Sets in completed measure space equals additive sets

Question

I‘m trying to prove that the sets in the completed (finite) measure space $(X,\mathcal{E},\mu)$ are the additive sets.
Definitions:
We have defined a set $A\in\mathcal{P}(X)$ as additive if: $$\mu^{*}(E)= \mu^{*}(E\cap A)+\mu^{*}(E\cap A^C)\quad \forall E\in\mathcal{P}(X)$$ Where $\mu^{*}$ is the outer measure (defined with the classic infimum definition).
The completion of a $\sigma$-Algebra is defined as: $$\mathcal{E}_\mu=\{A\in\mathcal{P}(X)\mid \exists B,C\in \mathcal{E}: A\Delta B\subset C, \mu(C)=0\}$$ Proof: I think I‘ve shown one direction with the following hint: $$\forall A\in\mathcal{P}(X)\:\exists A\subset B\in\mathcal{E}:\mu^{*}(A)=\mu(B)$$ (It would be nice if sombody can show me how I can prove this hint) $$\mu(B)=\mu ^{*}(B)= \underbrace{\mu^{*}(B\cap A)}_{=\mu^{*}(A)}+\mu^{*}(B\cap A^C)= \mu(B)+\mu^{*}(B\cap A^C) $$ Which implies that $\mu^{*}(B\cap A^C)=0$. Therefore we get: $$\exists A\Delta B \subset C\in\mathcal{E}:\mu(C)=\mu^{*}(A\Delta B)= \mu^{*}(\underbrace{(A\cap B^C)}_{=\emptyset}\cup (A^C\cap B))=0$$ But now I have no idea how I can show that all elements in $\mathcal{E}_\mu$ are additive.