[proofplan] We use the Hahn Decomposition to split $X$ into a positive set $P$ and a negative set $N$ for the signed measure $\nu$. The restrictions $\nu^+ := \nu(\cdot \cap P)$ and $\nu^- := -\nu(\cdot \cap N)$ are nonnegative measures satisfying $\nu = \nu^+ - \nu^-$ and $\nu^+ \perp \nu^-$. Uniqueness follows from the observation that any other such decomposition induces an alternative Hahn decomposition, and the Hahn Decomposition Theorem guarantees that positive and negative sets are essentially unique. [/proofplan] [step:Construct the Hahn decomposition of $X$ with respect to $\nu$] Since $\nu$ is a signed measure on $(X, \mathcal{A})$, the Hahn Decomposition Theorem provides a partition $X = P \cup N$ with $P \cap N = \varnothing$, where $P$ is a positive set for $\nu$ (meaning $\nu(A) \ge 0$ for every $A \in \mathcal{A}$ with $A \subseteq P$) and $N$ is a negative set for $\nu$ (meaning $\nu(A) \le 0$ for every $A \in \mathcal{A}$ with $A \subseteq N$). [guided] The starting point is the Hahn Decomposition Theorem, which asserts that for any signed measure $\nu$ on $(X, \mathcal{A})$, there exists a measurable partition $X = P \cup N$ with $P \cap N = \varnothing$ such that $P$ is a positive set and $N$ is a negative set for $\nu$. A **positive set** $P$ means that every measurable subset of $P$ has nonnegative $\nu$-measure — not merely that $\nu(P) \ge 0$, but that the positivity holds for all sub-measurable sets. Similarly for $N$. Why do we need the Hahn Decomposition rather than simply defining $\nu^+(A) = \sup \{ \nu(B) : B \subseteq A, B \in \mathcal{A} \}$? The supremum definition does work and gives the same result, but it requires a separate argument to show countable additivity. The Hahn Decomposition provides a direct geometric construction: once we have $P$ and $N$, we can define $\nu^+$ and $\nu^-$ as restrictions, and countable additivity is immediate. [/guided] [/step] [step:Define $\nu^+$ and $\nu^-$ as restrictions and verify they are nonnegative measures] Define two set functions on $(X, \mathcal{A})$: \begin{align*} \nu^+: \mathcal{A} &\to [0, \infty], \quad A \mapsto \nu(A \cap P), \\ \nu^-: \mathcal{A} &\to [0, \infty], \quad A \mapsto -\nu(A \cap N). \end{align*} We verify that $\nu^+$ is a nonnegative measure. Since $P$ is a positive set, for every $A \in \mathcal{A}$ the set $A \cap P \subseteq P$ satisfies $\nu(A \cap P) \ge 0$, so $\nu^+$ takes values in $[0, \infty]$. The empty set gives $\nu^+(\varnothing) = \nu(\varnothing \cap P) = \nu(\varnothing) = 0$. For countable additivity, let $(A_k)_{k=1}^\infty$ be a sequence of pairwise disjoint sets in $\mathcal{A}$. Then $(A_k \cap P)_{k=1}^\infty$ are pairwise disjoint sets in $\mathcal{A}$, and by countable additivity of the signed measure $\nu$: \begin{align*} \nu^+\left(\bigcup_{k=1}^\infty A_k\right) = \nu\left(\bigcup_{k=1}^\infty (A_k \cap P)\right) = \sum_{k=1}^\infty \nu(A_k \cap P) = \sum_{k=1}^\infty \nu^+(A_k). \end{align*} The argument for $\nu^-$ is symmetric: since $N$ is a negative set, $\nu(A \cap N) \le 0$ for every $A \in \mathcal{A}$, so $\nu^-(A) = -\nu(A \cap N) \ge 0$. The verification of $\nu^-(\varnothing) = 0$ and countable additivity follows identically. [guided] We define $\nu^+$ and $\nu^-$ by restricting $\nu$ to the positive and negative parts of the Hahn decomposition. The key point is that the positivity of $P$ guarantees $\nu(A \cap P) \ge 0$ for every measurable $A$, not just for $A = P$ itself. This is precisely the content of "$P$ is a positive set" — every measurable subset of $P$ has nonnegative $\nu$-measure. Without this stronger condition (if we only knew $\nu(P) \ge 0$), the restriction $\nu^+$ could take negative values and would fail to be a nonnegative measure. For countable additivity, we use that intersecting with a fixed set preserves disjointness: if $A_k \cap A_j = \varnothing$ for $k \neq j$, then $(A_k \cap P) \cap (A_j \cap P) = \varnothing$. The countable additivity of $\nu^+$ then reduces to the countable additivity of $\nu$ itself, applied to the disjoint sequence $(A_k \cap P)_{k=1}^\infty$. [/guided] [/step] [step:Verify the decomposition $\nu = \nu^+ - \nu^-$] For every $A \in \mathcal{A}$, write $A = (A \cap P) \cup (A \cap N)$, a disjoint union. By the additivity of $\nu$: \begin{align*} \nu(A) = \nu(A \cap P) + \nu(A \cap N) = \nu^+(A) + (-\nu^-(A)) = \nu^+(A) - \nu^-(A). \end{align*} [/step] [step:Verify mutual singularity $\nu^+ \perp \nu^-$] The measures $\nu^+$ and $\nu^-$ are mutually singular. The pair $(P, N)$ witnesses this: for every $A \in \mathcal{A}$, \begin{align*} \nu^+(A \cap N) &= \nu((A \cap N) \cap P) = \nu(\varnothing) = 0, \\ \nu^-(A \cap P) &= -\nu((A \cap P) \cap N) = -\nu(\varnothing) = 0. \end{align*} Since $\nu^+$ is concentrated on $P$ (vanishes on all subsets of $N$) and $\nu^-$ is concentrated on $N$ (vanishes on all subsets of $P$), we have $\nu^+ \perp \nu^-$. [guided] Two measures $\alpha$ and $\beta$ are mutually singular ($\alpha \perp \beta$) if there exists a measurable partition $X = S \cup S^c$ with $\alpha(S^c) = 0$ and $\beta(S) = 0$. Here the partition is $X = P \cup N$: the measure $\nu^+$ assigns zero to every measurable subset of $N$ (since $\nu^+(B) = \nu(B \cap P)$ and $B \subseteq N$ implies $B \cap P = \varnothing$), and $\nu^-$ assigns zero to every measurable subset of $P$ by the symmetric argument. In particular, $\nu^+(N) = 0$ and $\nu^-(P) = 0$, so $(P, N)$ is the witnessing partition for $\nu^+ \perp \nu^-$. [/guided] [/step] [step:Prove uniqueness of the decomposition] Suppose $\nu = \mu_1 - \mu_2$ is another decomposition with $\mu_1, \mu_2 \ge 0$ and $\mu_1 \perp \mu_2$. Let $X = Q \cup Q^c$ be a measurable partition witnessing $\mu_1 \perp \mu_2$, so that $\mu_1(Q^c) = 0$ and $\mu_2(Q) = 0$. We claim $Q$ is a positive set and $Q^c$ is a negative set for $\nu$. For any $A \in \mathcal{A}$ with $A \subseteq Q$: \begin{align*} \nu(A) = \mu_1(A) - \mu_2(A) = \mu_1(A) \ge 0, \end{align*} since $A \subseteq Q$ implies $\mu_2(A) \le \mu_2(Q) = 0$. Similarly, for $A \subseteq Q^c$: \begin{align*} \nu(A) = \mu_1(A) - \mu_2(A) = -\mu_2(A) \le 0, \end{align*} since $A \subseteq Q^c$ implies $\mu_1(A) \le \mu_1(Q^c) = 0$. Thus $(Q, Q^c)$ is also a Hahn decomposition for $\nu$. By the essential uniqueness of the Hahn decomposition, the symmetric difference $P \triangle Q$ is a $\nu$-null set: $|\nu|(P \triangle Q) = 0$. For any $A \in \mathcal{A}$: \begin{align*} \mu_1(A) &= \mu_1(A \cap Q) = \nu(A \cap Q) = \nu(A \cap P) + \nu(A \cap (Q \setminus P)) \\ &= \nu(A \cap P) = \nu^+(A), \end{align*} where $\nu(A \cap (Q \setminus P)) = 0$ because $Q \setminus P \subseteq P \triangle Q$ is a $|\nu|$-null set, and $|\nu|(E) = 0$ implies $\nu(E) = 0$. The equality $\mu_2 = \nu^-$ follows from $\mu_2 = \mu_1 - \nu = \nu^+ - \nu = \nu^-$. [guided] Uniqueness requires showing that any decomposition $\nu = \mu_1 - \mu_2$ with $\mu_1 \perp \mu_2$ must coincide with $\nu^+ - \nu^-$. The strategy is to show that the witnessing partition for $\mu_1 \perp \mu_2$ is itself a Hahn decomposition for $\nu$, and then appeal to the essential uniqueness statement in the Hahn Decomposition Theorem. Let $X = Q \cup Q^c$ witness $\mu_1 \perp \mu_2$. For any measurable $A \subseteq Q$, the mutual singularity gives $\mu_2(A) = 0$ (since $\mu_2$ vanishes on all subsets of $Q$), so $\nu(A) = \mu_1(A) - 0 = \mu_1(A) \ge 0$. This shows $Q$ is a positive set. By the symmetric argument, $Q^c$ is a negative set. Hence $(Q, Q^c)$ is a Hahn decomposition for $\nu$. The Hahn Decomposition Theorem guarantees that any two Hahn decompositions $(P, N)$ and $(Q, Q^c)$ differ only on a $|\nu|$-null set: the symmetric difference $P \triangle Q$ satisfies $|\nu|(P \triangle Q) = 0$. This means that for any measurable $A$, the set $A \cap (P \triangle Q)$ has $|\nu|$-measure zero, so $\nu(A \cap (P \triangle Q)) = 0$ (since $|\nu(E)| \le |\nu|(E)$ for all $E$). Now compute $\mu_1(A)$. Since $\mu_1$ is concentrated on $Q$, we have $\mu_1(A) = \mu_1(A \cap Q)$. On the set $A \cap Q$, the measure $\mu_2$ vanishes, so $\mu_1(A \cap Q) = \nu(A \cap Q)$. We decompose $A \cap Q = (A \cap P) \cup (A \cap (Q \setminus P))$, where $Q \setminus P \subseteq P \triangle Q$ is a $|\nu|$-null set. Therefore $\nu(A \cap (Q \setminus P)) = 0$, giving $\mu_1(A) = \nu(A \cap P) = \nu^+(A)$. Since $\mu_2 = \mu_1 - \nu = \nu^+ - \nu = \nu^-$, uniqueness is established. [/guided] [/step]