Let $\mu_n^*$ be a sequence of outer measures $\mu_n^* : P(X) \to[0,\infty].$ Prove that $\mu^*=\sup_n \mu_n^*$ is a outer measure too.
If $\mu_n^*$ is a outer measure, then I can conclude that for every $n \in \mathbb{N}$ $$\mu_n^*\left(\bigcup A_k\right) \le \sum_k \mu_n^*(A_k)$$ holds. Do I have to prove that the inequality still holds if $$\sup \mu_n^*\left(\bigcup A_k\right) \le \sup\sum_k \mu_n^*(A_k)$$ What's here the solution?
On
HINT: the inequality $$\sup_{n\ge 1} \mu_n^*\left(\bigcup_{k\ge 1} A_k\right) \le \sup_{n\ge 1}\sum_{k\ge 1} \mu_n^*(A_k)\tag1$$ follows trivially from $$\mu_n^*\left(\bigcup_{k\ge 1} A_k\right) \le \sum_{k\ge 1} \mu_n^*(A_k)\tag2$$ But you need to prove that $$\sup_{n\ge 1}\mu_n^*\left(\bigcup A_{k\ge 1}\right) \le \sum_{k\ge 1} \sup_{n\ge 1}\mu_n^*(A_k)\tag3$$
For this task you can prove first that for arbitrary real-valued sequences $(a_n),\, (b_n)$ $$\sup_{n\ge 1}(a_n+b_n)\le\sup_{n\ge 1} a_n+\sup_{n\ge 1} b_n\tag4$$ from what it follows that $$\sup_{n\ge 1} \sum_{k\ge 1} a_{k,n}\le \sum_{k\ge 1}\sup_{n\ge 1} a_{k,n}\tag5$$ for an arbitrary non-negative real-valued double sequence $(a_{k,n})$. And from here the answer is clear.
Fix $A_k$, note that $\sup_n\sum\mu_n^*(A_k) = \sum \sup_n\mu_n^*(A_k)$ since the measure is positive. Furthermore, $(\forall n, a_n \leq b_n) \implies \sup_na_n \leq \sup_nb_n$. Therefore you get your result.