Let $F_1:\mathbb{R}\to\mathbb{C}$ and $F_2:\mathbb{R}\to\mathbb{C}$ be functions of bounded variation on $\mathbb{R}$. In order to prove the bounded variation of their convolution defined as $F(x)=\int_{-\infty}^\infty F_1(x-\xi)dF_2(\xi)$, Kolmogorov and Fomin (p. 451 here) write$$|F(x_1)-F(x_2)|=\Bigg|\int_{-\infty}^\infty(F_1(x_1-\xi)-F_1(x_2-\xi))dF_2(\xi)\Bigg|\le$$$$\le\int_{-\infty}^\infty|F_1(x_1-\xi)-F_1(x_2-\xi)|d(\text{var}F_2(\xi))$$where the integrals are all to be intended as Lebesgue-Stieltjes integrals. It is the first time the book uses notation $\text{var}F_2(\xi)$, but I guess it means total variation $V_{-\infty}^{\xi}[F_2]$. Does anybody know a proof of the integral inequality? (I assume that it is valid in general for any $f_2$, although Kolmgorov-Fomin's sometimes states things valid only under some unsaid hypothesis, therefore I would not be amazed if someone gave a counterexample). From that inequality the following one derives:$$V[F]\le V[F_1]V[F_2].$$How can it be proved?
I think -please correct me if I am wrong- that all real valued functions of bounded variation can be written as $\Phi(x)=V_a^x[\Phi]-g(x)$ where both $V_a^x[\Phi]$ and $g$ are non decreasing function of bounded variation and, therefore, for $\Phi:[a,b]\to\mathbb{C}$ of bounded variation we have$\Phi(x)=V_a^x[\text{Re}\Phi]-g_R(x)+i(V_a^x[\text{Im}\Phi]-g_I(x))$ where $g_R$, $g_I$ also are non decreasing function of bounded variation. Therefore I would say that the Lebesgue-Stieltjes integral can be defined as $\int_a^b f(x) d\Phi(x)=\int_{[a,b]}fd\mu_{V_a^x[\text{Re}\Phi]}-\int_{[a,b]}fd\mu_{g_R}+i\int_{[a,b]}fd\mu_{V_a^x[\text{Im}\Phi]}-i\int_{[a,b]}fd\mu_{g_I}$ where the $\mu_{F}$ are Lebesgue-Stieltjes measures. This is, in essence, all that Kolmogorov-Fomin's says about the Lebesgue-Stieltjes integral. How can it be seen that those two inequalities hold?