Convergence of Riemann-Stieltjes sums

Let $f$ be Riemann-Stieltjes integrable with respect $G$ (increasing function). My definition for Riemann-Stieltjes integration is: for every $\epsilon$ there is a partition $\mathcal{P}_\epsilon$ such that when $\mathcal{P}_\epsilon \subset \mathcal{P}$ then $\left|S(\mathcal{P},f,G, \{t_i\}) - \int_a^bf dG \right| < \epsilon$ for any set of tags $\{t_i\}$.

For the Riemann integral and it is true that the integral is the limit of Riemann sums: $\lim_{\|\mathcal{P}\| \to 0}S(\mathcal{P},f)= \int_a^bf(x)dx$. Is this true for the Riemann-Stieltjes integral?

I know the integral may not exist when $f$ and $G$ are discontinuous at the same point, and I think I would need to add the condition that prevents this as well as $f$ is integrable and $G$ is increasing.

• You have $g$ instead of $G$ in one place (typo?). Your statement (including $G$ and $f$ not having discontinuity at the same point) is correct. – herb steinberg Aug 25 '18 at 0:16

When the Riemann-Stieltjes integral exists, there is no guarantee that it is the limit of sums as the partition norm tends to $0$ without stronger conditions than what you assume. In this way, it is unlike the Riemann integral.

Also, your statement that the integral will fail to exist when $f$ and $G$ have common points of discontinuity is inaccurate. The integral does not exist if the functions are both discontinuous from the right or both discontinuous from the left.

Consider the following counterexample where the integral exists but the limit of sums does not. Notice that both integrand and integrator are discontinuous at $x = 1/2$, although not both from the left or right.

$$f(x) = \begin{cases}0, \quad 0 \leqslant x < 1/2 \\1, \quad1/2 \leqslant x \leqslant 1 \end{cases}\\ G(x) = \begin{cases}0, \quad 0 \leqslant x \leqslant 1/2 \\1, \quad1/2 < x \leqslant 1 \end{cases}$$

With partition $P = (0,1/2,1)$ we have $U(P,f,G) = L(P,f,G) = 1$. Since upper and lower Darboux integrals satisfy

$$1 = L(P,f,G) \leqslant \underline{\int}_0^1 f \, dG \leqslant \overline{\int}_0^1 f \, dG \leqslant U(P,f,G) = 1,$$

the integral exists with $\displaystyle\int_0^1 f \,dG = 1$.

However, for a sequence of partitions $P_n = \left(0,\frac{1}{2n},\ldots, \frac{1}{2} - \frac{1}{2n} , \frac{1}{2} + \frac{1}{2n}, \ldots ,1\right)$, we have $\|P_n\| = 1/n \to 0$ where tags can be chosen such that $S(P_n,f,G)$ converges either to $0$ or $1$ and there is no unique limit.

• Thank you. I did'nt realize the part about common discontinuity. What are the stronger conditions where the limit of sums works? – WoodWorker Aug 27 '18 at 16:44
• You're welcome. Here you can see that the limit of sums equal the RS integral if the integrator is monotone (or of bounded variation) and either integrator or integrand is continuous. – RRL Aug 27 '18 at 17:53
• Also this answer provides a more detailed proof. – RRL Aug 27 '18 at 19:56