 Solution

From now on $|\mu |(X)<\infty$, so all signed quantities below are finite. Since $|\mu |(E)\ge |\mu (E)|$, the set functionis 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 $|\mu |(M_j)\to s$. Thenis ${\mu }^{+}$-null by countable subadditivity, and monotonicity gives $|\mu |(M_{-})\ge s$. The definition of $s$ gives the reverse inequality. Hence $M_{-}$ attains the maximum, rather than merely approximating it.