# The Unapologetic Mathematician

## Integrating Across a Jump

In the discussion of necessary conditions for Riemann-Stieltjes integrability we saw that when the integrand and integrator are discontinuous from the same side of the same point, the integral can’t exist. But how close can we come to that situation? It turns out that as long as one of the two functions is continuous from each side, then things generally work out.

Specifically, let’s consider jump discontinuities. These are especially useful to understand, since they’re the only sort a function of bounded variation can have. So let’s say we take an interior point $c\in\left(a,b\right)$ and define as simple a function $\alpha$ as we can with a jump there. We let it be constant on either side, with $\alpha(x)=\alpha(a)$ for $x\in\left[a,c\right)$ and $\alpha(x)=\alpha(b)$ for $x\in\left(c,b\right]$. We’ll let $\alpha(c)$ be anything at all. Generally it will be discontinuous from both sides at $c$, but if $\alpha(c)=\alpha(a)$ then we’ll have continuity from the left, and similarly on the right. Of course, we could have $\alpha(a)=\alpha(b)$, with $c$ being the only point with a different value for $\alpha$.

Now let’s let $f$ be any other function on $\left[a,b\right]$. We know that we can’t let it be discontinuous from the left at $c$ if $\alpha$ is, or from the right either, so let’s assume it’s continuous from the left or the right at $c$ to satisfy these conditions, but put no other assumptions on it. I assert that $f$ is then integrable with respect to $\alpha$ on $\left[a,b\right]$, and has the value

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=f(c)\left(\alpha(c^+)-\alpha(c^-)\right)$

Where $\alpha(c^+)$ is the limit of $\alpha$ as we approach $c$ from the right. In our situation this will be $\alpha(b)$, but we’ll telegraph a bit by writing it like this. Similarly, $\alpha(c^-)$ is the limit of $\alpha$ as we approach $c$ from the left, which here is $\alpha(a)$.

To see that this is the case, take a tagged partition ${x}$, and if it doesn’t already contain $c$ just refine it by throwing the new point in. Now every term in the Riemann-Stieltjes sum

$\displaystyle f_{\alpha,x}=\sum\limits_{i=1}^nf(t_i)\left(\alpha(x_i)-\alpha(x_{i-1})\right)$

is zero except for the one on either side of $c=x_k$. We can then write the difference $f_{\alpha,x}-f(c)\left(\alpha(c^+)-\alpha(c^-)\right)$ as

$\displaystyle\left(f(t_{k-1})-f(c)\right)\left(\alpha(c)-\alpha(x_{k-1})\right)+\left(f(t_k)-f(c)\right)\left(\alpha(x_{k+1})-\alpha(c)\right)$

Since either $f$ or $\alpha$ is continuous from the left, the first term above must go to zero. Similarly, because at least one is continuous from the right, the second term must also go to zero. Thus the sums $f_{\alpha,x}$ converge to the value asserted.

March 26, 2008 Posted by | Analysis, Calculus | 3 Comments