Let $(X, \mathcal X, \mu)$ be a finite measure space with $\mu(X)=1$. Let $\mu^*$ and $\mu_*$ be the outer and inner measures, respectively, induced by $\mu$.
A simple fact is
For all $A,B \in X$, if $\mu(A \triangle B)=0$, then $\mu(A)=\mu(B)$.
I am trying to generalize something like this to non-measurable sets.
Is it true that if $\mu^*(A \triangle B)=0$, then $A$ and $B$ have the same inner and outer measures?
I have been able to show that in this case the outer measures of $A$ and $B$ must be the same, but I don't see that it holds for their inner measures.
In case it doesn't hold for the inner measures, is there another way of generalizing the simple fact above for non-measurable sets?
You already know that $\mu^\star(A\triangle B)=0$ implies that $\mu^\star(A)=\mu^\star(B)$. Now apply this to $A^c$ and $B^c$, together with the nice identity $A\triangle B=A^c\triangle B^c$, to conclude that $\mu^\star(A\triangle B)=0$ also implies that $\mu^\star(A^c)=\mu^\star(B^c)$, which implies that $\mu_\star(A)=\mu_\star(B)$ by subtracting both sides from $1$.
There are many ways to prove the key identity $A\triangle B=A^c\triangle B^c$, one approach is to observe that $$ A\triangle B=(A\cap B^c)\cup (A^c\cap B), $$ and replacing $A,B$ by their complements just switches the sets in the left group of parentheses with those in the right group - so the union remains unchanged.