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?


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 :

  1. $K$ compact implies $m(K) < \infty$

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

  3. For $U$ open, $m(U) = \sup \{m(K): K \subseteq U, K \text{ is compact}\}$


1 Answer 1


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.


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