Hahn decomposition theorem

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 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 PΔP' and NΔN' 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 can be expressed as a difference of two positive measures μ⁺ and μ^-, at least one of which is finite; μ⁺ and μ^- are called the positive and negative part of μ, respectively. 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 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^-. The pair (μ⁺, μ^-) is called the Jordan decomposition (or sometimes Hahn–Jordan decomposition) of μ.

Proof

Since the signed measure μ cannot include both +∞ and −∞ amongst its values, it will be assumed, for the sake of definiteness, that −∞ is not included. Let F be the set {μ(B)|B is a negative set for μ}, which is not empty since ∅ is a negative set for μ. Let I be the infimum of that set, and let B₁, B₂, ..., B_n, ... be a sequence of negative sets for which I is the limit of the sequence μ(B_n).

Let N be the union of the sets B_n. Notice that every measurable subset A of N can be written as the union of a sequence of disjoint measurable sets, every one of which is contained in some B_n; we can see this by taking

$A=\bigcup _{n}\left(A\cap B_{n}-\bigcup _{m<n}B_{m}\right).$

Therefore A has negative measure, and thus N is a negative set. Hence I ≤ μ(N) ≤ μ(B_n) holds for each positive integer n, and therefore I = μ(N). Furthermore, since μ does not take the value −∞, μ(N) must be finite.

Let P = N^C. To check that P is a positive set, suppose there is a measurable A ⊆ P such that μ(A) < 0. Let

$\varepsilon _{1}:=\sup\{\mu (E)|E\in \Sigma \quad \mathrm {and} \quad E\subseteq A\}.$

By construction, ε₁ ≥ μ(∅) = 0. Also, ε₁ is finite since 0 > μ(A) > −∞. Let A₁ be a measurable subset of A that satisfies

$\mu (A_{1})\geq {\varepsilon _{1} \over 2}.$

Then, construct inductively a sequence ε₂, ε₃, ..., ε_n, ... of nonnegative finite reals, and another A₂, A₃, ..., A_n, ... of measurable sets, defining

$\varepsilon _{n}:=\sup \left\{\mu (E)\left|E\in \Sigma \quad \mathrm {and} \quad E\subseteq A-\bigcup _{i=1}^{n-1}A_{i}\right.\right\},$

and then choosing a measurable set A_n such that

$A_{n}\subseteq A-\bigcup _{i=1}^{n-1}A_{i}$

and

$\mu (A_{n})\geq {\varepsilon _{n} \over 2}.$

Now let A_∞ be the union of the sets A_n, and B:=A-A_∞. Since the sets A_n are disjoint and satisfy μ(A_n) ≥ 0, then μ(A_∞) ≥ 0, and therefore,

$0>\mu (A)=\mu (A_{\infty })+\mu (B)\geq \mu (B).$

From the construction of ε_n, every measurable subset E of B must satisfy μ(E) ≤ ε_n for every positive integer n. But, since the A_n are disjoint sets,

$\sum _{n=1}^{\infty }\varepsilon _{n}=2\sum _{n=1}^{\infty }{\varepsilon _{n} \over 2}\leq 2\sum _{n=1}^{\infty }\mu (A_{n})=2\mu (A_{\infty })<\infty ,$

and therefore,

$\lim _{n\to \infty }\varepsilon _{n}=0.$

Then, every measurable subset E of B must satisfy μ(E) ≤ 0, which proves that B is a negative set, and therefore N∪B is. But then, since N and B are disjoint,

$\mu (N\cup B)=\mu (N)+\mu (B)<\mu (N)=I,$

contradicting the definition of I. This is impossible, so P must be a positive set.

The only remaining task is to show the essential uniqueness of P and N. Let (P', N') be another Hahn decomposition of μ. Then, since both are partitions of X,

$(P\Delta P^{\prime })\cup (N\Delta N^{\prime })=(P\cap N^{\prime })\cup (N\cap P^{\prime }).$

Since P is a positive set, P ∩ N' must be, too; but since N' is a negative set, P ∩ N' must also be a negative set, and therefore, a μ-null set. The same reasoning applies to N ∩ P'. Therefore, both PΔP' and NΔN' are null sets.

Q.E.D.

External links

• Hahn decomposition theorem at PlanetMath.