Is every regular Borel outer measure topologically additive? If $m$ is a regular Borel outer measure is it true that $m$ is topologically additive?
If so what is a proof or a counterexample?
Definitions:
Topologically Additive: $X$ is a topological space, $m$ is a outer measure. 
$m$ is topologically additive iff $S,T \subseteq X$ separated by neighborhoods implies $m(S \cup T) = m(S)+ m(T)$.
Borel: $X$ topological space, $m$ outer measure on $X$. $m$ is Borel iff every open set is measurable
Regular: $X$ topological space, $m$ outer measure on X. $m$ is regular iff :


*

*$K$ compact implies $m(K) < \infty$

*$m(S) = \inf \{m(U) \colon S \subseteq U, U \text{ is open} \}$

*For $U$ open, $m(U) = \sup \{m(K): K \subseteq U, K \text{ is compact}\}$
 A: Yes, any regular Borel outer measure $m$ is topologically additive. Indeed, take any $S,T \subseteq X$ such that $S\subseteq U$ and $T\subseteq V$ for some disjoint open subsets $U$ and $V$ of $X$. Take any real $c>m(S\cup T)$ (if any such $c$ exists). Then, by part 2 of the regularity condition, there is some open subset $W$ of $X$ such that $W\supseteq S\cup T$ and  $m(W)<c$. Let now $U_1:=U\cap W$ and $V_1:=V\cap W$. Then $S\subseteq U_1$, $T\subseteq V_1$, and the sets $U_1$ and $V_1$ are open. 
Since $m$ is Borel and the restriction of $m$ to measurable sets is countably additive (see e.g. \url{https://en.wikipedia.org/wiki/Outer_measure}), one has 
$m(U_1\cup V_1)=m(U_1)+m(V_1)$. So, 
\begin{equation}
 c>m(W)\ge m(U_1\cup V_1)=m(U_1)+m(V_1)\ge m(S)+m(T), 
\end{equation}
for any $c>m(S\cup T)$. So, 
\begin{equation}
 m(S\cup T)\ge m(S)+m(T). 
\end{equation}
On the other hand, any outer measure is subadditive, by definition. So, the proof is complete. 
Note: Parts 1 and 3 of the regularity condition were not used here. 
