Caratheodory's criterion states the following: Let $\mu$ be an outer measure on $\mathbb{R}^n$. If for all sets $A,B\subseteq \mathbb{R}^n$, we have $\mu (A\cup B) = \mu(A) + \mu(B)$ whenever $\text{dist}(A,B)>0$, then $\mu$ is a Borel measure.
I am wondering wheather there is a converse to this theorem. That is, if $\mu$ is a Borel measure, then for any set $A,B\subseteq \mathbb{R}^n$, $\text{dist}(A,B) > 0$ implies $\mu(A\cup B) = \mu(A) + \mu(B)$. I am aware that if we additionally assume that for every $A\subseteq \mathbb{R}^n$, there exists a Borel set $B$ s.t. $A \subseteq B$ and $\mu(A) = \mu(B)$, then the converse is indeed true.
Without this additional assumption, I believe the converse statement is false. Yet I struggled to come up with a counter-example. I would appreciate some thoughts on wheather the converse (without the additional assumption of $\mu$ being Borel regular) is indeed false.
The converse is true. Notice that since $\text{dist}(A,B)>0$, we have $\overline{A}\cap\overline{B}=\emptyset$. Since $\mathbb{R}^n$ is a metric space, it is normal (in fact $T_6$); and so we can find an open set $U$ such that $A\subseteq U$ and $A\cap B=\emptyset$. Then, as $\mu$ is a Borel measure, \begin{align*} \mu(A\cup B) &= \mu((A\cup B)\setminus U) + \mu((A\cup B)\cap U) \\ &=\mu((A\setminus U)\cup (B\setminus U)) + \mu((A\cap U)\cup (B\cap U))\\ &=\mu(B)+\mu(A) \end{align*}
In fact, this argument works for metric spaces in general, and I believe the forward direction is also true in general.