# The Jordan decomposition theorem for signed measures

If $\nu$ is a signed measure there exists unique positive measures $\nu^{+}$ and $\nu^{-}$ s.t. $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$

Would appreciate if someone can guide me through the proof.

Proof: let $X=P\cup N$ be Hahn decomposition for $\nu$, define $\nu^{+}(E):=\nu(E\cap P)$ and $\nu^{-}(E):=-\nu(E\cap N)$ then both $\nu^{-}$ and $\nu^{+}$ are positive. Also $\nu=\nu^{+}-\nu^{-}$ and $\nu^{+}\perp\nu^{-}$

1) why is $\nu=\nu^{+}-\nu^{-}$ by the above definition we have $\nu(E)=\nu^{+}(E)-\nu^{-}(E)=\nu(E\cap P)+\nu(E\cap N)$ for some $E\in\mathcal{M}$ and $E\neq \emptyset$

2) Also why $\nu^{+}\perp\nu^{-}$? Is it because $P$ is null for $\nu^{-}$ and $N$ is null for $\nu^{+}$? and $P\cup N=X$,$P\cap N=\emptyset$

As for the uniqueness part. Let $\nu=\mu^{+}-\mu^{-}$ be another decomposition. Let $E,F$ be measurable s.t. $E\cup F=X$ and $F\cap E=\emptyset$ be another Hahn decompotion. Also $\mu^{+}(F)=\mu^{-}(E)=0$. Then $P\triangle E$ is $\nu$-null. Why is it $\nu$-null?

We need to show that $\nu^{-}=\mu^{-}$ so we pick $A\in\mathcal{M}$ then $\mu^{+}(A)=\mu^{+}(A\cap E)=\nu(A\cap E)=\nu(A\cap P)=\nu^{+}(A)$ How do we get the second equality? Does the third equality follows from the fact that $P\triangle E$ is $\nu$-null?

I think the idea of the proof is: Find a potential $\nu^+,\nu^-$ positive using Hahn dec; Via the construction it follows mutual singularity; (showing that is is unique is intuitively a consequence of $P\triangle E$) Get new pair of positive measures s.t. $\nu=\mu^+-\mu^-$, mutual singularity $\Rightarrow$ Hahn dec, but any difference in the two positive (negative) Hahn sets is a null set.
1) I think you have to do the steps the other way around using a Hahn dec. to obtain your two measures (one positive and one negative) which are a candidate to be proved to be the unique pair: $$\nu(E)=\nu(E\cap (P\cup N))=\nu(E\cap P) +\nu(E\cap N)$$ 2) Yes, $\forall A\subset N$ $$\nu^+(A)=\nu(A\cap P) = 0$$ Similar steps for $\nu^-$.
3.?1) $P\triangle E$ is $\nu$-null is a result from Hahn Decomposition Thm when you have two Hahn dec for the same signed measure.
3.?2) $$\nu(A\cap E)=\nu^+(A\cap E)-\nu^-(A\cap E)=\nu^+(A\cap E)+\nu(A\cap E\cap F)$$ but $E\cap F=\emptyset$ because $E,F$ is a Hahn dec for $\nu$.
3.?3)I do not see it the way I prove it, $$\nu(A\cap E\cap (P\cup N))= \nu(A\cap E\cap P)+\nu(A\cap E\cap N)$$ but the $E$ is positive and $N$ negative, then $$\nu(A\cap E\cap N)\geq 0 \text{ because }A\cap E\cap N\subset E$$ $$\nu(A\cap E\cap N)\leq 0 \text{ because }A\cap E\cap N\subset N$$ $$\Rightarrow \nu(A\cap E\cap N)=0.$$ Similarly $$\nu(A\cap P\cap (E\cup F))=\nu(A\cap P\cap E)+\nu(A\cap P\cap F)=\nu(A\cap P\cap E)$$ and so $$\nu(A\cap P)=\nu(A\cap E)$$