes c. This partition is cero almost everywhere because $h=0$ for every point except c.
But how can I relate this to the $g$-integrability of $f$ to make $|S(f,P,h)-\int_a ^ bf|<\epsilon$?