Hahn decomposition theorem

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

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

for every ${\displaystyle E\in \Sigma }$ and any Hahn decomposition ${\displaystyle (P,N)}$ of ${\displaystyle \mu }$.

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 ${\displaystyle (\mu ^{+},\mu ^{-})}$ of a finite signed measure ${\displaystyle \mu }$, one has

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

which is not allowed for ${\displaystyle \mu }$. Therefore, ${\displaystyle P}$ is a positive set.

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

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

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]