In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space ( X , Σ Σ --> ) {\displaystyle (X,\Sigma )} and any signed measure μ μ --> {\displaystyle \mu } defined on the σ σ --> {\displaystyle \sigma } -algebra Σ Σ --> {\displaystyle \Sigma } , there exist two Σ Σ --> {\displaystyle \Sigma } -measurable sets, P {\displaystyle P} and N {\displaystyle N} , of X {\displaystyle X} such that:
Moreover, this decomposition is essentially unique, meaning that for any other pair ( P ′ , N ′ ) {\displaystyle (P',N')} of Σ Σ --> {\displaystyle \Sigma } -measurable subsets of X {\displaystyle X} fulfilling the three conditions above, the symmetric differences P △ △ --> P ′ {\displaystyle P\triangle P'} and N △ △ --> N ′ {\displaystyle N\triangle N'} are μ μ --> {\displaystyle \mu } -null sets in the strong sense that every Σ Σ --> {\displaystyle \Sigma } -measurable subset of them has zero measure. The pair ( P , N ) {\displaystyle (P,N)} is then called a Hahn decomposition of the signed measure μ μ --> {\displaystyle \mu } .
A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure μ μ --> {\displaystyle \mu } defined on Σ Σ --> {\displaystyle \Sigma } has a unique decomposition into a difference μ μ --> = μ μ --> + − − --> μ μ --> − − --> {\displaystyle \mu =\mu ^{+}-\mu ^{-}} of two positive measures, μ μ --> + {\displaystyle \mu ^{+}} and μ μ --> − − --> {\displaystyle \mu ^{-}} , at least one of which is finite, such that μ μ --> + ( E ) = 0 {\displaystyle {\mu ^{+}}(E)=0} for every Σ Σ --> {\displaystyle \Sigma } -measurable subset E ⊆ ⊆ --> N {\displaystyle E\subseteq N} and μ μ --> − − --> ( E ) = 0 {\displaystyle {\mu ^{-}}(E)=0} for every Σ Σ --> {\displaystyle \Sigma } -measurable subset E ⊆ ⊆ --> P {\displaystyle E\subseteq P} , for any Hahn decomposition ( P , N ) {\displaystyle (P,N)} of μ μ --> {\displaystyle \mu } . We call μ μ --> + {\displaystyle \mu ^{+}} and μ μ --> − − --> {\displaystyle \mu ^{-}} the positive and negative part of μ μ --> {\displaystyle \mu } , respectively. The pair ( μ μ --> + , μ μ --> − − --> ) {\displaystyle (\mu ^{+},\mu ^{-})} is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of μ μ --> {\displaystyle \mu } . The two measures can be defined as
for every E ∈ ∈ --> Σ Σ --> {\displaystyle E\in \Sigma } and any Hahn decomposition ( P , N ) {\displaystyle (P,N)} of μ μ --> {\displaystyle \mu } .
Note that the Jordan decomposition is unique, while the Hahn decomposition is only essentially unique.
The Jordan decomposition has the following corollary: Given a Jordan decomposition ( μ μ --> + , μ μ --> − − --> ) {\displaystyle (\mu ^{+},\mu ^{-})} of a finite signed measure μ μ --> {\displaystyle \mu } , one has
for any E {\displaystyle E} in Σ Σ --> {\displaystyle \Sigma } . Furthermore, if μ μ --> = ν ν --> + − − --> ν ν --> − − --> {\displaystyle \mu =\nu ^{+}-\nu ^{-}} for a pair ( ν ν --> + , ν ν --> − − --> ) {\displaystyle (\nu ^{+},\nu ^{-})} of finite non-negative measures on X {\displaystyle X} , then
The last expression means that the Jordan decomposition is the minimal decomposition of μ μ --> {\displaystyle \mu } into a difference of non-negative measures. This is the minimality property of the Jordan decomposition.
Proof of the Jordan decomposition: For an elementary proof of the existence, uniqueness, and minimality of the Jordan measure decomposition see Fischer (2012).
Preparation: Assume that μ μ --> {\displaystyle \mu } does not take the value − − --> ∞ ∞ --> {\displaystyle -\infty } (otherwise decompose according to − − --> μ μ --> {\displaystyle -\mu } ). As mentioned above, a negative set is a set A ∈ ∈ --> Σ Σ --> {\displaystyle A\in \Sigma } such that μ μ --> ( B ) ≤ ≤ --> 0 {\displaystyle \mu (B)\leq 0} for every Σ Σ --> {\displaystyle \Sigma } -measurable subset B ⊆ ⊆ --> A {\displaystyle B\subseteq A} .
Claim: Suppose that D ∈ ∈ --> Σ Σ --> {\displaystyle D\in \Sigma } satisfies μ μ --> ( D ) ≤ ≤ --> 0 {\displaystyle \mu (D)\leq 0} . Then there is a negative set A ⊆ ⊆ --> D {\displaystyle A\subseteq D} such that μ μ --> ( A ) ≤ ≤ --> μ μ --> ( D ) {\displaystyle \mu (A)\leq \mu (D)} .
Proof of the claim: Define A 0 := D {\displaystyle A_{0}:=D} . Inductively assume for n ∈ ∈ --> N 0 {\displaystyle n\in \mathbb {N} _{0}} that A n ⊆ ⊆ --> D {\displaystyle A_{n}\subseteq D} has been constructed. Let
denote the supremum of μ μ --> ( B ) {\displaystyle \mu (B)} over all the Σ Σ --> {\displaystyle \Sigma } -measurable subsets B {\displaystyle B} of A n {\displaystyle A_{n}} . This supremum might a priori be infinite. As the empty set ∅ ∅ --> {\displaystyle \varnothing } is a possible candidate for B {\displaystyle B} in the definition of t n {\displaystyle t_{n}} , and as μ μ --> ( ∅ ∅ --> ) = 0 {\displaystyle \mu (\varnothing )=0} , we have t n ≥ ≥ --> 0 {\displaystyle t_{n}\geq 0} . By the definition of t n {\displaystyle t_{n}} , there then exists a Σ Σ --> {\displaystyle \Sigma } -measurable subset B n ⊆ ⊆ --> A n {\displaystyle B_{n}\subseteq A_{n}} satisfying
Set A n + 1 := A n ∖ ∖ --> B n {\displaystyle A_{n+1}:=A_{n}\setminus B_{n}} to finish the induction step. Finally, define
As the sets ( B n ) n = 0 ∞ ∞ --> {\displaystyle (B_{n})_{n=0}^{\infty }} are disjoint subsets of D {\displaystyle D} , it follows from the sigma additivity of the signed measure μ μ --> {\displaystyle \mu } that
This shows that μ μ --> ( A ) ≤ ≤ --> μ μ --> ( D ) {\displaystyle \mu (A)\leq \mu (D)} . Assume A {\displaystyle A} were not a negative set. This means that there would exist a Σ Σ --> {\displaystyle \Sigma } -measurable subset B ⊆ ⊆ --> A {\displaystyle B\subseteq A} that satisfies μ μ --> ( B ) > 0 {\displaystyle \mu (B)>0} . Then t n ≥ ≥ --> μ μ --> ( B ) {\displaystyle t_{n}\geq \mu (B)} for every n ∈ ∈ --> N 0 {\displaystyle n\in \mathbb {N} _{0}} , so the series on the right would have to diverge to + ∞ ∞ --> {\displaystyle +\infty } , implying that μ μ --> ( D ) = + ∞ ∞ --> {\displaystyle \mu (D)=+\infty } , which is a contradiction, since μ μ --> ( D ) ≤ ≤ --> 0 {\displaystyle \mu (D)\leq 0} . Therefore, A {\displaystyle A} must be a negative set.
Construction of the decomposition: Set N 0 = ∅ ∅ --> {\displaystyle N_{0}=\varnothing } . Inductively, given N n {\displaystyle N_{n}} , define
as the infimum of μ μ --> ( D ) {\displaystyle \mu (D)} over all the Σ Σ --> {\displaystyle \Sigma } -measurable subsets D {\displaystyle D} of X ∖ ∖ --> N n {\displaystyle X\setminus N_{n}} . This infimum might a priori be − − --> ∞ ∞ --> {\displaystyle -\infty } . As ∅ ∅ --> {\displaystyle \varnothing } is a possible candidate for D {\displaystyle D} in the definition of s n {\displaystyle s_{n}} , and as μ μ --> ( ∅ ∅ --> ) = 0 {\displaystyle \mu (\varnothing )=0} , we have s n ≤ ≤ --> 0 {\displaystyle s_{n}\leq 0} . Hence, there exists a Σ Σ --> {\displaystyle \Sigma } -measurable subset D n ⊆ ⊆ --> X ∖ ∖ --> N n {\displaystyle D_{n}\subseteq X\setminus N_{n}} such that
By the claim above, there is a negative set A n ⊆ ⊆ --> D n {\displaystyle A_{n}\subseteq D_{n}} such that μ μ --> ( A n ) ≤ ≤ --> μ μ --> ( D n ) {\displaystyle \mu (A_{n})\leq \mu (D_{n})} . Set N n + 1 := N n ∪ ∪ --> A n {\displaystyle N_{n+1}:=N_{n}\cup A_{n}} to finish the induction step. Finally, define
As the sets ( A n ) n = 0 ∞ ∞ --> {\displaystyle (A_{n})_{n=0}^{\infty }} are disjoint, we have for every Σ Σ --> {\displaystyle \Sigma } -measurable subset B ⊆ ⊆ --> N {\displaystyle B\subseteq N} that
by the sigma additivity of μ μ --> {\displaystyle \mu } . In particular, this shows that N {\displaystyle N} is a negative set. Next, define P := X ∖ ∖ --> N {\displaystyle P:=X\setminus N} . If P {\displaystyle P} were not a positive set, there would exist a Σ Σ --> {\displaystyle \Sigma } -measurable subset D ⊆ ⊆ --> P {\displaystyle D\subseteq P} with μ μ --> ( D ) < 0 {\displaystyle \mu (D)<0} . Then s n ≤ ≤ --> μ μ --> ( D ) {\displaystyle s_{n}\leq \mu (D)} for all n ∈ ∈ --> N 0 {\displaystyle n\in \mathbb {N} _{0}} and[clarification needed]
which is not allowed for μ μ --> {\displaystyle \mu } . Therefore, P {\displaystyle P} is a positive set.
Proof of the uniqueness statement: Suppose that ( N ′ , P ′ ) {\displaystyle (N',P')} is another Hahn decomposition of X {\displaystyle X} . Then P ∩ ∩ --> N ′ {\displaystyle 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 ∩ ∩ --> P ′ {\displaystyle N\cap P'} . As
this completes the proof. Q.E.D.
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