# Hahn decomposition theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space $(X,\Sigma )$ and any signed measure $\mu$ defined on the $\sigma$-algebra $\Sigma$, there exist two $\Sigma$-measurable sets, $P$ and $N$, of $X$ such that:
* $P\cup N=X$ and $P\cap N=\varnothing$.
* For every $E\in \Sigma$ such that $E\subseteq P$, one has $\mu (E)\geq 0$, i.e., $P$ is a positive set for $\mu$.
* For every $E\in \Sigma$ such that $E\subseteq N$, one has $\mu (E)\leq 0$, i.e., $N$ is a negative set for $\mu$.
Moreover, this decomposition is essentially unique, meaning that for any other pair $(P',N')$ of $\Sigma$-measurable subsets of $X$ fulfilling the three conditions above, the symmetric differences $P\triangle P'$ and $N\triangle N'$ are $\mu$-null sets in the strong sense that every $\Sigma$-measurable subset of them has zero measure. The pair $(P,N)$ is then called a Hahn decomposition of the signed measure $\mu$.

## Jordan measure decomposition

A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure $\mu$ defined on $\Sigma$ has a unique decomposition into a difference $\mu =\mu ^{+}-\mu ^{-}$ of two positive measures, $\mu ^{+}$ and $\mu ^{-}$, at least one of which is finite, such that ${\mu ^{+}}(E)=0$ for every $\Sigma$-measurable subset $E\subseteq N$ and ${\mu ^{-}}(E)=0$ for every $\Sigma$-measurable subset $E\subseteq P$, for any Hahn decomposition $(P,N)$ of $\mu$. We call $\mu ^{+}$ and $\mu ^{-}$ the positive and negative part of $\mu$, respectively. The pair $(\mu ^{+},\mu ^{-})$ is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of $\mu$. The two measures can be defined as
    ${\mu ^{+}}(E):=\mu (E\cap P)\qquad {\text{and}}\qquad {\mu ^{-}}(E):=-\mu (E\cap N)$
for every $E\in \Sigma$ and any Hahn decomposition $(P,N)$ of $\mu$.
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 $(\mu ^{+},\mu ^{-})$ of a finite signed measure $\mu$, one has
    ${\mu ^{+}}(E)=\sup _{B\in \Sigma ,~B\subseteq E}\mu (B)\quad {\text{and}}\quad {\mu ^{-}}(E)=-\inf _{B\in \Sigma ,~B\subseteq E}\mu (B)$
for any $E$ in $\Sigma$. Furthermore, if $\mu =\nu ^{+}-\nu ^{-}$ for a pair $(\nu ^{+},\nu ^{-})$ of finite non-negative measures on $X$, then
    $\nu ^{+}\geq \mu ^{+}\quad {\text{and}}\quad \nu ^{-}\geq \mu ^{-}.$
The last expression means that the Jordan decomposition is the minimal decomposition of $\mu$ 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 $\mu$ does not take the value $-\infty$ (otherwise decompose according to $-\mu$). As mentioned above, a negative set is a set $A\in \Sigma$ such that $\mu (B)\leq 0$ for every $\Sigma$-measurable subset $B\subseteq A$.
Claim: Suppose that $D\in \Sigma$ satisfies $\mu (D)\leq 0$. Then there is a negative set $A\subseteq D$ such that $\mu (A)\leq \mu (D)$.
Proof of the claim: Define $A_{0}:=D$. Inductively assume for $n\in \mathbb {N} _{0}$ that $A_{n}\subseteq D$ has been constructed. Let
    $t_{n}:=\sup(\{\mu (B)\mid B\in \Sigma ~{\text{and}}~B\subseteq A_{n}\})$
denote the supremum of $\mu (B)$ over all the $\Sigma$-measurable subsets $B$ of $A_{n}$. This supremum might a priori be infinite. As the empty set $\varnothing$ is a possible candidate for $B$ in the definition of $t_{n}$, and as $\mu (\varnothing )=0$, we have $t_{n}\geq 0$. By the definition of $t_{n}$, there then exists a $\Sigma$-measurable subset $B_{n}\subseteq A_{n}$ satisfying
    $\mu (B_{n})\geq \min \!\left(1,{\frac {t_{n}}{2}}\right).$
Set $A_{n+1}:=A_{n}\setminus B_{n}$ to finish the induction step. Finally, define
    $A:=D{\Bigg \backslash }\bigcup _{n=0}^{\infty }B_{n}.$
As the sets $(B_{n})_{n=0}^{\infty }$ are disjoint subsets of $D$, it follows from the sigma additivity of the signed measure $\mu$ that
    $\mu (D)=\mu (A)+\sum _{n=0}^{\infty }\mu (B_{n})\geq \mu (A)+\sum _{n=0}^{\infty }\min \!\left(1,{\frac {t_{n}}{2}}\right)\geq \mu (A).$
This shows that $\mu (A)\leq \mu (D)$. Assume $A$ were not a negative set. This means that there would exist a $\Sigma$-measurable subset $B\subseteq A$ that satisfies $\mu (B)>0$. Then $t_{n}\geq \mu (B)$ for every $n\in \mathbb {N} _{0}$, so the series on the right would have to diverge to $+\infty$, implying that $\mu (D)=+\infty$, which is a contradiction, since $\mu (D)\leq 0$. Therefore, $A$ must be a negative set.
Construction of the decomposition: Set $N_{0}=\varnothing$. Inductively, given $N_{n}$, define
    $s_{n}:=\inf(\{\mu (D)\mid D\in \Sigma ~{\text{and}}~D\subseteq X\setminus N_{n}\}).$
as the infimum of $\mu (D)$ over all the $\Sigma$-measurable subsets $D$ of $X\setminus N_{n}$. This infimum might a priori be $-\infty$. As $\varnothing$ is a possible candidate for $D$ in the definition of $s_{n}$, and as $\mu (\varnothing )=0$, we have $s_{n}\leq 0$. Hence, there exists a $\Sigma$-measurable subset $D_{n}\subseteq X\setminus N_{n}$ such that
    $\mu (D_{n})\leq \max \!\left({\frac {s_{n}}{2}},-1\right)\leq 0.$
By the claim above, there is a negative set $A_{n}\subseteq D_{n}$ such that $\mu (A_{n})\leq \mu (D_{n})$. Set $N_{n+1}:=N_{n}\cup A_{n}$ to finish the induction step. Finally, define
    $N:=\bigcup _{n=0}^{\infty }A_{n}.$
As the sets $(A_{n})_{n=0}^{\infty }$ are disjoint, we have for every $\Sigma$-measurable subset $B\subseteq N$ that
    $\mu (B)=\sum _{n=0}^{\infty }\mu (B\cap A_{n})$
by the sigma additivity of $\mu$. In particular, this shows that $N$ is a negative set. Next, define $P:=X\setminus N$. If $P$ were not a positive set, there would exist a $\Sigma$-measurable subset $D\subseteq P$ with $\mu (D)<0$. Then $s_{n}\leq \mu (D)$ for all $n\in \mathbb {N} _{0}$ and
    $\mu (N)=\sum _{n=0}^{\infty }\mu (A_{n})\leq \sum _{n=0}^{\infty }\max \!\left({\frac {s_{n}}{2}},-1\right)=-\infty ,$
which is not allowed for $\mu$. Therefore, $P$ is a positive set.
Proof of the uniqueness statement: Suppose that $(N',P')$ is another Hahn decomposition of $X$. Then $P\cap N'$ is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to $N\cap P'$. As
    $P\triangle P'=N\triangle N'=(P\cap N')\cup (N\cap P'),$
this completes the proof. Q.E.D.