# Is the outer measure of $A\cup B$ equal to the sum of their outer measures if $A\cap B=\varnothing$?

I understand that Lebesgue outer measure on $\mathbb R$ is not countably additive. But if there are two disjoint sets, does the outer measure of their union equal the sum of their outer measure? Can someone give me a counterexample?

Use transfinite induction to construct $2$ (or $2^{\aleph_0}$) disjoint subsets $A_i$ of the unit interval $[0,1]$, such that each $A_i$ has nonempty intersection with every uncountable closed subset of $[0,1]$. (These are called Bernstein sets.) Then each $A_i$ has outer measure $1$, as does the union of all the $A_i$.

Take $\Omega = \{a, b\}$. Define $$\mu^*(A) = \left\{ \begin{array}{ll} 0, & A = \emptyset \\ 1, & A \neq \emptyset \end{array} \right.$$ This is an outer measure.

In this case, $$\mu^*(\{a,b\}) = 1 \neq 2 = \mu^*(\{a\}) + \mu^*(\{b\}).$$

• Maybe I wasn't specific enough. I'm only considering Lebesgue outer measure regarding subsets of real numbers. Do you have counterexamples on the real line? – Badoe Sep 30 '13 at 3:08
• @Badoe: Maybe you can try something like this... Consider the Lebesgue outer measure in $[0,1]$. Take a non measurable set $A$. Since it is non measurable, it's inner measure must not be equal its outer measure. But maybe it means that $\lambda^*(A) + \lambda^*(A^c) \neq 1$. I am not sure... – André Caldas Sep 30 '13 at 3:22
• I guess you have to be more specific about what you admit as sets. There's a lot of riff-raff in $\mathcal{P}(\mathbb{R})$ that's really not worth considering, whereas if you admit docile (measurable) sets $E$ which satisfy $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ you get disjoint additivity immediately. Your counter-example will require a choice, Mr. Anderson. – snar Sep 30 '13 at 4:16
• @snarski: I don't think I have to be that specific about what is a set when writing a comment!!! What is the problem with a "choice"? We make choices all the time. You, for instance, have made the particular choice of making a comment to my comment! :-P – André Caldas Sep 30 '13 at 23:48
• @AndréCaldas Sorry, I was trying to be funny. What I mean is this: very roughly, you have two types of $E \subset \mathbb{R}$: measurable w.r.t. "Lebesgue" outer measure and not measurable. Thus, when you say "two disjoint sets" it is either trivial or impossible that $m_*(A \cup B) = m_*(A) + m_*(B)$. The pun about "choice" is that I think you need (something equivalent to) the axiom of choice to construct these sets. Maybe you know all this? Does this address your question at all? – snar Oct 1 '13 at 2:35

Countably additive holds only If the sets $\{ E_n \}$ are measurable and pairwise disjoint: That is. If $\{ E_n \}_{n\geq 1}$ are measurable, then

$$\mu^{*} ( \bigcup E_n ) = \sum \mu^*(E_n)$$

• I am not asking for countably additivity. Does the equality hold for just 2 disjoint sets (whether measurable or not)? If not, what is a possible counterexample? Thank you. – Badoe Sep 30 '13 at 2:45
• THe answer is NO. But I cannot think of a counter example right now – ILoveMath Sep 30 '13 at 2:45

For a non-measurable set, the outer measure will necessarily be positive. Therefore it suffices to consider a non-Lebesgue measurable subset $A$ of the circle, and let $B$ be its complement in the circle. Then additivity must fail for this pair of sets for the following example.

Consider a set $A$ of representatives for equivalence classes under the action of rational rotations on the circle (the usual example of a non-measurable set). Recall that the circle is a countable union of the sets $A_\alpha$ obtained by rotating $A$ by a rational angles $\alpha$.

Let $\epsilon=\mu(A)>0$ where $\mu$ is the outer measure. Let $N=1+\lfloor\frac{1}{\epsilon}\rfloor$. Now take $N$ rational translates of $A$, namely $A_1,A_2,\ldots,A_N$. These are all disjoint by construction. If $\mu$ satisifed finite additivity, then the total outer measure would be greater than $1$: $$\mu(A_1\cup\cdots\cup A_N)=N\cdot\epsilon>1,$$ yielding the desired contradiction. This shows that the outer measure is not even finitely additive.

• This makes no sense. If the set is not measurable, it has NO Lebesgue measure. Therefore, it cannot have an outer measure greater then something that does not exist. – André Caldas Oct 2 '13 at 15:46
• Thanks, this was written in too much of a hurry. I clarified what I had in mind above. – Mikhail Katz Oct 3 '13 at 8:18