I had a question about the additivity property of the outer measure.
Can someone provide an example of a disjoint union of sets which doesn't have an outer measure equal to the sum of the outer measure of each set (in $\mathbb{R}^n$)? That is:
$m_*(E_1\bigcup E_2)\neq m_*(E_1) + m_*(E_2)$, where $E_1\bigcap E_2=\emptyset$ and $d(E_1,E_2)$ possibly equal to zero.
The theorem states this holds in general if $d(E_1,E_2)> 0$, but I can't find a counterexample where it doesn't if $d(E_1,E_2)=0$.
Thanks!
Best Answer
First, a clarification on terminology. The property that you are mentioning is what makes an outer measure a metric outer measure. Lebesgue measure is a metric outer measure, so we know that $m_*(E_1\bigcup E_2)= m_*(E_1) + m_*(E_2)$, where $E_1\bigcap E_2=\emptyset$ must be true if $d(E_1,E_2)>0$.
Now, to your question. An outer measure is additive on the $\sigma$-algebra of measurable sets $\mathcal{M}$. In fact, if $E_1, E_2 \in \mathcal{M}$, we have $$ m_*(E_1\cup E_2)=m_*((E_1\cup E_2)\cap E_1)+m_*((E_1\cup E_2)\cap E_1^c)= m_*(E_1) + m_*(E_2), $$ using the definition of measurable set. So we need to look for a counterexample among non-measurable sets. Note that Solovay proved in 1970 that the existence of a non-measurable set is possible only if we assume the Axiom of Choice, so there won't be a trivial answer to your question.
The most famous construction of a non-measurable set is the Vitali set. As in this answer by JDH, we can construct a Vitali set $V$ in such a way that is contained in $[0,a]$, for any $a \in (0,1)$. Also, since the inner measure of $V$ is zero, we have that $m_*([0,a]\setminus V)=a$ and then $[0,a]\setminus V$ is a non-measurable set of measure $a$. Note that we must have $m_*(V)>0$, because a set is measurable if and only if its outer and inner measures coincide, and we know the latter to be zero. Then we have, by setting $E_1=V$ and $E_2= [0,a]\setminus V$, $$ a=m_*([0,a])=m_*(E_1 \cup E_2) < m_*(E_1)+m_*(E_2)=a + m_*(V). $$