# A Vitali set is non-measurable, direct proof, without using countable additivity

I am teaching a measure theory class, where we are in the process of constructing Lebesgue measure on $\mathbb{R}$ via the usual Caratheodory outer measure construction.

As motivation, we began by constructing a Vitali set $V \subset [0,1)$ which has the property that $\bigcup_{q \in \mathbb{Q} \cap [0,1)} V \oplus q = [0,1)$, where $\oplus$ is the usual "addition mod 1", and the sets $V \oplus q$ are pairwise disjoint. This led us to conclude that Lebesgue measure cannot measure every set; i.e. there is no measure defined on all sets which is countably additive, translation invariant, and has $m([0,1)) = 1$.

We have also constructed Lebesgue outer measure $m^*$ in the usual way, by defining $m^*(A) = \inf\left\{\sum_{i} b_i - a_i : A \subseteq \bigcup_i [a_i, b_i]\right\}$, and proved that it is countably subadditive, translation invariant, and that $m^*([a,b]) = b-a$.

Now the typical next step is to define a set $E$ to be measurable if for every $A \subset \mathbb{R}$ we have $m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$. We will of course show that when $m^*$ is restricted to the measurable sets, it is countably additive. It would then follow, indirectly, that the Vitali set $V$ cannot have been measurable.

But since this is a lot of work, I would like to start by proving directly that $V$ is not measurable; i.e. find a set $A$ such that $m^*(A) < m^*(A \cap V) + m^*(A \cap V^c)$. This should help to motivate the definition of "measurable".

I presume $A = [0,1)$ should work, so that we should try to prove $1 < m^*(V) + m^*([0,1) \setminus V)$. Of course it is clear from countable subadditivity that we must have $m^*(V) > 0$ and $m^*([0,1) \setminus V) > 0$. But I don't immediately see how to prove the sum exceeds 1.

So in short:

Is there a simple proof that $m^*(V) + m^*([0,1) \setminus V) > 1$, using only the basic properties of outer measure $m^*$, and in particular not using the countable additivity of Lebesgue measure?

I also saw Outer Measure of the complement of a Vitali Set in [0,1] equal to 1. It's hard to follow without the textbook in question, but it seems to use the fact that open sets are measurable, and that $m^*(A) = \inf\{m^*(U) : A \subset U, U \text{ open}\}$. Again, I would like to avoid that if possible.

• It's a shame you picked Vitali sets and outer measures for this endgame. Working with inner measure and Bernstein sets would have been much easier, I think. – Asaf Karagila Sep 3 '15 at 23:45
• @AsafKaragila: I might get to teach the class again in 2017... :-) – Nate Eldredge Sep 3 '15 at 23:47
• @AsafKaragila Can you construct Bernstein sets without the machinery of well-orderings, which areprobablynot familiar to someone just learning about Lebesgue measure? – Michael Greinecker Sep 4 '15 at 0:35
• @Michael: Sure. But I doubt they'll understand it, though. :-P But you can probably do it with Zorn's llama, although I doubt it's going to be as simple as constructing a Vitali set. What about free ultrafilters on $\omega$? Those are relatively easy to manufacture, and it's pretty easy to show they cannot be measurable. Although here you'd need to use the fact that $2^\Bbb N$ and $[0,1]$ are Borel isomorphic. So it's probably not a great example either. How about a Hamel basis, then? (Ah, trick question! Hamel bases can be Lebesgue measurable!) – Asaf Karagila Sep 4 '15 at 0:42

Let $n$ be such that $m^*(V)>1/n$, choose $n$ distinct rationals $q_1,\dots,q_n\in\mathbb{Q}/\mathbb{Z}$, and write $V_k=\bigcup_{i=1}^k V\oplus q_i$. Then if $V\oplus q_k$ is measurable, taking $A=V_k$ we find that $$m^*(V_k)=m^*(V\oplus q_k)+m^*(V_{k-1}).$$ Thus if $V$ were measurable, we could conclude by translation-invariance of $m^*$ and induction that $m^*(V_k)=km^*(V)$ for each $k$. In particular, $m^*(V_n)=nm^*(V)>1$, which is a contradiction.
• How do you start the induction? In particular, how do you show that $m^*(V_F)\geq |F|m^*(V)$ when you only have subadditivity? – Michael Greinecker Sep 4 '15 at 0:33
• @MichaelGreinecker: I was rewriting the answer while you commented, but: you apply measurability of $V_G$ for $G\subset F$ with $A=V_F$. – Eric Wofsey Sep 4 '15 at 0:44
• Is this a back-door proof that $m^*$ is finitely additive on measurable sets? Anyway, that's a significant improvement, because it's certainly a lot less work than countable additivity. – Nate Eldredge Sep 4 '15 at 0:48
• Yes, additivity of $m^*$ on finite disjoint unions of measurable sets is essentially immediate from the definition of "measurable" by this argument. – Eric Wofsey Sep 4 '15 at 0:52