Let $\mu ^{*}$ be a arbitrary outer measure in a set $X$ and $A,B \subset X$, the following inequality is necessarily satisfied? $$\mu ^*(A \cup B) + \mu^{*}(A \cap B) \leq \mu^{*}(A)+\mu ^{*}(B)$$
I know that if $A$ (or $B$) is $\mu ^{*}-$ measurable then the equality holds but i try to show that supposing that $A,B$ are not $\mu^{*}-$ measurables.
I proceded by cases:
Case $1$: $A \subset B$ in this case $A \cup B=B$ and $A \cap B=A$ then $$\mu ^*(A \cup B) + \mu^{*}(A \cap B) =\mu ^*( B) + \mu^{*}(A) $$ Case $2$: $A \cap B =\emptyset$ in this case $\mu^{*}(A \cap B)=0$ and for subadditivity $$\mu ^{*}(A \cup B) + \mu^{*}(A \cap B)=\mu^{*}(A \cup B)+0\leq \mu^{*}(A)+\mu ^{*}(B)$$ Case $3$: $A \cap B = \emptyset$ in this case i have problems, intuitively I think it could be true but the term $\mu^{*}(A \cap B)$ is the problem, i search counterexamples but i not sure.
Any hint or help i will be very grateful.
Counterexample. Let $X=\{x,y,z\}$ be a $3$-element set. For $A\subseteq X$ define $$\mu^*(A)=\begin{cases} 0\text{ if }A=\varnothing,\\ 1\text{ if }\varnothing\ne A\ne X,\\ 2\text{ if }A=X.\\ \end{cases}$$ Then $\mu^*$ is an outer measure. If $A=\{x,y\}$ and $B=\{y,z\}$, then $\mu^*(A\cup B)+\mu^*(A\cap B)=2+1=3$, and $\mu^*(A)+\mu^*(B)=1+1=2$.