The Unapologetic Mathematician

Mathematics for the interested outsider

Outer Lebesgue Measure and Content

Before I begin, I’d like to mention something in passing about Lebesgue measure. It’s pronounced “luh-bayg”. The “e” is a long “a”, the “s” is completely silent, and the “gue” is like in “analogue”. Moving on…

There is, as we might expect, a relationship between outer Lebesgue measure and Jordan content, some aspects of which we will flesh out now.

First off, if S\mathbb{R}^n is a bounded subset of n-dimensional Euclidean space, then we have \overline{m}(S)\leq\overline{c}(S). Indeed, if it’s bounded, then we can put it into a box [a,b], and choose a partition P of this box. List out the n-dimensional intervals of P which contain points of \mathrm{Cl}(S) as I_k=[a_k,b_k] for 1\leq k\leq m. Then by definition we have

\displaystyle\overline{J}(P,S)=\sum\limits_{k=1}^m\mathrm{vol}(I_k)

Now given an \epsilon>0, define the open n-dimensional intervals C_k=(a_k-\frac{\epsilon}{2m},b_k+\frac{\epsilon}{2m}). These form a Lebesgue covering C of S for which

\displaystyle\mathrm{vol}(C)=\sum\limits_{k=1}^m\mathrm{vol}(C_k)=\overline{J}(P,S)+\epsilon

Thus \overline{m}(S)\leq\overline{J}(P,S)+\epsilon, and passing to the infimum we find \overline{m}(S)\leq\overline{c}(S)+\epsilon. Since \epsilon is arbitrary, we have \overline{m}(S)\leq\overline{c}(S).

Secondly, if S\subseteq\mathbb{R}^n is bounded, and T\subseteq S is a compact subset, then \overline{c}(T)\leq\overline{m}(S). Start by putting S into a box [a,b], and take some \epsilon>0. We can find a Lebesgue covering C of S so that \mathrm{vol}(C)<\overline{m}(S)+\epsilon, and this will also cover T. Since T is compact, we can pick out a finite collection C' of open intervals which still manage to cover T. Finally, we can choose a partition P of [a,b] so that the corners of each interval in C' are partition points. Given all of these choices, we find

\displaystyle\overline{c}(T)\leq\overline{J}(P,T)\leq\mathrm{vol}(C')\leq\mathrm{vol}(C)<\overline{m}(S)+\epsilon

And since \epsilon is arbitrary we conclude \overline{c}(T)\leq\overline{m}(S).

Finally, putting these two results together we can see that if T\subseteq\mathbb{R}^n is a compact set, then \overline{c}(T)=\overline{m}(T).

December 11, 2009 Posted by | Analysis, Measure Theory | 8 Comments

   

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