Jordan decomposition of a signed measure into unique mutually singular positive parts

For the general existence assertion, obtain a Hahn partition from [L1]; this invokes AC through the Hahn theorem's optimizing choices and its positive-subset lemma. For the conditional assertion start with the supplied partition instead. Define ${\nu }^{+}(E)=\nu (E\cap P)$ and ${\nu }^{-}(E)=-\nu (E\cap N)$. The signs follow from Positive, negative, and null sets for a signed measure, and countable additivity and the empty-set value follow from those of $\nu$, so these are positive measures by [L3].

At least one of ${\nu }^{+}(X)=\nu (P)$ and ${\nu }^{-}(X)=-\nu (N)$ is finite, since otherwise $\nu$ would take both infinite signs. By positivity the same side is finite on every measurable $E$. Thus the difference is always defined, and finite additivity gives $\nu (E)={\nu }^{+}(E)-{\nu }^{-}(E)$. The defining formulas give ${\nu }^{+}(N)={\nu }^{-}(P)=0$, hence mutual singularity by [L2].

Suppose $\nu =\mu -\eta$ with positive mutually singular measures, with the displayed difference defined on every measurable set. Take one witnessing partition $P^{\mathrm{\prime }},N^{\mathrm{\prime }}$ from [L2]. On subsets of $P^{\mathrm{\prime }}$ one has $\nu =\mu \ge 0$, and on subsets of $N^{\mathrm{\prime }}$ one has $\nu =-\eta \le 0$. Thus this is another Hahn partition. A measurable subset of $P\setminus P^{\mathrm{\prime }}$ lies in positive $P$ and negative $N^{\mathrm{\prime }}$, so has value zero; the same holds on $P^{\mathrm{\prime }}\setminus P$. Splitting between these differences shows they are null sets. This comparison uses only the supplied partitions, not Hahn existence or further AC.

For every measurable $E$, $\mu (E)=\mu (E\cap P^{\mathrm{\prime }})=\nu (E\cap P^{\mathrm{\prime }})=\nu (E\cap P)={\nu }^{+}(E)$: the first equality uses vanishing on $N^{\mathrm{\prime }}$, and the middle change of sets discards only the null differences from step 3.1. Similarly $\eta (E)=-\nu (E\cap N)={\nu }^{-}(E)$. There is no subtraction of two infinite quantities. Hence the parts are unique; steps 1.1–4.1 also prove the choice-free conditional assertion because AC was used only to obtain the initial arbitrary Hahn partition.