The Unapologetic Mathematician

Mathematics for the interested outsider

Riemann Integration

Before continuing with methods of antidifferentiation, let’s consider another geometric problem: integration. Here’s an example:

An area to be integrated

We’ve got a function whose graph is drawn in red, and we want to find the area contained between the graph, the x-axis, and the two blue lines at x=3 and x=7. We’ll approximate this by cutting up this interval into n pieces and choosing a sample point t_i in each piece, like so:

Approximating the integral

Now we’ve just got a bunch of rectangles, and we can add up their areas to get


where f(x_i) is the value of the function at the ith sample point, and \Delta_i is the width of the ith strip. Now as we cut the strips thinner and thinner, our stairstep-like approximation to the function should get closer and closer to the real function, and our approximation to the area we’re interested in should get better and better.

So how can we formalize this process? First, let’s take an interval \left[a,b\right] and think about how to cut it up the strips. We do this by picking a collection of points a=x_0<x_1<...<x_{n-1}<x_n=b. We get a bunch of smaller intervals \left[x_{i-1},x_i\right], and in each one we pick some t_i. This structure we call a “tagged partition” of the interval \left[a,b\right]. We define the “mesh” of a partition to be its thickest subinterval, \max\limits_{1\leq i\leq n}(x_i-x_{i-1}), and we’ll want to somehow take this down to zero.

We can now see that the collection of all the tagged partitions of an interval form a directed set! We say that a tagged partition y=((y_0,...,y_m),(s_1,...,s_m)) is a “refinement” of a tagged partition x=((x_0,...,x_n),(t_1,...,t_n)) if every partition point x_i is one of the y_j, and every tag t_i is one of the s_j. That is, we get from x to y by splitting up some of the slices of x and adding new tags to the new slices. Then we define x\preceq y if y is a refinement of x. This makes the collection of tagged partitions into a partially-ordered set.

To show that this is a directed set, consider any two tagged partitions x=((x_0,...,x_n),(t_1,...,t_n)) and y=((y_0,...,y_m),(s_1,...,s_m)), and make a new partition by using all the partition points from each one. Now look at each slice in the new partition. It can’t have more than one t tag or s tag, so it has either zero, one, or two distinct tags. If it has no tags, add one. If it has one tag, do nothing. If it has two distinct tags, split it between them (notice how we’re using the topology of \mathbb{R} to say we can make this split). At the end, we’ve got a new partition that refines both of x and y. And thus we have a directed set.

Now if we have a function f on \left[a,b\right], we can get a net on this directed set. Given any tagged partition x=((x_0,...,x_n),(t_1,...,t_n)), we define the “Riemann sum”

\displaystyle f_x=\sum\limits_{i=1}^nf(t_i)(x_i-x_{i-1})

Finally, we say that the function f is “Riemann integrable” if this net converges to a limit s, and in this case we define the “Riemann integral” of f:

\displaystyle\int\limits_a^b f(x)dx=s

which is, at last, the area under the curve as we set out to find.

January 29, 2008 Posted by | Analysis, Calculus | 17 Comments