# 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

1. […] Content and Boundaries As a first exercise working with Jordan content, let’s consider how it behaves at the boundary of a […]

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2. […] now for our condition. The function is integrable if and only if the Jordan content for every […]

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3. […] to a very usable condition to determine integrability, so we need to further refine the idea of Jordan content. We want to get away from two restrictions in our definition of Jordan […]

Pingback by Outer Lebesgue Measure « The Unapologetic Mathematician | December 10, 2009 | Reply

4. […] is, as we might expect, a relationship between outer Lebesgue measure and Jordan content, some aspects of which we will flesh out […]

Pingback by Outer Lebesgue Measure and Content « The Unapologetic Mathematician | December 11, 2009 | Reply

5. […] mask we’ll use is the characteristic function of , which I’ve mentioned before. I’ll actually go more deeply into them in a bit, but for now we’ll recall that […]

Pingback by Integrals Over More General Sets « The Unapologetic Mathematician | December 22, 2009 | Reply

6. […] bounds on the integral in terms of the Jordan content of . Incidentally, here is serving a similar role to the integrator in the integral mean value […]

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7. […] but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries […]

Pingback by Integrals are Additive Over Regions « The Unapologetic Mathematician | December 30, 2009 | Reply

8. […] so I’m hoping an example will help illustrate what I mean. Let’s calculate the Jordan content of a three-dimensional sphere of radius , centered at the origin. This will also help cement the […]

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9. […] let be a Jordan measurable subset of , and let be defined and continuous on . Then we have the change of variables […]

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10. […] is injective and continuously differentiable on an open region , and is a compact, connected, Jordan measurable subset of , then we […]

Pingback by The Geometric Interpretation of the Jacobian Determinant « The Unapologetic Mathematician | January 8, 2010 | Reply

11. I need to show that a ball of radius 1 centered at 0 is a Jordan region. Any suggestions? I know that I can just show that it can be covered by a rectangle and I need to show that the boundary has volume zero. But I don’t know how to do this! Thanks

Comment by Brianne | March 24, 2010 | Reply