# The Unapologetic Mathematician

## Non-Lebesgue Measurable Sets

I need to make up for missing a post earlier this week…

The most important observation about the fact that Lebesgue measurable sets might not be all of $P(X)$ is sort of tautological: it means that there may be subsets of the real line which are not Lebesgue measurable. That is, sets for which it is impossible to give a sense of “how much space they take up”, in a way compatible with the length of an interval.

And we can show that such sets do, in fact, exist. At least, we can build them if we have use of the axiom of choice. This might seem like a reason not to use the axiom of choice, but remember that Zorn’s lemma — which is equivalent to the axiom of choice — was essential when we needed to show that every vector space has a basis, or Tychonoff’s theorem, or that exact sequences of vector spaces split. So it’s sort of a mixed bag. In practice, most working mathematicians seem to be willing to accept the existence of non-Lebesgue measurable sets in order to gain the above benefits.

So, first a lemma: if $\xi$ is irrational, then the set $A$ of all numbers of the form $n+m\xi$ with $n$ and $m$ any integers is dense in the real line. That is, every open interval $U$ contains at least one point of $A$. The same is true for the set $B$ where we restrict $n$ to be even, and for the set $C$ where we restrict $n$ to be odd. Note $A$ (and, incidentally, $B$) is actually a subgroup of the additive group of real numbers.

For every integer $i$ there is some unique integer $n_i$ so that $0\leq n_i+i\xi<1$; we will write $x_i=n_i+i\xi$. If $U$ is an open interval, there is some positive integer $k$ with $\mu(U)>\frac{1}{k}$. Picking out the first $k+1$ numbers $x_1,\dots,x_{k+1}$, there must be some pair $x_i$ and $x_j$ with $\lvert x_i-x_j\rvert<\frac{1}{k}$ (or else they wouldn’t all fit in the interval $\left[0,1\right)$). But then some multiple of $x_i-x_j$ must land within $U$, as we asserted. For $B$, we can do the same using the interval $\left[0,2\right)$, and for $C$ we can use the fact that $C=B+1$.

Now, I say that there exists at least one set $E_0$ which is not Lebesgue measurable. To show this, we consider the quotient group $\mathbb{R}/A$. That is, we use an equivalence relation $x\sim y$ if $x-y\in A$. This divides up the real numbers into a disjoint union of equivalence classes under this relation, and the axiom of choice allows us to build a set $E_0$ by picking exactly one point from each equivalence class. This is the set we will show is not Lebesgue measurable.

Suppose $F$ is a Borel set contained in $E_0$. The difference set $D(F)$ contains no point of $A$, since if this happened we’d have two points in $E_0$ picked from the same $\sim$-equivalence class. But we just saw that any open interval contains a point of $A$, and so our result from last time shows that $F$ must have outer measure zero — if it had positive outer measure then $D(F)$ would contain an open interval. And so if $E_0$ is Lebesgue measurable then its Lebesgue measure must be zero.

Now if $a_1$ and $a_2$ are distinct elements of $A$, then $E_0+a_1$ and $E_0+a_2$ must be disjoint. As we let $a$ range over the countable number of values in $A$, the sets $E_0+a$ then form a countable disjoint cover of $\mathbb{R}$. But each of the $E_0+a$ is just a translation of $E_0$, and so each one must have the same measure. And then since Lebesgue measure is countably additive, we must have

$\displaystyle\bar{\mu}\left(\mathbb{R}\right)=\bar{\mu}\left(\bigcup\limits_{a\in A}E_0+a\right)=\sum\limits_{a\in A}\bar{\mu}(E_0+a)=\sum\limits_{a\in A}0=0$

But this is clearly nonsense.

We can do even better, actually, in our efforts to find bizarre sets. There exists a subset $M\subseteq\mathbb{R}$ so that for every Lebesgue measurable set $E$ we have both

$\displaystyle\mu_*(M\cap E)=0$
$\displaystyle\mu^*(M\cap E)=\bar{\mu}(E)$

That is, no matter what Lebesgue measurable set we pick, its intersection with $M$ is so weird that no set of positive Lebesgue measure can fit inside it, and yet $E$ itself is the smallest Lebesgue measurable sets that can contain it.

To find this set, write $A=B\cup C$ from our lemma and take $E_0$ to the set we just constructed. Define $M=E_0+B$ — the set of sums of points in $E_0$ and points in $B$. If $F$ is a Borel set contained in $M$, then $D(F)$ can’t contain any point of $C$ (using a similar argument to that from earlier). And so we must have $\mu_*(M)=0$.

On the other hand, we just saw that $E_0+A=\mathbb{R}$, and thus

$\displaystyle M^c=E_0+C=E_0+(B+1)=M+1$

And so $\mu_*(M^c)=0$ as well. If $E$ is any Lebesgue measurable set, the monotonicity of $\mu_*$ gives us $\mu_*(M\cap E)=\mu_*(M^c\cap E)=0$. And then an earlier result tells us that

$\displaystyle\bar{\mu}(E)=\mu_*(M^c\cap E)+\mu^*((M^c)^c\cap E)=0+\mu^*(M\cap E)=\mu^*(M\cap E)$

April 24, 2010 Posted by | Analysis, Measure Theory | 7 Comments