# The Unapologetic Mathematician

## The Riemann-Stieltjes Integral III

Last Friday we explained the change of variables formula for Riemann integrals by using Riemann-Stieltjes integrals. Today let’s push it a little further and prove a change of variables formula for Riemann-Stieltjes integrals.

We start with a function $f:\left[a,b\right]$ which we assume to be Riemann-Stieltjes integrable by the function $\alpha$. Now, instead of the full generality we used before, let’s just let $g:\left[c,d\right]\rightarrow\left[a,b\right]$ be a strictly increasing continuous function with $g(c)=a$ and $g(d)=b$. Define $h$ and $\beta$ to be the composite functions $h(x)=f(g(x))$ and $\beta(x)=\alpha(g(x))$. Then $h$ is Riemann-Stieltjes integrable by $\beta$ on $\left[c,d\right]$, and we have the equality

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\int\limits_{\left[c,d\right]}hd\beta$

For decreasing functions we get almost the exact same thing, so you should figure out the parallel statement and proof yourself.

Since $g$ is strictly increasing, it must be one-to-one, and it’s onto by assumption. In fact, $g$ is an explicit homeomorphism of the intervals $\left[a,b\right]$ and $\left[c,d\right]$, and its inverse $g^{-1}:\left[a,b\right]\rightarrow\left[c,d\right]$ is also a strictly increasing continuous function. We can now use $g$ and its inverse to set up a bijection between partitions of $\left[a,b\right]$ and $\left[c,d\right]$: if $a=x_0 is a partition, then $c=g^{-1}(x_0) is a partition, and vice versa. Further, refinements of partitions of one side correspond to refinements of partitions on the other side.

So if we’re given an $\epsilon>0$ then there’s some partition $y_\epsilon$ of $\left[a,b\right]$ so that for any finer partition $y$ we have $|f_{\alpha,y}-\int_{\left[a,b\right]}f,d\alpha|<\epsilon$. Let $x_\epsilon=g^{-1}(y_\epsilon)$ be the corresponding partition of $\left[c,d\right]$, and let $x$ be a partition of $\left[c,d\right]$ finer than it. Then it’s easily verified that the Riemann-Stieltjes sum $h_{x,\beta}$ is equal to the Riemann-Stieltjes sum $f_{g(x),\alpha}$. Everything else follows quickly from here.