= Solution
A <subset> $E\subseteq\mathbb R$ is a <Lebesgue measurable set> when it satisfies <Carathéodory's criterion>: for every $T\subseteq\mathbb R$,
$$
m^*(T)=m^*(T\cap E)+m^*(T\setminus E).
$$
Suppose $m^*(A)=0$ 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 $m^*(B)=m^*(T\cap B)=0$, while $m^*(T\setminus B)\leq m^*(T)$. Consequently
$$
m^*(T\cap B)+m^*(T\setminus B)\leq m^*(T).
$$
The reverse inequality is <countable subadditivity of Lebesgue outer measure> applied to $T=(T\cap B)\cup(T\setminus B)$. Equality follows, even if $m^*(T)=\infty$. Therefore \b[$A$ 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.
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