Let $(X, \mathcal A, \mu)$ be a complete measure space and $\mu^*$ be the outer measure induced by $\mu$ i.e. for any $E \subseteq X,$ $$\mu^*(E) : = \inf \left \{\mu (G)\ |\ G \supseteq E,\ G \in \mathcal A \right \}.$$ Let $A \in \mathcal A$ and $B \subseteq X$ be such that $A \cap B = \varnothing.$ Then can we always say the following $:$ $$\mu^*(A \cup B) = \mu^*(A) + \mu^*(B) = \mu (A) + \mu^*(B).$$
I don't think that it is true since $\mu^*$ is not necessarily finitely additive. I am thinking of finding an example of a $\sigma$-algebra containing both $A$ and $A \cup B$ but not $B.$ But I didn't succeed in fetching a single one with that property. Then we might take measures similar to counting measure to arrive at a contradiction. Can anybody come up with a counter-example?
Thanks for reading.
It is true.
Since $A\in \mathcal{A}$, $A$ is $\mu$-measurable, which is to say that:
$\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$, for $E\subseteq X$
Now let $E=A\cup B:$
$\mu^*(A\cup B)=\mu^*((A\cup B)\cap A)+\mu^*((A\cup B)\setminus A)=\mu^*(A)+\mu^*(B)=\mu(A)+\mu^*(B)$