Lebesgue-Stieltjes Integral of Singleton

Question

I have two functions $f,g\in\textit{BV}(K,\mathbb R)$, where $\textit{BV}(K,\mathbb R)$ is the set of all functions mapping $K\subset\mathbb R$ to $\mathbb R$ that are of bounded variation. The function $f$ is, in addition, bounded and continuous except at a countable set $D\subset K$. I have to compute the Lebesgue-Stieltjes integral $$\int_K f\,\mathrm dg.$$ Question: Is the following manipulation valid: $$\int_Kf\,\mathrm dg = \int_{K\setminus D}f\,\mathrm dg + \int_Df\,\mathrm dg.$$ I believe yes, as we could define a measure $\mu$ that satisfies $\mu((a,b]) = g(b) - g(a-)$ for all left-half open intervals $(a,b]$. Then $$\int_Kf\,\mathrm dg = \int_Kf\,\mathrm d\mu$$ and a basic property of Lebesgue integrals yields $$\int_Kf,\mathrm d\mu = \int_Af\,\mathrm d\mu + \int_Bf\,\mathrm d\mu,$$ where $A\cup B = K$ and $A\cap B=\emptyset$. Using this property again and again should yield $$\int_{D}f\,\mathrm dg = \sum_{d\in D}\int_{\{d\}} f\,\mathrm dg$$ (see Prove the Countable additivity of Lebesgue Integral.). This brings me to the question stated in the title. For the usual Lebesgue measure $\nu$ on $\mathbb R$ we have that $$\int_{\{d\}}f\,\mathrm d\nu = 0.$$ But since the Lebesgue-Stieltjes measure is slightly different, the Lebesgue-Stieltjes integral of a point is not necessarily zero. Eventually, I am interested in bounded the integral by $$\int_{\{d\}}f\,\mathrm dg \leq \sup_{x\in K}\vert f(x)\vert (g(d) - g(d-)).$$ Is my reasoning correct? In particular, is the Lebesgue-Stieltjes integral of a signle point well-defined?