# The Unapologetic Mathematician

## Jordan Content

We’ve got Riemann’s condition, which is necessary and sufficient to say that a given function is integrable on a given interval. In one dimension, this really served us well, because it let us say that continuous functions were integrable. We could break a piecewise continuous function up into intervals on which it was continuous, and thus integrable, and this got us almost every function we ever cared about.

But in higher dimensions it’s not quite so nice. The region on which a function is continuous may well be irregular, and it often is for many functions we’re going to be interested. We need a more robust necessary and sufficient condition than Riemann’s. And towards this end we need to introduce a few concepts from measure theory. I want to say at the outset, though, that this will not be a remotely exhaustive coverage of measure theory (yet). Mostly we need to build up the concept of a set in a Euclidean space having Lebesgue measure zero, and the related notion of Jordan content.

So let’s say we’ve got a set bounded $S\subseteq\mathbb{R}^n$. Put it inside an $n$-dimensional box — an interval $[a,b]$. Then partition this interval with some partition $P\in\mathcal{P}[a,b]$ like we did when we set up higher-dimensional Riemann sums. Then we just count up all the subintervals in the partition $P$ that are completely contained in the interior $\mathrm{int}(S)$ and add their volumes together. This is the volume of some collection of boxes which is completely contained within $S$, and we call this volume $\underline{J}(P,S)$.

Any reasonable definition of “volume” would have to say that since $S$ contains this collection of boxes, the volume of $S$ must be greater than the volume $\underline{J}(P,S)$. Further, as we refine $P$ any box which was previously contained in $S$ will still have all of its subdivisions contained in $S$, but boxes which were only partially contained in $S$ may now have subdivisions completely contained in $S$. And so if $P'$ is a refinement of $P$, then $\underline{J}(P',S)\geq\underline{J}(P,S)$. That is, we have a monotonically increasing net. We then define the “lower Jordan content” of $S$ by taking the supremum

$\displaystyle\underline{c}(S)=\sup\limits_{P\in\mathcal{P}[a,b]}\underline{J}(P,S)$

sort of like how the lower integral is the supremum of all the lower Riemann sums. In fact, this is the lower integral of the characteristic function $\chi_S$, which is defined to be ${1}$ on elements of $S$ and ${0}$ elsewhere. If a subinterval contains any points not in $S$ the lowest sample will be ${0}$, but otherwise the sample must be ${1}$, and the lower Riemann sum for the subdivision $P$ is $L_P(\chi_S)=\underline{J}(P,S)$.

It should be clear, by the way, that this doesn’t really depend on the interval $[a,b]$ we start with. Indeed, if we started with a different interval $[a',b']$ that also contains $S$, then we can immediately partition each one so that one subinterval is the intersection with the other one, and this intersection contains all of $S$ anyway. Then we can throw away all the other subintervals in these initial partitions because they don’t touch $S$ at all, and the rest of the calculations proceed exactly as before, and we get the same answer in either case.

Similarly, given a partition $P$ of an interval $[a,b]$ containing a set $S$ we could count up the volumes of all the subintervals that contain any point of the closure $\mathrm{Cl}(S)$ at all. This gives the volume of a collection of boxes that together contains all of $S$, and we call this total volume $\overline{J}(P,S)$. Again, since the whole of $S$ is contained in this collection of boxes, the volume of $S$ will be less than $\overline{J}(P,S)$. And this time as we refine the partition we may throw out subdivisions which no longer touch any point of $S$, so this net is monotonically decreasing. We define the “upper Jordan content” of $S$ by taking the infimum

$\displaystyle\overline{c}(S)=\inf\limits_{P\in\mathcal{P}[a,b]}\overline{J}(P,S)$

Again, this doesn’t depend on the original interval $[a,b]$ containing $S$, and for the same reason. As before, we find that an upper Riemann sum for the characteristic function of $S$ is $U_P(\chi_S)=\overline{J}(P,S)$, so the upper Jordan content is the upper integral of $\chi_S$.

If $S$ is well-behaved, we will find $\overline{c}(S)=\underline{c}(S)$. In this case we say that $S$ is “Jordan measurable”, and we define the Jordan content $c(S)$ to be this common value. By Riemann’s condition, we find that

$\displaystyle c(S)=\int\limits_{[a,b]}\chi_S(x)\,dx$

December 3, 2009 Posted by | Analysis, Measure Theory | 11 Comments