A set $A\subseteq X$ is $\textbf{$\mu-$measurable }$ if $\mu(E)=\mu(E\cap A)+\mu(E\backslash A)$ for every $E\subseteq X$.
Note: $\mu$ is an outer measure.
Problem:
The collection $M_\mu(X)$ of $\mu$-measurable sets form a $\sigma$-algebra on $X$.
My attempt:
Clearly, $\varnothing$ is $\mu$-measurable. Let $E\subseteq X$ be arbitrary. Then $\mu(E\cap X)+\mu(E\backslash X)=\mu(E)$. Hence $X$ is $\mu$-measurable.
Suppose $A$ is $\mu$-measurable. I must show that $X\backslash A$ is likewise $\mu$-measurable. Let $E\subseteq X$, then $\mu(E\cap X\backslash A)$ $+$ $\mu(E\cap A)$. Since $A$ is $\mu$-measurable, it follows that $\mu(E\cap A)=\mu(E)- \mu(E\cap X\backslash A)$ and so $\mu(E\cap X\backslash A) + \mu(E\cap A)=\mu(E)$. Hence $X\backslash A$ is likewise $\mu$-measurable.
How do I show the last condition? May I have hints? Note: I just started learning measure theory.