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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).

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Well... since the Lebesgue measure is $\sigma$-additive, you'll have to look at sets which are not Lebesgue measurable. – arjafi Nov 4 '12 at 13:13
@ArthurFischer, so basically, it would have to be something with non-measureable sets? – BYS2 Nov 4 '12 at 13:24
Yes. My comment was meant to point you in the right direction. – arjafi Nov 4 '12 at 13:30
up vote 4 down vote accepted

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 $$\sum_{q \in Q} \mu^* ( V_k ) = + \infty.$$

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Just to make sure: the lebesgue outer measure is sigma-additive in the case where the sets are something like (1,3) and (4,6) right? basically the most elementary sets that are taught in like school :p – BYS2 Nov 5 '12 at 12:44
This shows that outer measure is not countable additive. How can we obtain counter-example for finite additivity? – Groups Sep 7 '14 at 4:16

A Bernstein set, $A\subseteq[0,1]$ has the property that both $A$ and $[0,1]\setminus A$ have outer measure $1$.

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