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Jordan measure decomposition

A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure μ {\displaystyle \mu } defined on Σ {\displaystyle \Sigma } has a unique decomposition into a difference μ = μ + − μ − {\displaystyle \mu =\mu ^{+}-\mu ^{-}} of two positive measures, μ + {\displaystyle \mu ^{+}} and μ − {\displaystyle \mu ^{-}} , at least one of which is finite, such that μ + ( E ) = 0 {\displaystyle {\mu ^{+}}(E)=0} for every Σ {\displaystyle \Sigma } -measurable subset E ⊆ N {\displaystyle E\subseteq N} and μ − ( E ) = 0 {\displaystyle {\mu ^{-}}(E)=0} for every Σ {\displaystyle \Sigma } -measurable subset E ⊆ P {\displaystyle E\subseteq P} , for any Hahn decomposition ( P , N ) {\displaystyle (P,N)} of μ {\displaystyle \mu } . We call μ + {\displaystyle \mu ^{+}} and μ − {\displaystyle \mu ^{-}} the positive and negative part of μ {\displaystyle \mu } , respectively. The pair ( μ + , μ − ) {\displaystyle (\mu ^{+},\mu ^{-})} is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of μ {\displaystyle \mu } . The two measures can be defined as

μ + ( E ) := μ ( E ∩ P ) and μ − ( E ) := − μ ( E ∩ N ) {\displaystyle {\mu ^{+}}(E):=\mu (E\cap P)\qquad {\text{and}}\qquad {\mu ^{-}}(E):=-\mu (E\cap N)}

for every E ∈ Σ {\displaystyle E\in \Sigma } and any Hahn decomposition ( P , N ) {\displaystyle (P,N)} of 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 − ) {\displaystyle (\mu ^{+},\mu ^{-})} of a finite signed measure μ {\displaystyle \mu } , one has

μ + ( E ) = sup B ∈ Σ ,   B ⊆ E μ ( B ) and μ − ( E ) = − inf B ∈ Σ ,   B ⊆ E μ ( B ) {\displaystyle {\mu ^{+}}(E)=\sup _{B\in \Sigma ,~B\subseteq E}\mu (B)\quad {\text{and}}\quad {\mu ^{-}}(E)=-\inf _{B\in \Sigma ,~B\subseteq E}\mu (B)}

for any E {\displaystyle E} in Σ {\displaystyle \Sigma } . Furthermore, if μ = ν + − ν − {\displaystyle \mu =\nu ^{+}-\nu ^{-}} for a pair ( ν + , ν − ) {\displaystyle (\nu ^{+},\nu ^{-})} of finite non-negative measures on X {\displaystyle X} , then

ν + ≥ μ + and ν − ≥ μ − . {\displaystyle \nu ^{+}\geq \mu ^{+}\quad {\text{and}}\quad \nu ^{-}\geq \mu ^{-}.}

The last expression means that the Jordan decomposition is the minimal decomposition of 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).

Proof of the Hahn decomposition theorem

Preparation: Assume that μ {\displaystyle \mu } does not take the value − ∞ {\displaystyle -\infty } (otherwise decompose according to − μ {\displaystyle -\mu } ). As mentioned above, a negative set is a set A ∈ Σ {\displaystyle A\in \Sigma } such that μ ( B ) ≤ 0 {\displaystyle \mu (B)\leq 0} for every Σ {\displaystyle \Sigma } -measurable subset B ⊆ A {\displaystyle B\subseteq A} .

Claim: Suppose that D ∈ Σ {\displaystyle D\in \Sigma } satisfies μ ( D ) ≤ 0 {\displaystyle \mu (D)\leq 0} . Then there is a negative set A ⊆ D {\displaystyle A\subseteq D} such that μ ( A ) ≤ 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

t n := sup ( { μ ( B ) ∣ B ∈ Σ   and   B ⊆ A n } ) {\displaystyle t_{n}:=\sup(\{\mu (B)\mid B\in \Sigma ~{\text{and}}~B\subseteq A_{n}\})}

denote the supremum of μ ( B ) {\displaystyle \mu (B)} over all the 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 ( ∅ ) = 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 {\displaystyle \Sigma } -measurable subset B n ⊆ A n {\displaystyle B_{n}\subseteq A_{n}} satisfying

μ ( B n ) ≥ min ( 1 , t n 2 ) . {\displaystyle \mu (B_{n})\geq \min \!\left(1,{\frac {t_{n}}{2}}\right).}

Set A n + 1 := A n ∖ B n {\displaystyle A_{n+1}:=A_{n}\setminus B_{n}} to finish the induction step. Finally, define

A := D \ ⋃ n = 0 ∞ B n . {\displaystyle A:=D{\Bigg \backslash }\bigcup _{n=0}^{\infty }B_{n}.}

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 {\displaystyle \mu } that

μ ( D ) = μ ( A ) + ∑ n = 0 ∞ μ ( B n ) ≥ μ ( A ) + ∑ n = 0 ∞ min ( 1 , t n 2 ) ≥ μ ( A ) . {\displaystyle \mu (D)=\mu (A)+\sum _{n=0}^{\infty }\mu (B_{n})\geq \mu (A)+\sum _{n=0}^{\infty }\min \!\left(1,{\frac {t_{n}}{2}}\right)\geq \mu (A).}

This shows that μ ( A ) ≤ 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 {\displaystyle \Sigma } -measurable subset B ⊆ A {\displaystyle B\subseteq A} that satisfies μ ( B ) > 0 {\displaystyle \mu (B)>0} . Then t n ≥ 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 ( D ) = + ∞ {\displaystyle \mu (D)=+\infty } , which is a contradiction, since 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

s n := inf ( { μ ( D ) ∣ D ∈ Σ   and   D ⊆ X ∖ N n } ) . {\displaystyle s_{n}:=\inf(\{\mu (D)\mid D\in \Sigma ~{\text{and}}~D\subseteq X\setminus N_{n}\}).}

as the infimum of μ ( D ) {\displaystyle \mu (D)} over all the 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 ( ∅ ) = 0 {\displaystyle \mu (\varnothing )=0} , we have s n ≤ 0 {\displaystyle s_{n}\leq 0} . Hence, there exists a Σ {\displaystyle \Sigma } -measurable subset D n ⊆ X ∖ N n {\displaystyle D_{n}\subseteq X\setminus N_{n}} such that

μ ( D n ) ≤ max ( s n 2 , − 1 ) ≤ 0. {\displaystyle \mu (D_{n})\leq \max \!\left({\frac {s_{n}}{2}},-1\right)\leq 0.}

By the claim above, there is a negative set A n ⊆ D n {\displaystyle A_{n}\subseteq D_{n}} such that μ ( A n ) ≤ 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

N := ⋃ n = 0 ∞ A n . {\displaystyle N:=\bigcup _{n=0}^{\infty }A_{n}.}

As the sets ( A n ) n = 0 ∞ {\displaystyle (A_{n})_{n=0}^{\infty }} are disjoint, we have for every Σ {\displaystyle \Sigma } -measurable subset B ⊆ N {\displaystyle B\subseteq N} that

μ ( B ) = ∑ n = 0 ∞ μ ( B ∩ A n ) {\displaystyle \mu (B)=\sum _{n=0}^{\infty }\mu (B\cap A_{n})}

by the sigma additivity of 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 {\displaystyle \Sigma } -measurable subset D ⊆ P {\displaystyle D\subseteq P} with μ ( D ) < 0 {\displaystyle \mu (D)<0} . Then s n ≤ μ ( D ) {\displaystyle s_{n}\leq \mu (D)} for all n ∈ N 0 {\displaystyle n\in \mathbb {N} _{0}} and

μ ( N ) = ∑ n = 0 ∞ μ ( A n ) ≤ ∑ n = 0 ∞ max ( s n 2 , − 1 ) = − ∞ , {\displaystyle \mu (N)=\sum _{n=0}^{\infty }\mu (A_{n})\leq \sum _{n=0}^{\infty }\max \!\left({\frac {s_{n}}{2}},-1\right)=-\infty ,}

which is not allowed for 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

P △ P ′ = N △ N ′ = ( P ∩ N ′ ) ∪ ( N ∩ P ′ ) , {\displaystyle P\triangle P'=N\triangle N'=(P\cap N')\cup (N\cap P'),}

this completes the proof. Q.E.D.

• Billingsley, Patrick (1995). Probability and Measure -- Third Edition. Wiley Series in Probability and Mathematical Statistics. New York: John Wiley & Sons. ISBN 0-471-00710-2.
• Fischer, Tom (2012). "Existence, uniqueness, and minimality of the Jordan measure decomposition". arXiv:1206.5449 [math.ST].

External links

• Hahn decomposition theorem at PlanetMath.
• "Hahn decomposition", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• "Jordan decomposition (of a signed measure)", Encyclopedia of Mathematics, EMS Press, 2001 [1994]