= Solution
If $S\subseteq M_-$, then $\mu^+(S)=0$, so $\mu(S)=-|\mu|(S)\leq0$. 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 \b[$\mu(S)\geq0$ 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 gives
$$
\boxed{\mu^+(E)=\int_E\mathbf1_{M_+}\,d|\mu|,\qquad \frac{d\mu^+}{d|\mu|}=\mathbf1_{M_+}\quad |\mu|\text{-a.e.}}
$$
In particular $\mu^+\ll|\mu|$, and the exhibited <Radon-Nikodym derivative> verifies the <Radon-Nikodym theorem> for this pair directly.
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