Hahn decomposition theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space ${\displaystyle (X,\Sigma )}${\displaystyle (X,\Sigma )} and any signed measure ${\displaystyle \mu }${\displaystyle \mu } defined on the ${\displaystyle \sigma }${\displaystyle \sigma }-algebra ${\displaystyle \Sigma }${\displaystyle \Sigma }, there exist two ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable sets, ${\displaystyle P}${\displaystyle P} and ${\displaystyle N}${\displaystyle N}, of ${\displaystyle X}${\displaystyle X} such that:

1. ${\displaystyle P\cup N=X}${\displaystyle P\cup N=X} and ${\displaystyle P\cap N=\varnothing }${\displaystyle P\cap N=\varnothing }.
2. For every ${\displaystyle E\in \Sigma }${\displaystyle E\in \Sigma } such that ${\displaystyle E\subseteq P}${\displaystyle E\subseteq P}, one has ${\displaystyle \mu (E)\geq 0}${\displaystyle \mu (E)\geq 0}, i.e., ${\displaystyle P}${\displaystyle P} is a positive set for ${\displaystyle \mu }${\displaystyle \mu }.
3. For every ${\displaystyle E\in \Sigma }${\displaystyle E\in \Sigma } such that ${\displaystyle E\subseteq N}${\displaystyle E\subseteq N}, one has ${\displaystyle \mu (E)\leq 0}${\displaystyle \mu (E)\leq 0}, i.e., ${\displaystyle N}${\displaystyle N} is a negative set for ${\displaystyle \mu }${\displaystyle \mu }.

Moreover, this decomposition is essentially unique, meaning that for any other pair ${\displaystyle (P',N')}${\displaystyle (P',N')} of ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subsets of ${\displaystyle X}${\displaystyle X} fulfilling the three conditions above, the symmetric differences ${\displaystyle P\triangle P'}${\displaystyle P\triangle P'} and ${\displaystyle N\triangle N'}${\displaystyle N\triangle N'} are ${\displaystyle \mu }${\displaystyle \mu }-null sets in the strong sense that every ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset of them has zero measure. The pair ${\displaystyle (P,N)}${\displaystyle (P,N)} is then called a Hahn decomposition of the signed measure ${\displaystyle \mu }${\displaystyle \mu }.

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

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

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

${\displaystyle {\mu ^{+}}(E)=\sup _{B\in \Sigma ,~B\subseteq E}\mu (B)\quad {\text{and}}\quad {\mu ^{-}}(E)=-\inf _{B\in \Sigma ,~B\subseteq E}\mu (B)}${\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 ${\displaystyle E}${\displaystyle E} in ${\displaystyle \Sigma }${\displaystyle \Sigma }. Furthermore, if ${\displaystyle \mu =\nu ^{+}-\nu ^{-}}${\displaystyle \mu =\nu ^{+}-\nu ^{-}} for a pair ${\displaystyle (\nu ^{+},\nu ^{-})}${\displaystyle (\nu ^{+},\nu ^{-})} of finite non-negative measures on ${\displaystyle X}${\displaystyle X}, then

${\displaystyle \nu ^{+}\geq \mu ^{+}\quad {\text{and}}\quad \nu ^{-}\geq \mu ^{-}.}${\displaystyle \nu ^{+}\geq \mu ^{+}\quad {\text{and}}\quad \nu ^{-}\geq \mu ^{-}.}

The last expression means that the Jordan decomposition is the minimal decomposition of ${\displaystyle \mu }${\displaystyle \mu } into the 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 ${\displaystyle \mu }${\displaystyle \mu } does not take the value ${\displaystyle -\infty }${\displaystyle -\infty } (otherwise decompose according to ${\displaystyle -\mu }${\displaystyle -\mu }). As mentioned above, a negative set is a set ${\displaystyle A\in \Sigma }${\displaystyle A\in \Sigma } such that ${\displaystyle \mu (B)\leq 0}${\displaystyle \mu (B)\leq 0} for every ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset ${\displaystyle B\subseteq A}${\displaystyle B\subseteq A}.

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

Proof of the claim: Define ${\displaystyle A_{0}:=D}${\displaystyle A_{0}:=D}. Inductively assume for ${\displaystyle n\in \mathbb {N} _{0}}${\displaystyle n\in \mathbb {N} _{0}} that ${\displaystyle A_{n}\subseteq D}${\displaystyle A_{n}\subseteq D} has been constructed. Let

${\displaystyle t_{n}:=\sup(\{\mu (B)\mid B\in \Sigma ~{\text{and}}~B\subseteq A_{n}\})}${\displaystyle t_{n}:=\sup(\{\mu (B)\mid B\in \Sigma ~{\text{and}}~B\subseteq A_{n}\})}

denote the supremum of ${\displaystyle \mu (B)}${\displaystyle \mu (B)} over all the ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subsets ${\displaystyle B}${\displaystyle B} of ${\displaystyle A_{n}}${\displaystyle A_{n}}. This supremum might a priori be infinite. As the empty set ${\displaystyle \varnothing }${\displaystyle \varnothing } is a possible candidate for ${\displaystyle B}${\displaystyle B} in the definition of ${\displaystyle t_{n}}${\displaystyle t_{n}}, and as ${\displaystyle \mu (\varnothing )=0}${\displaystyle \mu (\varnothing )=0}, we have ${\displaystyle t_{n}\geq 0}${\displaystyle t_{n}\geq 0}. By the definition of ${\displaystyle t_{n}}${\displaystyle t_{n}}, there then exists a ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset ${\displaystyle B_{n}\subseteq A_{n}}${\displaystyle B_{n}\subseteq A_{n}} satisfying

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

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

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

As the sets ${\displaystyle (B_{n})_{n=0}^{\infty }}${\displaystyle (B_{n})_{n=0}^{\infty }} are disjoint subsets of ${\displaystyle D}${\displaystyle D}, it follows from the sigma additivity of the signed measure ${\displaystyle \mu }${\displaystyle \mu } that

${\displaystyle \mu (D)=\mu (A)+\sum _{n=0}^{\infty }\mu (B_{n})\geq \mu (A)+\sum _{n=0}^{\infty }\min \!\left(1,{\frac {t_{n}}{2}}\right)\geq \mu (A).}${\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 ${\displaystyle \mu (A)\leq \mu (D)}${\displaystyle \mu (A)\leq \mu (D)}. Assume ${\displaystyle A}${\displaystyle A} were not a negative set. This means that there would exist a ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset ${\displaystyle B\subseteq A}${\displaystyle B\subseteq A} that satisfies ${\displaystyle \mu (B)>0}${\displaystyle \mu (B)>0}. Then ${\displaystyle t_{n}\geq \mu (B)}${\displaystyle t_{n}\geq \mu (B)} for every ${\displaystyle n\in \mathbb {N} _{0}}${\displaystyle n\in \mathbb {N} _{0}}, so the series on the right would have to diverge to ${\displaystyle +\infty }${\displaystyle +\infty }, implying that ${\displaystyle \mu (D)=+\infty }${\displaystyle \mu (D)=+\infty }, which is a contradiction, since ${\displaystyle \mu (D)\leq 0}${\displaystyle \mu (D)\leq 0}. Therefore, ${\displaystyle A}${\displaystyle A} must be a negative set.

Construction of the decomposition: Set ${\displaystyle N_{0}=\varnothing }${\displaystyle N_{0}=\varnothing }. Inductively, given ${\displaystyle N_{n}}${\displaystyle N_{n}}, define

${\displaystyle s_{n}:=\inf(\{\mu (D)\mid D\in \Sigma ~{\text{and}}~D\subseteq X\setminus N_{n}\}).}${\displaystyle s_{n}:=\inf(\{\mu (D)\mid D\in \Sigma ~{\text{and}}~D\subseteq X\setminus N_{n}\}).}

as the infimum of ${\displaystyle \mu (D)}${\displaystyle \mu (D)} over all the ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subsets ${\displaystyle D}${\displaystyle D} of ${\displaystyle X\setminus N_{n}}${\displaystyle X\setminus N_{n}}. This infimum might a priori be ${\displaystyle -\infty }${\displaystyle -\infty }. As ${\displaystyle \varnothing }${\displaystyle \varnothing } is a possible candidate for ${\displaystyle D}${\displaystyle D} in the definition of ${\displaystyle s_{n}}${\displaystyle s_{n}}, and as ${\displaystyle \mu (\varnothing )=0}${\displaystyle \mu (\varnothing )=0}, we have ${\displaystyle s_{n}\leq 0}${\displaystyle s_{n}\leq 0}. Hence, there exists a ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset ${\displaystyle D_{n}\subseteq X\setminus N_{n}}${\displaystyle D_{n}\subseteq X\setminus N_{n}} such that

${\displaystyle \mu (D_{n})\leq \max \!\left({\frac {s_{n}}{2}},-1\right)\leq 0.}${\displaystyle \mu (D_{n})\leq \max \!\left({\frac {s_{n}}{2}},-1\right)\leq 0.}

By the claim above, there is a negative set ${\displaystyle A_{n}\subseteq D_{n}}${\displaystyle A_{n}\subseteq D_{n}} such that ${\displaystyle \mu (A_{n})\leq \mu (D_{n})}${\displaystyle \mu (A_{n})\leq \mu (D_{n})}. Set ${\displaystyle N_{n+1}:=N_{n}\cup A_{n}}${\displaystyle N_{n+1}:=N_{n}\cup A_{n}} to finish the induction step. Finally, define

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

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

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

by the sigma additivity of ${\displaystyle \mu }${\displaystyle \mu }. In particular, this shows that ${\displaystyle N}${\displaystyle N} is a negative set. Next, define ${\displaystyle P:=X\setminus N}${\displaystyle P:=X\setminus N}. If ${\displaystyle P}${\displaystyle P} were not a positive set, there would exist a ${\displaystyle \Sigma }${\displaystyle \Sigma }-measurable subset ${\displaystyle D\subseteq P}${\displaystyle D\subseteq P} with ${\displaystyle \mu (D)<0}${\displaystyle \mu (D)<0}. Then ${\displaystyle s_{n}\leq \mu (D)}${\displaystyle s_{n}\leq \mu (D)} for all ${\displaystyle n\in \mathbb {N} _{0}}${\displaystyle n\in \mathbb {N} _{0}} and^{[clarification needed ]}

${\displaystyle \mu (N)=\sum _{n=0}^{\infty }\mu (A_{n})\leq \sum _{n=0}^{\infty }\max \!\left({\frac {s_{n}}{2}},-1\right)=-\infty ,}${\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 ${\displaystyle \mu }${\displaystyle \mu }. Therefore, ${\displaystyle P}${\displaystyle P} is a positive set.

Proof of the uniqueness statement: Suppose that ${\displaystyle (N',P')}${\displaystyle (N',P')} is another Hahn decomposition of ${\displaystyle X}${\displaystyle X}. Then ${\displaystyle P\cap N'}${\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 ${\displaystyle N\cap P'}${\displaystyle N\cap P'}. As

${\displaystyle P\triangle P'=N\triangle N'=(P\cap N')\cup (N\cap P'),}${\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].

• 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]

• Theorems in measure theory

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