I read in Kolmogorov-Fomin's Элементы теории функций и функционального анализа that (p. 372 here), if the Riemann-Stieltjes integrals $$\int_a^b f(x) d\Phi_1(x)\quad\text{ and }\quad\int_a^b f(x) d\Phi_2(x),$$ which are the same as Lebesgue-Stieltjes integrals for $f\in C[a,b]$, with respect to two functions of bounded variation $\Phi_1:[a,b]\to\mathbb{R}$ and $\Phi_2:[a,b]\to\mathbb{R}$, are equal for all $f\in C[a,b]$, then $\Phi_1$ and $\Phi_2$ coincide in every point where $\Phi_1-\Phi_2$ is continuous up to a constant (and I know that a function of bounded variation on $[a,b]$ is continuous except for countably many point).
That does not seem so trivial to me, although the book does not prove it. I know that if $\Phi_1-\Phi_2$ is constant except for a countable number of points contained in $(a,b)$ then $\forall f\in C[a,b]$ $\int_a^b f(x) d\Phi_1(x)=\int_a^b f(x) d\Phi_2(x)$, but I cannot prove the converse to myself. Could anybody prove this interesting fact? $\infty$ thanks!!!
EDIT: Corrected imprecision thanks to T.A.E.
I'm going to use the Riemann-Stieltjes integral to prove this. Assume that $\int_{a}^{b}fd\Phi=0$ for all $f\in C[a,b]$ and some $\Phi$ of bounded variation on $[a,b]$. Suppose $t \in [a,b)$ is a point of continuity of $\Phi$. For any $\epsilon \in [0,b-t)$, define $$ f_{t,\epsilon}(s)=\left\{\begin{array}{ll} 1, & a \le s \le t,\\ 1-\frac{1}{\epsilon}(s-t), & t < s \le t+\epsilon,\\ 0, & t+\epsilon < t \le b. \end{array}\right. $$ This function $f_{t,\epsilon}$ is continuous on $[a,b]$. The nice part about Riemann-Stieltjes is that one may always integrate by parts. Hence, $$ 0= \int_{a}^{b}f_{t,\epsilon}d\Phi = \int_{a}^{t}d\Phi +\int_{t}^{t+\epsilon}f_{t,\epsilon}d\Phi \\ = \Phi(t)-\Phi(a)+f_{t,\epsilon}\Phi|_{t}^{t+\epsilon}-\int_{t}^{t+\epsilon}\Phi df_{t,\epsilon}\\ = \Phi(t)-\Phi(a)-\Phi(t)+\frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(t)dt \\ = \frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(t)\,dt-\Phi(a). $$ Therefore, if $\Phi$ is continuous at $t$, $$ 0=\lim_{\epsilon\downarrow 0}\int_{a}^{b}f_{\epsilon,t}d\Phi = \lim_{\epsilon\downarrow 0}\frac{1}{\epsilon}\int_{t}^{t+\epsilon}\Phi(s)\,ds -\Phi(a) = \Phi(t)-\Phi(a). $$ It follows that $\Phi(t)=\Phi(a)$ for all points of continuity $t \in [a,b)$. Similarly, you can argue that $\Phi(t)=\Phi(b)$ for all points of continuity $t\in (a,b]$. (In fact, if you look carefully at the expressions, you see $\Phi(a)=\Phi(t+0)$ for all $t\in [a,b)$ and $\Phi(t-0)=\Phi(b)$ for all $t \in (a,b]$, but these identities can be deduced from the weaker statement, too.)