How $|S(x_n, \xi_n) - S(x'_n, \xi'_n)|$ being arbitrary small implies existence of Stieltjes-Riemann Integral?

Question

I am studying Multiplicative number theory I: Classical theory by Hugh L. Montgomery, Robert C. Vaughan (E-book). In Appendix A, Theorem A.1. states that $I = \int f dg$ exists if $f$ is continuous and $g$ is of bounded variation. Let $a=x_0 \le x_1 \le \dots \le x_N=b$ is a partition of $[a,b]$, and $S(x_n, \xi_n) = \sum_1^N f(\xi_n) (g(x_n)-g(x_{n-1}))$. The proof claims that for existence of $I$ it is sufficient to prove that for every $\epsilon$ there is a $\delta$ such that $|I - S(x_n, \xi_n)|<\epsilon$. It is clear up to this point. Then it is claimed that it is sufficient to prove that $|S(x_n, \xi_n) - S(x'_n, \xi'_n)|<2 \epsilon \text{Var(g)}$ for two arbitrary partitions $x_n$ and $x'_n$. I could follow all steps of the proof and I can see that $|S(x_n, \xi_n) - S(x'_n, \xi'_n)|<2 \epsilon \text{Var(g)}$ is true. But why $|S(x_n, \xi_n) - S(x'_n, \xi'_n)|<2 \epsilon \text{Var(g)}$ being true implies $|I - S(x_n, \xi_n)|<\epsilon$ to be true ??

Answer 1

Let me consider fixed $g$, so "Cauchy like" property for inequality $(A.2)$ can be formulated as following:

$\forall \varepsilon>0, \exists\delta >0,$ such that for $\forall$ partitions $x'_n,x''_n$ with mesh $< \delta$, holds $$|S(x'_n, \xi_n)-S(x''_n, \xi'_n)|<\varepsilon \quad(A.2)$$ where $\xi_n, \xi'_n$ are any from corresponding intervals.

Now, let's consider $\varepsilon_n=\frac{1}{n}, n\in \mathbb{N}$ and choose for any $n$ choose $\delta_n$, for which holds $(A.2)$. We are able consider $\delta_n$ monotone decreasing. Also, for any $n$ we can choose partition $x_n$ with mesh $\lambda_{x_n}<\delta_n$ and take $\xi_n$ from this partition intervals. Let's denote Riemann–Stieltjes sum for obtained values by $T_n$. For exactness we can write $\lambda_{x_n}$ as $\lambda_{T_n}$.

For given any $\forall \varepsilon>0$ we can find $N$ for which $\varepsilon_N < \varepsilon$. For any $k,m>N$ we have $\lambda_{T_m }<\delta_m<\delta_N$ and $\lambda_{T_k }<\delta_k<\delta_N$, so from $(A.2)$ holds $$|T_k - T_m|< \varepsilon$$ This gives, that $T_n$ converged and we can show, that its limit $T$ will be integral.

For this for $\forall \varepsilon>0$ we find $n$, such that $\varepsilon_n < \varepsilon$. For any $p>n$ we have $\lambda_{T_p }<\delta_p\leqslant \delta_n$, so for any partition $\{x\}$ with mesh $<\delta_n$ and corresponding any collection of $\xi$ we will have $$|T_p-S(\{x\},\xi)|< \varepsilon_n$$ Taking limit $p\to\infty$ in last inequality gives $$|T-S(\{x\},\xi)|\leqslant \varepsilon_n < \varepsilon$$ which mean, that $T$ is integral