Hahn decomposition theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that given a measurable space (X,Σ) and a signed measure μ defined on the σ-algebra Σ, there exist two measurable sets P and N in Σ such that:

1. P ∪ N = X and P ∩ N = ∅.
2. For each E in Σ such that E ⊆ P one has μ(E) ≥ 0; that is, P is a positive set for μ.
3. For each E in Σ such that E ⊆ N one has μ(E) ≤ 0; that is, N is a negative set for μ.

Moreover, this decomposition is essentially unique, in the sense that for any other pair (P', N') of measurable sets fulfilling the above three conditions, the symmetric differences P Δ P' and N Δ N' are μ-null sets in the strong sense that every measurable subset of them has zero measure. The pair (P,N) is called a Hahn decomposition of the signed measure μ.

Hahn–Jordan decomposition

A consequence of this theorem is the Jordan decomposition theorem, which states that every signed measure μ has a unique decomposition into a difference μ = μ⁺ − μ^– of two positive measures μ⁺ and μ^–, at least one of which is finite, such that μ⁺(E) = 0 if E ⊆ N and μ⁻(E) = 0 if E ⊆ P for any Hahn decomposition (P,N) of μ. μ⁺ and μ^– are called the positive and negative part of μ, respectively. The pair (μ⁺, μ^–) is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of μ. The two measures can be defined as

$\mu ^{+}(E):=\mu (E\cap P)\,$

and

$\mu ^{-}(E):=-\mu (E\cap N)\,$

for every E in Σ and any Hahn decomposition (P,N) of μ.

Note that the Hahn-Jordan decomposition is unique, while the Hahn decomposition is only essentially unique.

Proof of the Hahn-Jordan decomposition:

Existence: Let μ⁺ and μ⁻ be defined as in the statement of the theorem, that means μ⁺(E) = μ(E ∩ P) and μ⁻(E) = − μ(E ∩ N) for a Hahn decomposition (P,N) of μ and any E in Σ. It is an easy task to verify that both μ⁺ and μ^– are positive measures on the space (X,Σ), at least one of them is finite (since μ cannot take both +∞ and −∞ as values), and satisfy μ = μ⁺ − μ^–. We have to show that μ⁺(E) = 0 if E ⊆ N', and μ⁻(E) = 0 if E ⊆ P' for any (other) Hahn decomposition (P',N') of μ. Assume therefore E ⊆ N'. Observe that P = (P ∩ P') ∪ (P ∩ (P Δ P')) and (P ∩ P') ∩ (P ∩ (P Δ P')) = ∅. Now, μ⁺(E) = μ(E ∩ P) = μ(E ∩ (P ∩ P')) + μ(E ∩ (P ∩ (P Δ P'))) = 0 , as E ∩ (P ∩ P') = ∅, and E ∩ (P ∩ (P Δ P')) contained in P Δ P' and therefore a μ-null set. The statement for E ⊆ P' follows in analogy.

Uniqueness: Consider now an arbitrary Hahn-Jordan decomposition μ = μ⁺ − μ^– and an arbitrary Hahn decomposition (P,N) for μ. For any element E in the σ-algebra Σ for which E ⊆ P, one obtains μ(E) = μ⁺(E) − μ⁻(E) = μ⁺(E). Consider now an arbitrary element A in the σ-algebra Σ. One has μ⁺(A) = μ⁺(A ∩ P) + μ⁺(A ∩ N) = μ⁺(A ∩ P) = μ(A ∩ P). Likewise, μ⁻(A) = − μ(A ∩ N) can be shown. (μ⁺, μ⁻) is therefore uniquely determined by the choice of μ.///

Remarks Note that if two non-negative measures μ⁺ and μ⁻ with μ = μ⁺ − μ^– have the property μ⁺(E) = μ(E ∩ P) and μ⁻(E) = − μ(E ∩ N) for any E in Σ for one Hahn decomposition (P,N) of μ, then (μ⁺, μ^–) is the Hahn-Jordan decomposition. So, in search for a decomposition, the property does not have to be shown for all Hahn decompositions.

Proof: We assume to be given two (potentially distinct) Hahn decompositions (P₁,N₁) and (P₂,N₂) and two pairs of non-negative measures, (μ⁺₁, μ^–₁) and (μ⁺₂, μ^–₂), such that μ = μ⁺_i − μ^–_i and μ⁺_i(E) = 0 if E ⊆ N_i, and μ⁻_i(E) = 0 if E ⊆ P_i for i=1,2. Observe that P₁ = (P₁ ∩ P₂) ∪ (P₁ ∩ (P₁ Δ P₂)) and (P₁ ∩ P₂) ∩ (P₁ ∩ (P₁ Δ P₂)) = ∅. Consider now an arbitrary element A in the σ-algebra Σ. Denoting B = A ∩ (P₁ ∩ (P₁ Δ P₂)), we therefore obtain μ⁺₁(A) = μ⁺₁(A ∩ P₁) = μ⁺₁(A ∩ (P₁ ∩ P₂)) + μ⁺₁(B). B is contained in the (by essential uniqueness of the Hahn decomposition) μ-null set P₁ Δ P₂. Therefore, μ(B) = 0. Because B is also contained in the positive set P₁ of μ, one obtains 0 = μ(B) = μ⁺₁(B). Therefore, μ⁺₁(A) = μ⁺₁(A ∩ (P₁ ∩ P₂)). However, as A ∩ (P₁ ∩ P₂) is contained in the positive set P₁, μ⁺₁(A) = μ⁺₁(A ∩ (P₁ ∩ P₂)) = μ(A ∩ (P₁ ∩ P₂)). Likewise, one can show that μ⁺₂(A) = μ(A ∩ (P₁ ∩ P₂)), therefore proving μ⁺₁(A) = μ⁺₂(A). The case for identity of μ^–₁ and μ^–₂ follows in analogy. ///

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

$t_{n}=\sup\{\mu (B):B\in \Sigma ,\,B\subset A_{n}\}$

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

$\mu (B_{n})\geq \min\{1,t_{n}/2\}.$

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

$A=D\setminus \bigcup _{n=0}^{\infty }B_{n}.$

Since the sets (B_n)_n≥0 are disjoint subsets of D, it follows from the sigma additivity of the signed measure μ that

$\mu (A)=\mu (D)-\sum _{n=0}^{\infty }\mu (B_{n})\leq \mu (D)-\sum _{n=0}^{\infty }\min\{1,t_{n}/2\}.$

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

$s_{n}:=\inf\{\mu (D):D\in \Sigma ,\,D\subset X\setminus N_{n}\}.$

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

$\mu (D_{n})\leq \max\{s_{n}/2,-1\}\leq 0.$

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

$N=\bigcup _{n=0}^{\infty }A_{n}.$

Since the sets (A_n)_n≥0 are disjoint, we have for every B ⊆ N in Σ that

$\mu (B)=\sum _{n=0}^{\infty }\mu (B\cap A_{n})$

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

$\mu (N)=\sum _{n=0}^{\infty }\mu (A_{n})\leq \sum _{n=0}^{\infty }\max\{s_{n}/2,-1\}=-\infty ,$

which is not allowed for μ. Therefore, P is a positive set.

Proof of the uniqueness statement: Suppose that $(N',P')$ is another Hahn decomposition of $X$. Then $P\cap N'$ is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to $N\cap P'$. Since

$(P\,\triangle \,P')\cup (N\,\triangle \,N')=(P\cap N')\cup (N\cap P'),$

this completes the proof. Q.E.D.

External links

• Hahn decomposition theorem at PlanetMath.