 Solution

For $E\subseteq {\mathbb{R}}^n$, define the Lebesgue outer measure bywhere the boxes are bounded and $\mathrm{vol}(Q_j)={\prod }_r(b_{jr}-a_{jr})$. The empty covering handles the empty set. Monotonicity follows by inclusion of admissible covers. Taking covers within $\epsilon 2^{-j}$ 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 ${\mathcal{L}}^{\ast }$ 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 ${\mathcal{L}}^{\ast }(A)+{\mathcal{L}}^{\ast }(B)$. Taking the infimum and then using subadditivity givesThus the outer measure underlying Lebesgue measure is a metric outer measure, and in particular every Borel set is Lebesgue measurable.