Take $\mu^*_1,\mu^*_2$ be an outer measures on X, then prove that $$\nu^*(A)=\max\{ \mu^*_1(A),\mu^*_2(A) \},~A\in2^X$$ is an outer measure.
Let's remind definition of outer measure.
Let $X$ be any (non-empty) set. A function $\nu \colon X \longrightarrow [0,\infty]$ is an outer measure if
- $$ \nu^*(\emptyset)=0. $$
- $$ \nu^*(A) \leq \nu^*(B)$$ whenever $A\subseteq B \subseteq X$.
- $$ \nu^* \left( \cup_{n=1}^\infty A_n \right) \leq \sum_{n=1}^\infty\nu^* \left( A_n \right) $$ for any sequence $\left( A_n \right)_{n=1}^\infty$ of subsets of $X$.
Proof:
Step $0$: Take $A\subset X$. Then $$ \nu^*(A)= \max \left\{ \mu_1^*(A),\mu_2^* (A) \right\} \in [0, \infty] $$ because both $\mu_1^*(A)$ and $\mu_2^* (A)$ are in $[0, \infty]$.
Step $1$: $$ \nu^*(\emptyset)= \max \left\{ \mu_1^*(\emptyset),\mu_2^* (\emptyset) \right\}=0 $$ because $\mu_1^*(\emptyset)=0 = \mu_2^*(\emptyset) $.
Step $2$: Let $A$ and $B$ be any subsets of $X$ such that $A\subseteq B$. Let us assume without any loss of generality that $ \mu_1^* (A) \leq \mu_2^* (A)$. Then $$ \begin{align} \nu^*(A) &= \max \left\{ \mu_1^* (A), \mu_2^* (A) \right\} \\ &= \mu_2^*(A) \\ &\leq \mu_2^* (B) \\ &\leq \max \left\{ \mu_1^* (B), \mu_2^* (B) \right\} \\ &= \nu^*(B) \end{align} $$
Step $3$: Let $\left( A_n \right)_{n=1}^\infty$ be any sequence of subsets of $X$. Let us assume without any loss of generality that $$ \mu_1^* \left( \cup_{n=1}^\infty A_n \right) \leq \mu_2^* \left( \cup_{n=1}^\infty A_n \right). $$ Then $$ \begin{align} \nu^* \left( \cup_{n=1}^\infty A_n \right) &= \max \left\{ \mu_1^* \left( \cup_{n=1}^\infty A_n \right), \mu_2^* \left( \cup_{n=1}^\infty A_n \right) \right\} \\ &= \mu_2^* \left( \cup_{n=1}^\infty A_n \right) \\ &\leq \sum_{n=1}^\infty \mu_2^* \left(A_n \right) \\ &\leq \sum_{n=1}^\infty \max \left\{ \mu_1^* \left(A_n\right), \mu_2^* \left(A_n\right) \right\} \\ &= \sum_{n=1}^\infty \nu^* \left(A_n\right). \end{align} $$
$\square$
Should I correct something in my proof?