The Unapologetic Mathematician

Mathematics for the interested outsider

Darboux Integration

Okay, defining the integral as the limit of a net of Riemann sums is all well and good, but it’s a huge net, and it seems impossible to calculate with. We need a better way of getting a handle on these things. What we’ll use is a little trick for evaluating limits of nets that I haven’t mentioned yet: “cofinal sets”.

Given a directed set (D,\preceq), a directed subset S is cofinal if for every d\in D there is some s\in S with s\succeq d. Now watch what happens when we try to show that the limit of a net x_d is a point x. We need to find for every neighborhood U of x an index d_0 so that for every d\succeq d_0 we have x_d\in U. But if d_0 is such an index, then there is some s_0\in S above it, and every s\in S above that is also above d_0, and so x_s\in U. That is, if the limit over D exists, then the limit over S exists and has the same value.

Let’s give a cofinal set of tagged partitions by giving a rule for picking the tags that go with any partition. Then our net consists just of partitions of the interval \left[a,b\right], and the tags come for free. If the function f is Riemann-integrable, then the limit over this cofinal set will be the integral. Here’s our rule: in the closed subinterval \left[x_{i-1},x_i\right] pick a point t_i so that \lim\limits_{x\rightarrow t_i}f(x) is the supremum of the values of f in that subinterval. If the function is continuous it will attain a maximum at our tag, and if not it’ll get close or shoot off to infinity (if there is no supremum).

Why is this cofinal? Let’s imagine a tagged partition x=((x_0,...,x_n),(t_1,...,t_n)) where t_i is not chosen according to this rule. Then we can refine the partition by splitting up the ith strip in such a way that t_i is the maximum in one of the new strips, and choosing all the new tags according to the rule. Then we’ve found a good partition above the one we started with. Similarly, we can build another cofinal set by always choosing the tags where f approaches an infimum.

When we consider a partition x in the first cofinal set we can set up something closely related to the Riemann sums: the “upper Darboux sums”

\displaystyle U_x(f)=\sum\limits_{i=1}^n M_i(x_i-x_{i-1})

where M_i is the supremum of f(x) on the interval \left[x_{i-1},x_i\right], or infinity if the value of f is unbounded above here. Similarly, we can define the “lower Darboux sum”

\displaystyle L_x(f)=\sum\limits_{i=1}^n m_i(x_i-x_{i-1})

where now m_i is the infimum (or negative infinity). If the function is Riemann-integrable, then the limits over these cofinal sets both exist and are both equal to the Riemann integral. So we define a function to be “Darboux-integrable” if the limits of the upper and lower Darboux sums both exist and have the same value. Then the Darboux integral is defined to be this common value. Notice that if the function ever shoots off to positive or negative infinity we’ll get an infinite value for one of the terms, and we can never converge, so such functions are not Darboux-integrable.

We should notice here that given any partition x, the upper Darboux sum must be larger than any Riemann sum with that same partition, since no matter how we choose the tag t_i we’ll find that f(t_i)\leq M_i by definition. Similarly, the lower Darboux sum must be smaller than any Riemann sum on the same partition. Now let’s say that the upper and lower Darboux sums both converge to the same value s. Then given any neighborhood of s we can find a partition x_U so that every upper Darboux sum over a refinement of x_U is in the neighborhood, and a similar partition x_L for the lower Darboux sums. Choosing a common refinement x_R of both (which we can do because partitions form a directed set) both its upper and lower Darboux sums (and those of any of its refinements) will be in our neighborhood. Then we can choose any tags in x_R we want, and the Riemann sum will again be in the neighborhood. Thus a Darboux-integrable function is also Riemann-integrable.

So this new notion of Darboux-integrability is really the same one as Riemann-integrability, but it involves taking two limits over a much less complicated directed set. For now, we’ll just call a function which satisfies either of these two equivalent conditions “integrable” and be done with it, using whichever construction of the integral is most appropriate to our needs at the time.

January 30, 2008 Posted by | Analysis, Calculus, Orders | 3 Comments

   

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