Hahn Decomposition Theorem - Proof of The Hahn Decomposition Theorem

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 A_n ⊆ D has been constructed. Let

denote the supremum of μ(B) for all the measurable subsets B of A_n. This supremum might a priori be infinite. Since the empty set ∅ is a possible B in the definition of t_n and μ(∅) = 0, we have t_n ≥ 0. By definition of t_n there exists a B_n ⊆ A_n in Σ satisfying

Set A_n+1 = A_n \ B_n to finish the induction step. Define

Since the sets (B_n)_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 t_n ≥ 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 N_n, define

as the infimum of μ(D) for all the measurable subsets D of X \ N_n. This infimum might a priori be –∞. Since the empty set is a possible D and μ(∅) = 0, we have s_n ≤ 0. Hence there exists a D_n in Σ with D_n ⊆ X \ N_n and

By the claim above, there is a negative set A_n ⊆ D_n such that μ(A_n) ≤ μ(D_n). Define N_n+1 = N_n ∪ A_n to finish the induction step.

Define

Since the sets (A_n)_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 s_n ≤ 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.