I am studying the Lebesgue-Stieltjes integral from this PDF: https://www.math.utah.edu/~li/L-S%20integral.pdf.
In Theorem 8 the authors claim to use Dynkin's theorem in a way that I do not understand. The setting is the following. We are given a signed measure $\mu$ and a positive measure $\nu$, both defined on $\mathbb{R}$ (in the notations of the PDF they are called $dA$ and $dV$). Suppose that we know that $|\mu((a,b])|\leq\nu((a,b])$ for all semi-open intervals ($|\mu((a,b])|$ is just the absolute value of $\mu((a,b])$). The authors seem to claim that then we can apply Dynkin's theorem to get $|\mu(B)|\leq\nu(B)$ for all Borel sets $B\subset\mathbb{R}$.
My problem with the application of Dynkin's theorem is that, while I know that the intervals are a $\pi-$system, I can't see why the collection of $B$ such that $|\mu(B)|\leq\nu(B)$ is a $\lambda-$system, because it does not seem to be closed under complements. Any help or hint is much appreciated. Thanks!
Dynkin's theorem is not well-suited for inequalities. Another approach lies on regularity of finite measures on metric spaces. As continuous functions are uniform limit of left-continuous step functions, you have $$ \bigg| \int_{(s,t]} \varphi\,dA\bigg| \le \int_{(s,t]} |\varphi| \,dV $$ for all continuous function $\varphi$ on $[0,a]$. Take disjoint Borel sets $E_1,\ldots,E_n$ such that $(s,t]= \cup_{i=1}^n E_i$. Given $\varepsilon > 0$, one can find compact sets $K_i \subseteq E_i$ and open sets $O_i \supseteq E_i$ such that $|dA|(O_i \setminus K_i) \le \varepsilon/2^i$, so $$ \sum_i |dA(E_i)| \le \varepsilon + \sum_i |dA(K_i)| $$ Pick some continuous functions $\varphi_i$ with disjoint supports such that $0 \le \varphi_i \le 1$, $\varphi_i = 1$ on $K_i$ and $\operatorname{supp} \varphi_i \subset O_i$ (such functions exist by Urysohn's lemma). Using triangular inequality again $$ \sum_i |dA(E_i)| \le 2\varepsilon + \sum_i \bigg|\int_{(s,t]} \varphi_i\,dA\bigg| \le 2\varepsilon + \int_{(s,t]} \bigg(\sum_i \varphi_i\bigg)dV \le 2\varepsilon +dV(s,t] $$ Since $\varepsilon > 0$ and $E_1,\ldots,E_n$ were arbitrary, you get $|dA|(s,t] \le dV(s,t]$. The reverse inequality is obvious, thus $dV$ and $|dA|$ agree on sets of the form $(s,t]$ so on Borel sets by Dynkin's theorem (because semi-open intervals constitute a $\pi$-system generating the Borel $\sigma$-algebra, and both measures $dV$ and $|dA|$ are finite with same total mass).