Prove: $m(A)=0 \iff m^*(A\cup B)=m^*(B)$ for all $B\subset \mathbb{R}$
Where $m^*$ is an outer measure and $m$ is lebesgue measure
$\to:$ $$ m^*(A\cup B)\leq m^*(A)+m^*(B)$$ by subadditivity $$m(A)=m^*(A)=0$$ so $$m^*(A\cup B)\leq m^*(B)$$ now $$B\subseteq A\cup B$$ so $$m^*(B)\leq m^*(A\cup B)$$ by monotonic of the outer measure so $$m^*(A\cup B)=m^*(B)$$
$\leftarrow:$
We know that $0\leq m^*(A)$ by definition, how can I show that $m^*(A)\leq 0$? Can I conclude that if $m^*(A)=0$ then $m(A)=0$?
If the hypothesis is a for fixed $B$ we cannot conclude that $m(A)=0$, For example $m^{*}(A\cup B)=m^{*}(B)$ holds with $B =\mathbb R$. If this holds for all $B$ we can take $B=\emptyset $ to get $m^{*}(A)=0$ and hence $m(A)=0$.