Whether the set $E_2 \in B$?

Question

Show that if $\mu$ is complete then $E_1 \in B$ and $\mu(A\cap B)=0$ implies $E_2 \in B$. I tried in this way if $\mu$ is complete then $\mu(E1)=0$ (by the definition of completeness since $E_1 \in B$). Given that $\mu(E_1\backslash E_2 \cup E_2\backslash E_1)=0$, then $\mu(E1\backslash E2)=0$,so $\mu(E_1)=\mu(E_1 \cap E_2)=0$, similarly $\mu(E_2)=0$, can we say that $E_1$ is a null set (measure is 0 does not always specify null set right ? ) ? So considering that , $E_2 \subseteq E_1$, since subset of null set will have measure 0 going by the completeness definition. So $E_2 \in B$. Please explain.