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 $\overline{c}(T)=\overline{m}(T)$ .

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December 11, 2009 - Posted by John Armstrong | Analysis, Measure Theory

8 Comments »

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
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
Oh I missed that part. Sorry about the trouble.

NS

Comment by notedscholar | December 13, 2009 | Reply
sweet jesus, a blog where people talk abt math – recreationally?

Comment by none | December 13, 2009 | Reply
- Yes, and you know this already because I recognize your fake email address.
  
  Comment by John Armstrong | December 13, 2009 | Reply
  - oh snaps!!!
    
    lol,
    NS
    
    Comment by notedscholar | December 21, 2009 | Reply
[...] , we also have , and therefore have as [...]

Pingback by Lebesgue’s Condition « The Unapologetic Mathematician | December 15, 2009 | Reply
[...] 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 contained in this boundary, it will also have [...]

Pingback by Integrals Over More General Sets « The Unapologetic Mathematician | December 22, 2009 | Reply

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