4
$\begingroup$

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 :

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

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.