Example of strictly subadditive lebesgue outer measure One of the properties of the Lebesgue outer measure is that it is subadditive and not countably additive. In fact, I have read that even when the sets A_i  are disjoint, there is still generally strict inequality.
I was just wondering if anyone can give me an example of this, any set of disjoint real number intervals that I can think of always lead to a equality (hence countable additivity).
 A: Take a Vitali set in the unit interval $[0,1)$, and call it $V$.  (Define an equivalence relation on $[0,1)$ by $x \sim y$ iff $x - y \in \mathbb{Q}$, and let $V$ be obtained from the Axiom of Choice by choosing exactly one element from every $\sim$-equivalence class.)
By $Q$ I will denote $\mathbb{Q} \cap [0,1)$.  For each $q \in Q$ let $V_q$ denote the modulo $1$ shift of $V$ by $q$: $$V_q = \{ x + q : x \in V, x < 1-q \} \cup \{ x + q - 1 : x \in V , 1-q \leq x \}.$$  Note the following facts:


*

*$V_q \cap V_p = \emptyset$ if $p \neq q$.

*$\bigcup_{q \in Q} V_q = [0,1)$.


(Both of the above follow from the fact that $V$ contains exactly one element of every $\sim$-equivalence class.)
By the second fact above we have that $$\mu^* \left( \textstyle{\bigcup_{q \in Q}} V_q \right) = 1.$$
On the other hand, by the translation invariance of $\mu^*$ is follows that $\mu^* ( V_q ) = \mu^* ( V )$ for all $q$, and since $V$ is not Lebesgue measurable it follows that $\mu^* ( V ) > 0$, and therefore $$\sum_{q \in Q} \mu^* ( V_k ) =  + \infty.$$
A: A Bernstein set, $A\subseteq[0,1]$ has the property that both $A$ and $[0,1]\setminus A$ have outer measure $1$.
