 Completeness of Lebesgue measure

Every subset $B$ of a Lebesgue null set is a Lebesgue measurable set and is itself a null set. In fact the same conclusion follows directly from $m^{\ast }(B)=0$, without assuming measurability in advance. For any $A\subseteq \mathbb{R}$, monotonicity of Lebesgue outer measure gives $m^{\ast }(A\cap B)=0$ and $m^{\ast }(A\setminus B)\le m^{\ast }(A)$. ThereforeThe reverse inequality follows from countable subadditivity of Lebesgue outer measure, applied to the union $A=(A\cap B)\cup (A\setminus B)$. This is Carathéodory's criterion, proving measurability. If $B$ is a subset of an outer-null $N$, its outer measure is zero by the same monotonicity. Thus the restriction of outer measure to measurable sets is a complete measure.