If $A$ is a Lebesgue measurable set such that $m^*(A\triangle B)=0$, then $B$ is Lebesgue measurable.
I tried the problem as follows:
$m^*(A\cup B)=m^*((A\triangle B)\cup(A\cap B))\leq m^*(A\triangle B)+m^*(A\cap B)=m^*(A\cap B)$
Thus $m^*(A\cap B)=m^*(A\cup B)$ and so $m^*(A)=m^*(B)$. But I got stuck at this point. Please give any hint!
Notice that
$$B = (A \cap B) \cup (B - A).$$
We have that $B - A \subseteq A \triangle B$ so $m^*(B - A) =0$ and hence $B - A$ is measurable. Thus if we can show that $A \cap B$ is measurable, then we are done as we would have $B$ as the union of two measurable sets.
To show $A \cap B$ is measurable, write
$$A \cap B = A \cap (A - B)^c.$$
Because $A - B \subseteq A \triangle B$, we have $A - B$ is measurable and hence so is $(A-B)^c$. We are given that $A$ is measurable so $A \cap B$ is the intersection of two measurable sets, and thus is measurable itself.