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A Hahn decomposition of a signed measure partitions its space into measurable sets $P,N$ such that every measurable subset of $P$ has nonnegative measure and every measurable subset of $N$ has nonpositive measure. For finite total variation measure, set ${\mu }^{+}=(|\mu |+\mu )/2$ and choose a ${\mu }^{+}$-null set $N$ maximizing $|\mu |(N)$: a countable union of approximating null sets attains the supremum. If some subset of its complement has negative signed mass, successively remove disjoint subsets with positive mass at least half the supremum still available. Their masses tend to zero because their sum is bounded by total variation. The remainder has no positive-measure subset, has negative mass, and is ${\mu }^{+}$-null, contradicting maximality after adjoining it to $N$. On $P$ one has $|\mu |=\mu$ and on $N$ one has $|\mu |=-\mu$. Consequently $d{\mu }^{+}/d|\mu |=1_P$, an explicit Radon-Nikodym derivative.

For a signed measure define its total variation measure byThe value at the empty set is zero. For disjoint measurable sets $A_j$ with union $A$, any finite partition $E_r$ of $A$ givesTaking the supremum proves one inequality. In the reverse direction choose, for the first $k$ sets, finite partitions approaching their variation values. Combining these partitions and the remaining set $A\setminus {\bigcup }_{j\le k}{A}_j$ givesIf one term is infinite, its partition sums can instead be made arbitrarily large. Letting $\epsilon \downarrow 0$ and $k\to \infty$ proves countable additivity, including infinite values. Thus $|\mu |$ is a positive measure. This uses only countable additivity of $\mu$, without assuming a Hahn decomposition.

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.