Prove that in general $$\lambda^*(A)+\lambda^*(B)\geq\lambda^*(A\cup B)+\lambda^*(A\cap B)$$ and $$\lambda_*(A)+\lambda_*(B)\leq\lambda_*(A\cup B)+\lambda_*(A\cap B)$$
$\lambda^*$ is the outer measure defined as $\inf\left\{\lambda(G)|A\subset G \;is\;Open\right\}$and $\lambda_*$ is the inner measure defined as $\sup\left\{\lambda(K)|K\subset A \;is\;Compact\right\}$
We, a priori, know the countable subadditivity property of both inner(With disjoint sets) and outer measure(In general).
I tried expressing $A\cup B$ as disjoint union of $A\backslash B$ and $B$ and use the existing countable subadditivity property. But I get stuck at a point where I have to combine the $\lambda^*(A\cap B)$ with $\lambda^*(A\backslash B)$ which doesn't work out well.
Another attempt I tried, in case of outer measure, is to set $G_A\supset A$ and $G_B\supset B$, We can then write. $$\lambda(G_A)+\lambda(G_B)=\lambda(G_A\cup G_B)+\lambda(G_A \cap G_B)$$ And try to take Infimum both side but I am not sure how can I use it.
P.S. :The measure of Open set is defined in terms of union non-overlapping special rectangle(parallel to coordinate axes and closed) and compact set as a limit of open sets.