Bernstein set in [0,2]

Question

I have got a question to suggest an example of outer measure that is strictly not additive. I have thought about some special sets such as Bernstein set or Vitali set in an interval to suffice. However, can anyone tell me a more explicit way to describe such a set? Cause the question suggests that we should construct such an example to prove, however, what can I do is just to come up with the terminologies. e.g. can anyone give me an explicit form of Bernstein set in [0,2]? thanks

Answer 1

For $x,y\in\Bbb R$ let $x\sim y$ iff $x-y\in\Bbb Q$; $\sim$ is easily seen to be an equivalence relation, and for each $x\in\Bbb R$ the equivalence class of $x$ is $x+\Bbb Q=\{x+q:q\in\Bbb Q\}$. Each of these equivalence classes is dense in $\Bbb R$, and of course they are pairwise disjoint. Thus, there is a set $T\subseteq[0,1]$ that contains exactly one point from each equivalence class. Let $$S=\bigcup_{q\in\Bbb Q\cap[-1,1]}(q+T)\;;$$ $S$ is the union of countably many translates of $T$. If $p,q\in\Bbb Q\cap [-1,1]$, and $x\in(p+T)\cap(q+T)$, then there are $t_p,t_q\in T$ such that $x=p+t_p=q+t_q$; but then $t_p-t_q\in\Bbb Q$, so $t_p\sim t_q$, and by the choice of $T$ we must have $t_p=t_q$. Thus, these translates of $T$ are pairwise disjoint. Clearly $S\subseteq[-1,2]$, so $m^*(S)\le 3$.

• Show that $[0,1]\subseteq S$, so that $m^*(S)\ge 1$.

At this one can show that no matter what $m^*(T)$ is, the example violates countable additivity.

• Use countable subadditivity of $m^*$ to show that $m^*(T)\ne 0$. Conclude that $nm^*(T)>3$ for some positive integer $n$ and hence that $m^*$ cannot be finitely additive.

Answer 2

A Bernstein set in $[0,2]$ is a set $E \subset [0,2]$ such that: both the set $E$ and the complement $[0,2]\setminus E$ meet every nonempty compact set $K \subseteq [0,2]$. Such a set can be used to show that Lebesgue outer measure is not additive.