# The Unapologetic Mathematician

## Step Function Integrators

Now that we know how a Riemann-Stieltjes integral behaves where the integrand has a jump, we can put jumps together into more complicated functions. The ones we’re interested in are called “step functions” because their graphs look like steps: flat stretches between jumps up and down.

More specifically, let’s say we have a sequence of points

$\displaystyle a\leq c_1

and define a function $\alpha$ to be constant in each open interval $\left(c_{i-1},c_i\right)$. We can have any constant values on these intervals, and any values at the jump points. The difference $\alpha(c_k^+)-\alpha(c_k^-)$ we call the “jump” of $\alpha$ at $c_k$. We have to be careful here about the endpoints, though: if $c_1=a$ then the jump at $c_1$ is $\alpha(c_1^+)-\alpha(c_1)$, and if $c_n=b$ then the jump at $c_n$ is $\alpha(c_n)-\alpha(c_n^-)$. We’ll designate the jump of $\alpha$ at $c_k$ by $\alpha_k$.

So, as before, the function $\alpha$ may or may not be continuous from one side or the other at a jump point $c_k$. And if we have a function $f$ discontinuous on the same side of the same point, then the integral can’t exist. So let’s consider any function $f$ so that at each $c_k$, at least one of $\alpha$ or $f$ is continuous from the left, and at least one is continuous from the right. We can chop up the interval $\left[a,b\right]$ into chunks so that each one contains only one jump, and then the result from last time (along with the “linearity” of the integral in the interval) tells us that

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha=\sum\limits_{i=1}^nf(c_k)\alpha_k$

That is, each jump gives us the function at the jump point times the jump at that point, and we just add them all together. So finite weighted sums of function evaluations are just special cases of Riemann-Stieltjes integrals.

Here’s a particularly nice family of examples. Let’s start with any interval $\left[a,b\right]$ and some natural number $n$. Define a step function $\alpha_n$ by starting with $\alpha_n(a)=a$ and jumping up by $\frac{b-a}{n}$ at $a+\frac{1}{2}\frac{b-a}{n}$, $a+\frac{3}{2}\frac{b-a}{n}$, $a+\frac{5}{2}\frac{b-a}{n}$, and so on. Then the integral of any continuous function on $\left[a,b\right]$ gives

$\displaystyle\int\limits_{\left[a,b\right]}fd\alpha_n=\sum\limits_{i=1}^nf\left(a+(i-\frac{1}{2})\frac{b-a}{n}\right)\frac{b-a}{n}$

But notice that this is just a Riemann sum for the function $f$. Since $f$ is continuous, we know that it’s Riemann integrable, and so as $n$ gets larger and larger, these Riemann sums must converge to the Riemann integral. That is

$\displaystyle\lim\limits_{n\to\infty}\int\limits_{\left[a,b\right]}fd\alpha_n=\int\limits_{\left[a,b\right]}fdx$

But at the same time we see that $\alpha_n(x)$ converges to ${x}$. Clearly there is some connection between convergence and integration to be explored here.

March 27, 2008 - Posted by | Analysis, Calculus

## 1 Comment »

1. […] though, that we’ve seen a way to get finite sums before: using step functions as integrators. So let’s use the step function , which is defined for any real number as the largest […]

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