In Richard Bass' Real Analysis, I am struggling to understand part of the proof of Theorem 4.6 on page 28.
Theorem. If $\mu^∗$ is an outer measure on $X$, then the collection $\mathcal{A}$ of $\mu^∗$-measurable sets is a $\sigma$-algebra. If $\mu$ is the restriction of $\mu^∗$ to $\mathcal{A}$, then $\mu$ is a measure. Moreover, $\mathcal{A}$ contains all the null sets.
In the part of the proof where he shows $\mathcal{A}$ contains all the null sets, Bass writes
If $\mu^*(A)=0$ and $E\subset X$, then $$\mu^*(E\cap A)+\mu^*(E\cap A^c)=\mu^*(E\cap A^c)\leq \mu^*(E),$$ which shows $\mathcal{A}$ contains all the null sets.
Why does this show $\mathcal{A}$ contains all the null sets?
Ok, I get what means in this context $\mu^*$-measurable. A set $B$ is $\mu^*$-measurable if and only if
$$\mu^*(B\cap R)+\mu^*(B^\complement\cap R)\le\mu^*(R)\tag1$$ for any subset $R$ of the space.
Then note that $(1)$ holds for any $B$ such that $\mu^*(B)=0$, because any outer measure is increasing, that is, if $V\subset W$ then $\mu^*(V)\le\mu^*(W)$, consequently $\mu^*(B\cap R)\le\mu^*(B)=0$ and $\mu^*(B^\complement\cap R)\le\mu^*(R)$ because $(B^\complement\cap R)\subset R$, for any chosen $R$.