 Solution

A set $H$ is Lebesgue measurable when the Caratheodory measurability criterion holds:Subadditivity always gives the inequality with left side at most right side. For $H=\{x_i>t\}$, split each covering box $Q_j$ across the plane $x_i=t$. Its two pieces have total volume $|Q_j|$. The piece with a closed cut face can be enlarged by arbitrarily little into an open box; hence its outer measure is at most its corresponding volume, just as for the other piece. Subadditivity applied separately on the two sides consequently givesFor finite outer measure take a covering sum arbitrarily close to ${\mathcal{L}}^{\ast }(E)$; infinite outer measure follows already from subadditivity. Thus every coordinate half space is measurable.

Carathéodory's theorem says that the measurable sets form a sigma-algebra and the outer measure restricted to it is a complete measure. Coordinate half spaces generate all open boxes and then all open sets, using countably many rational-endpoint boxes. They therefore generate the Borel sigma-algebra. Hence every Borel set is Lebesgue measurable.