Cfrgtkky

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 $Esubset X$, then
$$mu^*(Ecap A)+mu^*(Ecap A^c)=mu^*(Ecap A^c)leq mu^*(E),$$
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

$$mu^*(Bcap R)+mu^*(B^complementcap R)lemu^*(R)tag1$$ 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 $Vsubset W$ then $mu^*(V)lemu^*(W)$, consequently $mu^*(Bcap R)lemu^*(B)=0$ and $mu^*(B^complementcap R)lemu^*(R)$ because $(B^complementcap R)subset R$, for any chosen $R$.