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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).

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What exactly is $m*$ here? How are you defining measurable? This is clearly false for the Borel sigma algebra, but is true for Lebesgue and m* the usual outer measure. – Chris Janjigian Feb 20 '13 at 16:36
I know there are Lebesgue nonmeasurable subsets of $[0,1]$ with Lebesgue outer measure $1$ (for an extreme example, see my comments to this question), so $A$ measurable and m*(A) = m*(B) does not imply that $B$ is measurable. – Dave L. Renfro Feb 20 '13 at 16:38

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.

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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 $m^*(N) = 0$.

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@Jonas: yes, I decided to remove them – Ilya Feb 20 '13 at 20:08
But if you have $m$ complete you can not have $m^*(N)=0$ and $N$ non-measurable. – tom Feb 20 '13 at 20:51
@tom: indeed, it is true. Though, in general measures are not complete – Ilya Feb 20 '13 at 20:54

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.

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