Past exam of the integration course of the University of Oxford / 2017 / 1 / a

If ${\sum }_n{m}^{\ast }(A_n)=\infty$, the required inequality is automatic. Otherwise each $A_n$ has finite Lebesgue outer measure. Fix $\epsilon >0$. By the defining infimum, for each $n\ge 1$ choose a countable interval cover $\{J_{n,k}:k\ge 1\}$ withThe family indexed by $(n,k)\in {\mathbb{N}}^2$ is countable and covers ${\bigcup }_n{A}_n$. Since all lengths are nonnegative, its total length is at most ${\sum }_n{m}^{\ast }(A_n)+\epsilon$. Taking the infimum over covers of the union and then letting $\epsilon \downarrow 0$ givesThis proves countable subadditivity of Lebesgue outer measure directly from interval covers, without assuming that any $A_n$ is measurable.

If the intervals $J_k$ cover $A$, their translates $J_k+a$ cover $A+a$ and have the same lengths. Every candidate total length in the definition of $m^{\ast }(A)$ is therefore also a candidate for $m^{\ast }(A+a)$. Taking infima gives $m^{\ast }(A+a)\le m^{\ast }(A)$. Apply the same argument to $A+a$ and the translation $-a$ to obtain the reverse inequality. HenceThis translation invariance of Lebesgue outer measure holds for every subset $A$, including non-measurable sets and sets of infinite outer measure.