# 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

1. You say that all three sets (A, B, C) are subgroups. I see that for A (all n) and B (all even n). But how is C (all odd n) an additive subgroup — it’s not even an additive group since it doesn’t contain 0.

Or do I totally misunderstand?

Thanks!

Comment by marshall | April 25, 2010 | Reply

2. Sorry, I misspoke in my haste to get out the catch-up post…

Comment by John Armstrong | April 25, 2010 | Reply

3. In case anyone is interested, the example you ended with is sometimes called a “maximally non-measurable set” or a “saturated nonmeasurable set”, and more examples and their properties can be found in the following posts:

Remarks on Bernstein sets

http://tinyurl.com/29sgs2b

http://tinyurl.com/27sqz2d [minor correction]

Now for something REALLY STRANGE. In 1917 Lusin and Sierpinski showed that the unit interval [0,1] can be partitioned into c = 2^(aleph_0) many pairwise disjoint sets each having Lebesgue outer measure 1. This shows a massive failure of additivity in the case of Lebesgue outer measure! And yes, they constructed c many such sets, not just uncountably many (which would give you “only” aleph_1 many such sets under CH).

Here’s their paper:

Nikolai N. Lusin and Waclaw Sierpinski, “Sur une décomposition d’un intervalle en une infinité non dénombrable d’ensembles non mesurables” [On a decomposition of an interval into a nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422-424.
[JFM 46.0294.01] [available on-line]

http://www.emis.de/cgi-bin/JFM-item?46.0294.01

Comment by Dave L. Renfro | April 26, 2010 | Reply

4. [...] is), this condition reduces to asking that . If, further, , then we ask that . As an example, the maximally nonmeasurable set we constructed is [...]

Pingback by Measurable Subspaces II « The Unapologetic Mathematician | April 28, 2010 | Reply

5. [...] to add the requirement that . Also, the converse of this theorem is definitely not true; if is a non-measurable set, then the function is not measurable even though the absolute value is [...]

Pingback by Composing Real-Valued Measurable Functions I « The Unapologetic Mathematician | May 4, 2010 | Reply

6. [...] now we can take a thick, non-Lebesgue measurable set whose intersection with is itself a non-Lebesgue measurable set . However, , and has Lebesgue [...]

Pingback by Composing Real-Valued Measurable Functions II « The Unapologetic Mathematician | May 5, 2010 | Reply

7. [...] During this long period from 1904 to 1962, many bizzare results such as the Banach Tarsky paradox, existence of non-Lebesgue measurable sets etc were found. This put a suspicion in people’s minds as to whether this axiom was [...]

Pingback by The axiom of choice | Notes on Mathematics | September 19, 2012 | Reply