# The Unapologetic Mathematician

## 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 $i$th 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