= Solution
For $E\subseteq\mathbb R^n$, define the <Lebesgue outer measure> by
$$
\mathscr L^*(E)=\inf\left\{\sum_{j=1}^{\infty}\operatorname{vol}(Q_j):E\subseteq\bigcup_jQ_j,\quad Q_j=\prod_{r=1}^n[a_{jr},b_{jr})\right\},
$$
where the boxes are bounded and $\operatorname{vol}(Q_j)=\prod_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. <Monotonicity> follows by inclusion of admissible covers. Taking covers within $\varepsilon2^{-j}$ of the outer <measures> of a sequence of sets proves <countable subadditivity>; when the sum is infinite the inequality is automatic. <Lebesgue measure> is the restriction of $\mathscr L^*$ to its Caratheodory-<measurable sets>.
If $d(A,B)=\delta>0$, subdivide each box of any countable cover of $A\cup B$ into finitely many half-open boxes of diameter smaller than $\delta$, preserving its total volume. No small box can meet both $A$ and $B$. The boxes meeting $A$ cover $A$, and the disjoint family meeting $B$ covers $B$, so the original cover costs at least $\mathscr L^*(A)+\mathscr L^*(B)$. Taking the <infimum> and then using subadditivity gives
$$
\boxed{\mathscr L^*(A\cup B)=\mathscr L^*(A)+\mathscr L^*(B).}
$$
Thus the <outer measure> underlying <Lebesgue measure> is a <metric outer measure>, and in particular every <Borel set> is Lebesgue measurable.
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