= Solution
From now on $|\mu|(X)<\infty$, so all signed quantities below are finite. Since $|\mu|(E)\geq|\mu(E)|$, the set function
$$
\mu^+(E)=\frac{|\mu|(E)+\mu(E)}2
$$
is nonnegative. <Countable additivity> follows from that of the two terms, so $\mu^+$ is a finite <positive measure>.
Set $s=\sup\{|\mu|(M):\mu^+(M)=0\}$. This is finite. Choose measurable $M_j$ with $\mu^+(M_j)=0$ and $|\mu|(M_j)\to s$. Then
$$
M_-=\bigcup_{j=1}^{\infty}M_j
$$
is $\mu^+$-null by <countable subadditivity>, and <monotonicity> gives $|\mu|(M_-)\geq s$. The definition of $s$ gives the reverse inequality. Hence \b[$M_-$ attains the maximum], rather than merely approximating it.
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