# The Unapologetic Mathematician

## Outer Lebesgue Measure

We’re still not quite 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 content.

First, we have some explicit box within which we make our calculations. We showed the choice of such a box is irrelevant, but it’s still inelegant to have to make the choice in the first place. It’s also annoying to have to deal with all the subintervals in the box that aren’t involved in the set at all. More importantly, we’re restricted to cutting the box up into a finite number of pieces. We can gain some flexibility if we allow a countably infinite number of pieces, and allow ourselves to take limits. Limits will also allow us to avoid having to take the closure of the set we’re really interested in. We can still deal with boundary points, because we can take get them in limits of open sets.

Okay, so let $S\subseteq\mathbb{R}^n$ be a set in $n$-dimensional space. A Lebesgue covering of $S$ is a countable collection of open $n$-dimensional intervals $C=\{I_1,I_2,\dots\}$ so that each point of $S$ lies in at least one of the intervals. If $\mathrm{vol}(I_k)$ is the $n$-dimensional volume of the $k$th interval, then the $n$-dimensional volume of the cover is defined as

$\displaystyle\mathrm{vol}(C)=\sum\limits_{k=1}^\infty\mathrm{vol}(I_k)$

as long as this infinite series converges, and define it to be $+\infty$ if the series diverges. We then define the outer Lebesgue measure as the infimum

$\displaystyle\overline{m}(S)=\inf\left\{\mathrm{vol}(C)\right\}$

over all Lebesgue covers $C$ of $S$. This may seem odd, since two intervals in a Lebesgue cover may overlap, and so that volume ends up getting counted twice, but it’s supposed to be an overestimate of the “volume” of $S$, and in practice more refined Lebesgue covers can shrink such overlaps down arbitrarily small.

We also allow for the possibility that there may be only finitely many intervals in our cover. In this case, the infinite series above is simply a finite sum. For example, if $S$ is bounded, we can put it into some large enough interval $[a,b]$, which is then a Lebesgue cover itself. We then find that $0\leq\overline{m}(S)\leq\mathrm{vol}([a,b])$.

We’re most concerned with the case when $\overline{m}(S)=0$, and we then say that $S$ is a set of measure zero. There is a corresponding notion of inner Lebesgue measure, which we will not describe (yet), and the inner measure is always less than the outer measure and greater than zero. Thus if the outer measure is zero, then the inner measure must be as well.

It should also be clear, by the way, that if $S\subseteq T$ then we have the inequality $\overline{m}(S)\leq\overline{m}(T)$.

December 10, 2009 - Posted by | Analysis, Measure Theory

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