 Solution

A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion: for every $T\subseteq \mathbb{R}$,Suppose $m^{\ast }(A)=0$ and let $B\subseteq A$, allowing $B=A$. The monotonicity of Lebesgue outer measure follows directly from covers: every cover of a larger set covers a smaller set. Thus $m^{\ast }(B)=m^{\ast }(T\cap B)=0$, while $m^{\ast }(T\setminus B)\le m^{\ast }(T)$. ConsequentlyThe reverse inequality is countable subadditivity of Lebesgue outer measure applied to $T=(T\cap B)\cup (T\setminus B)$. Equality follows, even if $m^{\ast }(T)=\infty$. Therefore $A$ and every subset $B\subseteq A$ are Lebesgue measurable sets of measure zero. This is the completeness of Lebesgue measure, obtained without presupposing measurability of the outer-null set.