The Unapologetic Mathematician

Mathematics for the interested outsider

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.

About these ads

March 26, 2008 - Posted by | Analysis, Calculus

3 Comments »

  1. Hi, Been following your introductory analysis posts with some interest, since they’ve mirrored the first couple of years of undergrad. analysis we did – but in a more logically consistent order (that’s not to say your choice of order would be simpler if the reader hadn’t seen most of it before!).

    Anyhow, I wanted to say keep up the good work and please don’t abandon the category theory posts. Also, some of these analysis concepts might be a bit more obvious to someone who hadn’t seen them before with a couple of pictures – a graph of something with a jump discontinuity is quite doable, and might make the text easier to understand: “You what?!, a(a) er… oh it’s that line there, ah now I understand” sort of thing.

    Comment by Rupert Swarbrick | March 27, 2008 | Reply

  2. A graph is quite doable if the only graphing software I had on here wasn’t so old and awkward to use. Eventually I’ll manage to go back and tweak, but for now it’s really more difficult than you’d expect.

    As for the category theory, I’ve got a ways to go before I can get back to that line.

    Comment by John Armstrong | March 27, 2008 | Reply

  3. [...] 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 [...]

    Pingback by Step Function Integrators « The Unapologetic Mathematician | March 14, 2009 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 391 other followers

%d bloggers like this: