# The Unapologetic Mathematician

## Lebesgue Measurable Sets

The attentive reader will note that in our study of Lebesgue measure we’ve defined it on some complete $\sigma$-algebra $\overline{\mathcal{S}}\subseteq P(X)$. In general, there’s no reason to believe that such a $\sigma$-algebra would be all of $P(X)$, and so there’s some interesting structure worth investigating. Our intuition is amazingly bad at conceiving of just how bizarre and perverse a “general” subset of the real line can be, and so the facts we learn about Lebesgue measurable sets seem sort of opaque when we first see them, but they will turn out to be useful.

First, let’s recall where they come from. There’s an outer measure $\mu^*$ associated with Lebesgue measure $\bar{\mu}$, and then there is the collection of sets measurable by $\mu^*$. These are the sets $E$ so that for any subset $A\subseteq\mathbb{R}$ we have $\displaystyle\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$

If $E$ is a Lebesgue measurable set of positive, finite measure, and if $\alpha\in\left[0,1\right)$ is a real number, then there exists some open interval $U$ so that $\displaystyle\bar{\mu}(E\cap U)>\alpha\mu(U)$

That is, Lebesgue measurable sets do a pretty good job of covering sections of the real line. If we think of $\alpha$ as a measure of efficiency, we can always find some interval so that $E$ covers at least that portion of it.

Let $\mathcal{U}$ is the class of open sets. We already know that $\displaystyle\bar{\mu}(E)=\inf\{\mu(U)\vert U\in\mathcal{U},E\subseteq U\}$

So we can find some open $U_0$ containing $E$ and satisfying $\alpha\mu(U_0)\leq\bar{\mu}(E)$. This will be the disjoint union of a sequence of open intervals $\{U_i\}$, and so it follows that $\displaystyle\alpha\sum\limits_{i=1}^\infty\mu(U_i)\leq\sum\limits_{i=1}^\infty\bar{\mu}(E\cap U_i)$

We must have $\mu(U_i)\leq\bar{\mu}(E\cap U_i)$ for at least one of these summands, and this gives us the $U$ we were looking for.

If $E$ is again a Lebesgue measurable set with positive measure, then there exists an open interval containing ${0}$ and entirely contained in the set $D(E)$ of “differences” in $E$. That is, the set defined by $\displaystyle D(E)=\{x-y\vert x,y\in E\}$

If $E$ contains an open interval $(a,b)$ itself, then this is trivially true. We can just take the desired interval to be $(a-b,b-a)$, which clearly falls into the difference set $D(E)$. But general Lebesgue measurable sets don’t contain open intervals. And yet this is not just a collection of scattered points, because then it would have measure zero. We have points clustered closely enough to fill up area, but not enough to actually fill out any open intervals. This is, in fact, supremely weird if you really think about it.

But even if $E$ doesn’t fill up an interval, it comes close! The above result tells us that we can find a $U$ so that $\displaystyle\bar{\mu}(E\cap U)\geq\frac{3}{4}\mu(U)$

Now pick some $x\in\left(-\frac{1}{2}\mu(U),\frac{1}{2}\mu(U)\right)$. Take the set $E\cap U$, slide it over by adding $x$ to every point in the set, and call the result $(E\cap U)+x$. The set $(E\cap U)\cup((E\cap U)+x)$ is now contained in an open interval — $U\cup(U+x)$ — whose length is less than $\frac{3}{2}\mu(U)$. If $E\cap U$ and $(E\cap U)+x$ were disjoint, then they’d have to have the same measure, and their union would have measure greater than $\frac{3}{2}\mu(U)$ and wouldn’t fit into this interval. So $E\cap U$ and $(E\cap U)+x$ must overlap a bit.

But this tells us that there’s a point $a\in(E\cap U)$ that’s equal to the point $b+x\in((E\cap U)+x)$. Thus $a-b=x$ is a point in $D(E)$. And so $D(E)$ contains the whole interval $\left(-\frac{1}{2}\mu(U),\frac{1}{2}\mu(U)\right)$, as we wanted.

April 23, 2010 Posted by | Analysis, Measure Theory | 1 Comment