 Solution

If $S\subseteq M_{-}$, then ${\mu }^{+}(S)=0$, so $\mu (S)=-|\mu |(S)\le 0$. Suppose instead that $S\cap M_{-}=\varnothing$ but $\mu (S)<0$. The preceding construction gives $R\subseteq S$ with ${\mu }^{+}(R)=0$ and $|\mu |(R)=-\mu (R)>0$. Then $M_{-}\cup R$ is ${\mu }^{+}$-null with variation strictly larger than that of $M_{-}$, a contradiction. Hence $\mu (S)\ge 0$ on the complement of $M_{-}$. This is a Hahn decomposition.

On $M_{+}=X\setminus M_{-}$ every partition has nonnegative signed measures, so $|\mu |(E)=\mu (E)={\mu }^{+}(E)$ for $E\subseteq M_{+}$. On $M_{-}$, ${\mu }^{+}$ vanishes. Splitting an arbitrary measurable set between these two sets givesIn particular ${\mu }^{+}\ll |\mu |$, and the exhibited Radon-Nikodym derivative verifies the Radon-Nikodym theorem for this pair directly.