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

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

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

Jordan measure decomposition

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).

Proof of the Hahn decomposition theorem

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^{[clarification needed]}

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

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.