On symmetric difference of the Lebesgue measure

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I found the following proposition in Measure, Integral and Probability from Marek Capinski and Ekkehard Kopp. It concerns the Lebesgue measure space ($X$, $M$, $m$).

If $A \in M $ and $m(A \triangle B) = 0$ then $B \in M$ and $m(A) = m(B)$.

I managed to prove that the measure of the set $A$ and the measure of the set $B$ are the same since

$m(A) = m(A \cup B) + m(A \cup B^c) = m(A \cup B) = m(B),$

but I seem to get stuck at proving that $B$ must be in $M$. Anyone that can provide me with a hint?

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Notice that $E:=A\setminus B$ and $F:=B\setminus A$ are both in $M$ for being a null set.

Now notice that we can write $B$ as $$B=[A\setminus E] \cup F$$ But then that means...