# How is the Riemann integral a special case of the Stieltjes integral?

From Rudin's Principles of mathematical analysis,

6.2 Definition

Let $$\alpha$$ be a monotonically increasing function on $$[a,b]$$. ... Corresponding to each partition $$P$$ of $$[a,b]$$, we write

$$\Delta \alpha_i = \alpha(x_i) - \alpha(x_{i-1}).$$

He then goes on to define the Riemann Stieltjes integral of $$f$$ with respect to $$\alpha$$, over the interval $$[a,b]$$.

The Riemann integral is then pointed out to be a special case of this when $$\alpha(x)=x$$.

With $$\alpha(x)=x$$, I understand $$\Delta x = x_i - x_{i-1}$$ to represent the directed magnitude of the "base of the approximating rectangle" that we then multiply by the value of $$f$$ taken somewhere within this interval, thus obtaining the area of an approximating rectangle.

I don't know where to begin to interpret the case where $$\alpha(x) \not\equiv x$$.

• Integrating with respect to a function of bounded variation is a Borel measure. It's a weight function: where a(x) is increasing faster, the more heavily the integral is weighted towards the area under that part of the curve. It's like a CDF, and the integral is weighted wrt to the corresponding density. The idea is that defining it this way gives you a clearer understanding of what things are integrable, because you can think outside the box a bit more; R-S is the natural way to show that all maps from C([a,b])->R are integrals. Jun 3, 2011 at 16:46

The idea is that we define a different way of measuring the width of the rectangle. Instead of using the difference of the two coordinates, we take the difference of the images of those two coordinates under the map $\alpha$. Basically, the idea is that the $(x_i-x_{i-1})$ is to be interpreted as the width or measure of the interval, a construction which leads more generally to the Lebesgue measure of a subset of the reals. The $\alpha(x_i)-\alpha(x_{i-1})$ leads to a different measure, starting with a different assignment of "size" to an interval.
Note that when $\alpha$ is continuously differentiable, then $\alpha(x_i) -\alpha(x_{i-1})$ is equal to $\alpha'(x) (x_i - x_{i-1})$ for $x \in [x_{i-1}, x_i]$ up to higher order terms. So in this case the Stieltjes integral with respect to $\alpha$ is the Riemann integral of the same function times $\alpha'$.