Suppose $E$ is a Lebesgue measurable set and let $F$ be a subset of $\mathbb{R}$ such that $$m^*(E\Delta F)=0$$ Show that $F$ is measurable.
I tried seeing Prove that $\mu$(E$\Delta$F) = 0 implies $\mu$(E) = $\mu$(F), however I believe I may not be able to use those results since outer measure is not finitely additive, while the measure is.
Next, I tried using the Caratheodory definition, i.e. showing for any set $A$, $m^*(A)\geq m^*(A\cap F)+m^*(A\cap F^c)$.
I managed to get till here:
$\begin{align} m^*(A)&=m^*(A\cup(E\Delta F))\ \ \ \text{since}\ m^*(E\Delta F)=0\\ &\geq m^*((A\cup F)\cup (A\cap F^c)) \end{align}$
Just one more step from the conclusion... However I believe I can't say that $m^*((A\cup F)\cup (A\cap F^c))\geq m^*(A\cap F)+m^*(A\cup F^c)$, since outer measure is not finitely additive.
Thanks for any help!