= Solution
A set $H$ is Lebesgue measurable when the <Caratheodory measurability criterion> holds:
$$
\mathcal L^*(E)=\mathcal L^*(E\cap H)+\mathcal L^*(E\setminus H)\quad\text{for every }E\subseteq\mathbb R^n.
$$
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 gives
$$
\mathcal L^*(E\cap H)+\mathcal L^*(E\setminus H)\leq\sum_j|Q_j|.
$$
For finite <outer measure> take a covering sum arbitrarily close to $\mathcal L^*(E)$; infinite <outer measure> follows already from subadditivity. Thus \b[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 \b[every <Borel set> is Lebesgue measurable].
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