If $A$ is measurable does $m^*(B)=m^*(A)$ imply that $B$ is measurable?

Question

I'm trying to prove that if $A$ is measurable and $m^*(A\Delta B)=0$ then $B$ is measurable. I've concluded that $m^*(B)=m^*(A)$, but I'm not sure if this is sufficient to show that $B$ is measurable.

EDIT: $m^*$ is the outer measure (Lebesgue measure when restricted to the family of measurable sets).

Answer 1

No. Given any nonmeasurable set $B$, the interval $[0,m^*(B)]$ is measurable and has outer measure equal to $m^*(B)$. See Dave L. Renfro's comment for a note about more interesting examples that may better answer why this method is not enough.

You can use $B=B\setminus A\cup (A\setminus (A\setminus B))$ to help solve the original problem.

Answer 2

Note that in general (when the measurable space and outer measure are arbitrary) even the original statement is false. Consider an example when $A$ is measurable and $B = A\cup N$ where $N$ is not a measurable set, but $m^*(N) = 0$.

Answer 3

Assuming that $m$ is Lebesgue measure. $m$ is complete than sets $A \Delta B$, $A\setminus B$,$B\setminus A$ are measurable.

So $B = (A\setminus A\Delta B) \cup (B\setminus A)$.

$B$ is union of two measurable sets so it is measurable.