Theorem 6.21 (Jordan decomposition). If $\nu$ is a signed measure on a measurable space $(X, \mathcal{A})$, then there exist unique measures $\nu^+, \nu^- \colon \mathcal{A} \to [0, \infty]$, one of which is finite, such that $$ \nu = \nu^+ - \nu^- \quad\text{and}\quad \nu^+ \perp \nu^- \,. $$
Proof. Let $X = P \cup N$ where $P$, $N$ are positive, negative sets for $\nu$. Then $$ \nu^+(A) = \nu(A \cap P) \,, \quad \nu^-(A) = -\nu(A \cap N) $$ is the required decomposition. The values of $\nu^{\pm}$ are independent of the choice of $P$, $N$ up to a $\nu$-null set, so the decomposition is unique.
Question 1: why are the values of $\nu^+$ and $\nu^-$ independent of the decomposition chosen? That is, for another decomposition of $X$ into a disjoint positive $P'$ and negative set $N'$, why do we have $$ \nu(A \cap P) = \nu(A \cap P') \,? $$
My thoughts: we have \begin{align*} \nu(A \cap P) &= \nu(A \cap P \cap P') + \nu(A \cap (P \setminus P')) \\ &\stackrel{?}{=} \nu(A \cap P \cap P') + \nu(A \cap (P' \setminus P)) \tag{$*$} \\ &= \nu(A \cap P') \end{align*} where the sets $P \setminus P'$ and $P' \setminus P$ are $\nu$-null. If we were dealing with a (positive) measure I could use monoticity to show $$ \nu(A \cap (P \setminus P')) = 0 = \nu(A \cap (P \setminus P')) $$ and prove the equality in $(*)$.
Question 2: even if $\nu^+$ and $\nu^-$ are independent of the decomposition chosen, how does that show that they are the unique singular measures with the property $$ \nu = \nu^+ - \nu^- \,? $$
Q1: You may write $A\cap(P-P')=A\cap P\cap N'$, then it's a subset of P with $\nu\ge 0 $; it's also a subset of N' with $\nu\le 0$; so $\nu(A\cap(P-P'))=0$.
Q2: If there is another decomposition: $\nu=\nu_1-\nu_2$, here $\nu_{1, 2}$ are singular measures. Then there exist disjointed $P'', N''$ satisfy $$X=P''\cup N'',\quad \nu_1(N'')=\nu_2(P'')=0.$$ We can prove that $P''$ is a positive set, $N''$ is a negative set, then go back to Q1.