Proof of The Hahn Decomposition Theorem
Preparation: Assume that μ does not take the value −∞ (otherwise decompose according to −μ). As mentioned above, a negative set is a set A in Σ such that μ(B) ≤ 0 for every B in Σ which is a subset of A.
Claim: Suppose that a set D in Σ satisfies μ(D) ≤ 0. Then there is a negative set A ⊆ D such that μ(A) ≤ μ(D).
Proof of the claim: Define A0 = D. Inductively assume for a natural number n that An ⊆ D has been constructed. Let
denote the supremum of μ(B) for all the measurable subsets B of An. This supremum might a priori be infinite. Since the empty set ∅ is a possible B in the definition of tn and μ(∅) = 0, we have tn ≥ 0. By definition of tn there exists a Bn ⊆ An in Σ satisfying
Set An+1 = An \ Bn to finish the induction step. Define
Since the sets (Bn)n≥0 are disjoint subsets of D, it follows from the sigma additivity of the signed measure μ that
This shows that μ(A) ≤ μ(D). Assume A were not a negative set. That means there exists a B in Σ which is a subset of A and satisfies μ(B) > 0. Then tn ≥ μ(B) for every n, hence the series on the right has to diverge to +∞, which means μ(A) = –∞, which is not allowed. Therefore, A must be a negative set.
Construction of the decomposition: Set N0 = ∅. Inductively, given Nn, define
as the infimum of μ(D) for all the measurable subsets D of X \ Nn. This infimum might a priori be –∞. Since the empty set is a possible D and μ(∅) = 0, we have sn ≤ 0. Hence there exists a Dn in Σ with Dn ⊆ X \ Nn and
By the claim above, there is a negative set An ⊆ Dn such that μ(An) ≤ μ(Dn). Define Nn+1 = Nn ∪ An to finish the induction step.
Define
Since the sets (An)n≥0 are disjoint, we have for every B ⊆ N in Σ that
by the sigma additivity of μ. In particular, this shows that N is a negative set. Define P = X \ N. If P were not a positive set, there exists a D ⊆ P in Σ with μ(D) < 0. Then sn ≤ μ(D) for all n and
which is not allowed for μ. Therefore, P is a positive set.
Proof of the uniqueness statement: Suppose that is another Hahn decomposition of . Then is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to . Since
this completes the proof. Q.E.D.
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