Countable subadditivity of Lebesgue outer measure ($m^{\ast }({\bigcup }_n{A}_n)\le {\sum }_n{m}^{\ast }(A_n)$)

If the sum of Lebesgue outer measures on the right is infinite, the inequality is automatic. Otherwise, fix $\epsilon >0$ and cover each $A_n$ by a countable family of intervals whose total length is less than $m^{\ast }(A_n)+\epsilon 2^{-n}$. These covers exist by the definition of infimum. The combined countable family covers ${\bigcup }_n{A}_n$ and has total length at most ${\sum }_n{m}^{\ast }(A_n)+\epsilon$. Take the infimum over covers of the union, then let $\epsilon$ tend to zero. No measurability assumption on the $A_n$ is needed.