How to use a simple function in Stieltjes integral to approximate Riemann integral?

Question

The question if very straightforward. However, I am not sure that I can write a very strict proof. The goal is to give a series of simple function $\beta_n$ such that $$\lim_{n \to \infty}\int_a^b fd\beta_n=\int_a^b fdx$$ I think the construction of the simple function has a very simple idea, namely for $\beta_n$, we simply cut the amount $b-a$ into $n$ pieces and assign those values with equal step to our simple function. Then $\beta_n$ will tend to function $f=x$ naturally. I just have some trouble in writing a nice and neat proof. I appreciate your help.

Answer 1

Consider the following step function as an integrator

$$\beta_n (x) = \sum_{k=1}^{n+1} x_{k-1}\chi_{[x_{k-1}, x_k)},$$

where $x_k = a + (b-a)k/n$ for $k = 1,2 ,\ldots,n$.

Then the Riemann-Stieltjes integral is

$$\int_a^bf\,d\beta_n = \sum_{k=1}^nf(x_k)[\beta_n(x_k+)-\beta_n(x_k-)]\\= \sum_{k=1}^nf(x_k)\frac{b-a}{n}= \sum_{k=1}^nf(x_k)(x_k-x_{k-1}).$$

Hence, the limit of the Riemann-Stieltjes integral reduces to the limit of a Riemann sum as the partition mesh tends to $0$ and

$$\lim_{n \to \infty}\int_a^bf \, d \beta_n = \lim_{n \to \infty}\sum_{k=1}^nf(x_k)(x_k-x_{k-1})= \int_a^bf(x) \, dx.$$