# The Unapologetic Mathematician

## The Riemann-Stieltjes Integral IV

Let’s do one more easy application of the Riemann-Stieltjes integral. We know from last Friday that when our integrator is continuously differentiable, we can reduce to a Riemann integral:

$\displaystyle\int\limits_{\left[a,b\right]}f(x)d\alpha(x)=\int\limits_{\left[a,b\right]}f(x)\alpha'(x)dx$

So where else have we seen derivatives as factors in integrands? Right! integration by parts! Here our formula says that

$\displaystyle\int\limits_a^bf(x)g'(x)dx+\int\limits_a^bf'(x)g(x)dx=f(b)g(b)-f(a)g(a)$

We can rewrite this using Riemann-Stieltjes integrals as

$\displaystyle\int\limits_{\left[a,b\right]}f(x)dg(x)+\int\limits_{\left[a,b\right]}g(x)df(x)=f(b)g(b)-f(a)g(a)$

So if $f$ and $g$ are both continuously differentiable, this formula gives back our rule for integration by parts. But we can prove this without making those assumptions. In fact, we just need to assume that one of the two integrals exists, and the existence of the other one (and the formula) will follow.

Let’s assume that $\int_{\left[a,b\right]}f,dg$ exists. That is, for every $\epsilon>0$ there is some tagged partition $x_\epsilon$ so that for every finer partition $x$ we have

$\displaystyle\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-f_{g,x}\right|<\epsilon$

Now let’s take any partition $x$ finer than $x_\epsilon$ and use it to set up the Rieman-Stieltjes sum

$\displaystyle g_{f,x}=\sum\limits_{i=1}^ng(t_i)f(x_i)-\sum\limits_{i=1}^ng(t_i)f(x_{i-1})$

We can also use this partition to rewrite $A=f(b)g(b)-f(a)g(a)$ as

$\displaystyle A=\sum\limits_{i=1}^nf(x_i)g(x_i)-\sum\limits_{i=1}^nf(x_{i-1})g(x_{i-1})$

So subtracting the one from the other we find

$\displaystyle A-g_{f,x}=\sum\limits_{i=1}^nf(x_i)(g(x_i)-g(t_i))+\sum\limits_{i=1}^nf(x_{i-1})(g(t_i)-g(x_{i-1}))$

But this is a Riemann-Stieltjes sum $f_{g,\bar{x}}$ for the partition $\bar{x}$ we get by throwing together all the $x_i$ and $t_i$ as partition points, and using $x_i$ as tags. This is a finer partition than $x_\epsilon$, and so we see that

$\displaystyle\left|\left(A-\int\limits_{\left[a,b\right]}f(x)dg(x)\right)-g_{f,x}\right|=\left|\int\limits_{\left[a,b\right]}f(x)dg(x)-\left(A-g_{f,x}\right)\right|<\epsilon$

whenever $x$ is a partition finer than $x_\epsilon$. This shows that the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and has the value we want.