$m^*(E)=\infty$ in Caratheodory's criterion for Lebesgue measurability

Question

I'm having trouble with the $m^*(E)=\infty$ case from this question. I've looked at the answers where $E$ is split into countably infinitely many bounded subsets, but I still cannot prove this case.

Answer 1

Covering $\mathbb R$ with countably many boxes $B_n$ we see that for every elementary set $A$, \begin{align}m^*(A\setminus (B_n\cap E))+m^*(A\cap(B_n\cap E))&\ge m^*(A)\tag{subadditivity}\\ &=m(A)\end{align} and \begin{align}m^*(A\setminus(B_n\cap E))+m^*(A\cap(B_n\cap E)) &=m^*(((A\cap B_n)\setminus E)\cup(A\setminus B_n))+m^*((A\cap B_n)\cap E))\\ &\le m^*((A\cap B_n)\setminus E)+m^*(A\setminus B_n)+m^*((A\cap B_n)\cap E))\tag{subadditivity}\\ &=m(A\cap B_n)+m(A\setminus B_n)\tag{$A\cap B_n$ and $A\setminus B_n$ are elementary}\\ &=m(A),\tag{additivity} \end{align} so $m^*(A\setminus (B_n\cap E))+m^*(A\cap(B_n\cap E))=m(A)$. By the finite case and the fact that countable unions of Lebesgue measurable sets are Lebesgue measurable, we conclude that $E=\bigcup_{n=1}^{\infty}(B_n\cap E)$ is Lebesgue measurable.