Jordan Content and Boundaries

As a first exercise working with Jordan content, let’s consider how it behaves at the boundary of a region.

I’ve used this term a few times when it’s been pretty clear from context, but let me be clear. We know about the interior and closure of a set, and particularly of a subset of $\mathbb{R}^n$ . The boundary of such a set will consist of all the points in the closure of the set, but not in its interior. That is, we have

$\mathrm{Cl}(S)=\mathrm{int}(S)\uplus\partial S$

That is, a point $x$ is in $\partial S$ if $x\in\mathrm{Cl}(S)$ , but any neighborhood of $x$ contains points not in $S$ .

So let’s put $S$ inside a box $[a,b]$ and partition the box with a partition $P$ . When we calculate $\overline{J}(P,S)$ , we include all the subintervals that we do when we calculate $\underline{J}(P,S)$ , along with some other subintervals which contain both points within $\mathrm{int}(S)$ and points not in $\mathrm{int}(S)$ . I say that each of these are exactly the subintervals which contain points in $\partial S$ . Indeed, if a subinterval contains a point of $\partial S$ it cannot be included when calculating $\underline{J}(P,S)$ , but must be included when calculating $\overline{J}(P,S)$ . Inversely, if a subinterval contains no point of $\partial S$ then it is either contained completely within $\mathrm{int}(S)$ — and is included in both calculations — or it is contained completely within the complement of $\mathrm{Cl}(S)$ — and is contained in neither computation. Thus we have the relation

$\displaystyle\overline{J}(P,S)=\underline{J}(P,S)+\overline{J}(P,\partial S)$

which we rewrite

$\displaystyle\overline{J}(P,\partial S)=\overline{J}(P,S)-\underline{J}(P,S)$

We can then pass to infima to find

$\displaystyle\overline{c}(\partial S)\geq\overline{c}(S)-\underline{c}(S)$

Check this carefully to see how the equality is weakened to an inequality.

On the other hand, given any $\epsilon>0$ we can pick partitions $P_1$ so that $\overline{J}(P_1,S)<\overline{c}(S)+\frac{\epsilon}{2}$ and $P_2$ so that $\underline{J}(P_2,S)>\underline{c}(S)-\frac{\epsilon}{2}$ . We let $P$ be a common refinement of $P_1$ and $P_2$ , which will then satisfy both of these inequalities. We find

$\displaystyle\overline{c}(S)\leq\overline{J}(P,\partial S)=\overline{J}(P,S)-\underline{J}(P,S)<\overline{c}(S)-\underline{c}(S)+\epsilon$

since $\epsilon$ is arbitrary, we find $\overline{c}(\partial S)\leq\overline{c}(S)-\underline{c}(S)$ .

Together with the previous result, we conclude that $\overline{c}(\partial S)=\overline{c}(S)-\underline{c}(S)$ . In particular, we find that $S$ is Jordan measurable if and only if $c(\partial S)=\overline{c}(\partial S)=0$ .

6 Comments »

A point x is in the bdry of S if any neighborhood of x contains points in S and pts not in S …

Comment by Kevin | December 5, 2009 | Reply
Sorry, something was left out of a revision.. I’ll fix it to something that’s equivalent to what you just said, KEvin.

Comment by John Armstrong | December 5, 2009 | Reply
[…] . We must ask that be Jordan measurable, because this will happen if and only if the boundary has zero Jordan content, and thus zero outer Lebesgue measure. Since the collection of new discontinuities must be […]

Pingback by Integrals Over More General Sets « The Unapologetic Mathematician | December 22, 2009 | Reply
[…] 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 in an integral will make no […]

Pingback by Integrals are Additive Over Regions « The Unapologetic Mathematician | December 30, 2009 | Reply
[…] notice how the last step of this proof echoes our earlier result that a set is Jordan measurable if and only if the Jordan content of its boundary is […]

Pingback by Using Measurable Covers and Kernels I « The Unapologetic Mathematician | April 12, 2010 | Reply
[…] The original motivation was a set like a boundary of a region; in the context of Jordan content we saw that such a set was […]

Pingback by Baire Sets « The Unapologetic Mathematician | August 13, 2010 | Reply

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