The Unapologetic Mathematician

Mathematics for the interested outsider

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.

December 4, 2009 - Posted by | Analysis, Measure Theory


  1. 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

  2. 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

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