# The Unapologetic Mathematician

## Sufficient Conditions for Integrability

Let’s consider some conditions under which we’ll know that a given Riemann-Stieltjes integral will exist. First off, we have a straightforward adaptation of our old result that continuous functions are Riemann integrable. Now I assert that any continuous function $f$ on an interval $\left[a,b\right]$ is Riemann-Stieltjes integrable over that interval with respect to any function $\alpha$ of bounded variation on the same interval. In particular, the function $\alpha(x)=x$ is clearly of bounded variation, and so we will recover our old result.

In fact, we can even adapt the old proof. The Heine-Cantor theorem says that the function $f$, being continuous on the compact interval $\left[a,b\right]$ is uniformly compact. As usual, we can assume that $\alpha$ is increasing on $\left[a,b\right]$. And now Riemann’s condition tells us to consider the difference

$\displaystyle U_{\alpha,x}(f)-L_{\alpha,x}(f)=\sum\limits_{i=1}^n\left(M_i(f)-m_i(f)\right)\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

We want this difference to go to zero as we choose finer and finer partitions $x$

By uniform continuity we can pick a small enough $\delta$ (depending only on $\epsilon$) so that when $|x-y|<\delta$ we’ll have $|f(x)-f(y)|<\frac{\epsilon}{V}$, where $V$ is the total variation $\alpha(b)-\alpha(a)$. Then picking a partition whose subintervals are thinner than $\delta$ makes it so that $M_i(f)-m_i(f)<\frac{\epsilon}{V}$, which we can then pull out of the sum. What remains sums to exactly the total variation $V$, and so the difference $U_{\alpha,x}(f)-L_{\alpha,x}(f)$ is below $\epsilon$, and our theorem holds.

Immediately from this result and integration by parts we come up with another set of sufficient conditions. If $\alpha$ is continuous and $f$ is of bounded variation on an interval $\left[a,b\right]$ then $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ over $\left[a,b\right]$. Since the integrator $\alpha(x)=x$ is also continuous, this tells us that any function of bounded variation is Riemann integrable!

Of course, these conditions are just sufficient. That is, if they hold then we know that the integral exists. However, if an integral exists, we can’t use these to conclude anything about either the integrand or the integrator. For that we need necessary conditions.

March 24, 2008 - Posted by | Analysis, Calculus