Signed Measures and Radon-Nikodym - Key Proof

ProofProof Sketch of the Hahn Decomposition Theorem

We outline the proof of the Hahn Decomposition Theorem for a signed measure $\nu$ .

Goal: Find disjoint sets $P, N$ with $P \cup N = X$ such that $\nu$ is non-negative on subsets of $P$ and non-positive on subsets of $N$ .

Proof (assuming $\nu$ takes no value $-\infty$ ):

Step 1: Define: $\alpha = \sup\{\nu(A) : A \in \mathcal{F}\}$

Since $\nu$ takes no value $-\infty$ , we have $\alpha \in [0, \infty]$ .

Step 2: Choose a sequence $\{A_n\}$ in $\mathcal{F}$ with $\nu(A_n) \to \alpha$ . Define: $P = \bigcup_{n=1}^{\infty} A_n$

Step 3: We claim $P$ is a positive set. Suppose not: there exists $E \subseteq P$ with $\nu(E) < 0$ .

Then for large $n$ , considering $A_n \cup E$ : $\nu(A_n \cup E) = \nu(A_n) + \nu(E \setminus A_n) \geq \nu(A_n) + \nu(E)$

If $\nu(E \setminus A_n) \geq 0$ , this gives $\nu(A_n \cup E) \geq \nu(A_n) + \nu(E) > \nu(A_n)$ , contradicting $\alpha = \sup \nu(A)$ as $n \to \infty$ .

Step 4: More carefully, one shows that $P$ can be chosen so that for any $E \subseteq P$ , we have $\nu(E) \geq 0$ . This requires an iterative construction that removes negative subsets.

Step 5: Set $N = P^c$ . We must show $\nu(E) \leq 0$ for all $E \subseteq N$ .

If $\nu(E) > 0$ for some $E \subseteq N$ , then $P \cup E$ would give $\nu(P \cup E) > \nu(P) = \alpha$ , contradicting the definition of $\alpha$ .

Step 6: Thus $(P, N)$ is a Hahn decomposition.

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Remark

Key insights in the proof:

Supremum argument: The proof uses the supremum of $\nu$ over all measurable sets to construct $P$ . This is a variational approach.
Countable unions: Taking $P = \bigcup A_n$ ensures $P$ "captures" the supremum in the limit, using countable additivity of $\nu$ .
Iterative refinement: The actual proof requires iteratively removing bad subsets from candidates for $P$ until a true positive set is achieved.
Non-uniqueness: The decomposition is not unique. If $(P, N)$ is a Hahn decomposition and $E$ is a null set (with $\nu^+(E) = \nu^-(E) = 0$ ), then $(P \cup E, N \setminus E)$ is also a Hahn decomposition.
Jordan decomposition: Once the Hahn decomposition is established, the Jordan decomposition $\nu = \nu^+ - \nu^-$ follows immediately by defining $\nu^+(A) = \nu(A \cap P)$ and $\nu^-(A) = -\nu(A \cap N)$ .