# The Unapologetic Mathematician

## 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

1. Hey John. Sorry I haven’t commented for a while. I was wondering, what is the (or a?) Lebesgue measure?

Thanks in advancement and happy holidays,
NS

Comment by notedscholar | December 13, 2009 | Reply

2. I haven’t defined it, and I don’t really intend to. I was pretty clear on this point when I started talking about Jordan content.

Comment by John Armstrong | December 13, 2009 | Reply

3. Oh I missed that part. Sorry about the trouble.

NS

Comment by notedscholar | December 13, 2009 | Reply

4. sweet jesus, a blog where people talk abt math – recreationally?

Comment by none | December 13, 2009 | Reply