Inmathematics, theHahn decomposition theorem, named after theAustrianmathematicianHans Hahn, states that for anymeasurable space${\displaystyle (X,\mathrm{\Sigma })}$ and anysigned measure${\displaystyle \mu }$ defined on the${\displaystyle \sigma }$-algebra${\displaystyle \mathrm{\Sigma }}$, there exist two${\displaystyle \mathrm{\Sigma }}$-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${\displaystyle E\in \mathrm{\Sigma }}$ such that${\displaystyle E\subseteq P}$, one has${\displaystyle \mu (E)\ge 0}$, i.e.,${\displaystyle P}$ is apositive set for${\displaystyle \mu }$.
3. For every${\displaystyle E\in \mathrm{\Sigma }}$ such that${\displaystyle E\subseteq N}$, one has${\displaystyle \mu (E)\le 0}$, i.e.,${\displaystyle N}$ is a negative set for${\displaystyle \mu }$.

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

A consequence of the Hahn decomposition theorem is theJordan decomposition theorem, which states that every signed measure${\displaystyle \mu }$ defined on${\displaystyle \mathrm{\Sigma }}$ has aunique 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${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle E\subseteq N}$ and${\displaystyle {\mu }^{-}(E)=0}$ for every${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle E\subseteq P}$, for any Hahn decomposition${\displaystyle (P,N)}$ of${\displaystyle \mu }$. We call${\displaystyle {\mu }^{+}}$ and${\displaystyle {\mu }^{-}}$ thepositive andnegative part of${\displaystyle \mu }$, respectively. The pair${\displaystyle ({\mu }^{+},{\mu }^{-})}$ is called aJordan decomposition (or sometimesHahn–Jordan decomposition) of${\displaystyle \mu }$. The two measures can be defined as

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

for every${\displaystyle E\in \mathrm{\Sigma }}$ 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

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

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

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

The last expression means that the Jordan decomposition is theminimal decomposition of${\displaystyle \mu }$ into a difference of non-negative measures. This is theminimality 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 seeFischer (2012).

Preparation: Assume that${\displaystyle \mu }$ does not take the value${\displaystyle -\mathrm{\infty }}$ (otherwise decompose according to${\displaystyle -\mu }$). As mentioned above, a negative set is a set${\displaystyle A\in \mathrm{\Sigma }}$ such that${\displaystyle \mu (B)\le 0}$ for every${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle B\subseteq A}$.

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

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

${\displaystyle t_n:=sup(\{\mu (B)\mid B\in \mathrm{\Sigma }\text{and}B\subseteq A_n\})}$

denote thesupremum of${\displaystyle \mu (B)}$ over all the${\displaystyle \mathrm{\Sigma }}$-measurable subsets${\displaystyle B}$ of${\displaystyle A_n}$. This supremum mighta 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${\displaystyle t_n\ge 0}$. By the definition of${\displaystyle t_n}$, there then exists a${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle B_n\subseteq A_n}$ satisfying

${\displaystyle \mu (B_n)\ge min\left(1,\frac{t_n}{2}\right).}$

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

${\displaystyle A:=D\backslash \bigcup _{n=0}^{\mathrm{\infty }}{B}_n.}$

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

${\displaystyle \mu (D)=\mu (A)+\sum _{n=0}^{\mathrm{\infty }}\mu (B_n)\ge \mu (A)+\sum _{n=0}^{\mathrm{\infty }}min\left(1,\frac{t_n}{2}\right)\ge \mu (A).}$

This shows that${\displaystyle \mu (A)\le \mu (D)}$. Assume${\displaystyle A}$ were not a negative set. This means that there would exist a${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle B\subseteq A}$ that satisfies${\displaystyle \mu (B)>0}$. Then${\displaystyle t_n\ge \mu (B)}$ for every${\displaystyle n\in {\mathbb{N}}_0}$, so theseries on the right would have to diverge to${\displaystyle +\mathrm{\infty }}$, implying that${\displaystyle \mu (D)=+\mathrm{\infty }}$, which is a contradiction, since${\displaystyle \mu (D)\le 0}$. Therefore,${\displaystyle A}$ must be a negative set.

Construction of the decomposition: Set${\displaystyle N_0=\varnothing }$. Inductively, given${\displaystyle N_n}$, define

${\displaystyle s_n:=inf(\{\mu (D)\mid D\in \mathrm{\Sigma }\text{and}D\subseteq X\setminus N_n\}).}$

as theinfimum of${\displaystyle \mu (D)}$ over all the${\displaystyle \mathrm{\Sigma }}$-measurable subsets${\displaystyle D}$ of${\displaystyle X\setminus N_n}$. This infimum mighta priori be${\displaystyle -\mathrm{\infty }}$. 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${\displaystyle s_n\le 0}$. Hence, there exists a${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle D_n\subseteq X\setminus N_n}$ such that

${\displaystyle \mu (D_n)\le max\left(\frac{s_n}{2},-1\right)\le 0.}$

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

${\displaystyle N:=\bigcup _{n=0}^{\mathrm{\infty }}{A}_n.}$

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

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

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${\displaystyle \mathrm{\Sigma }}$-measurable subset${\displaystyle D\subseteq P}$ with. Then${\displaystyle s_n\le \mu (D)}$ for all${\displaystyle n\in {\mathbb{N}}_0}$ and^{[clarification needed]}

${\displaystyle \mu (N)=\sum _{n=0}^{\mathrm{\infty }}\mu (A_n)\le \sum _{n=0}^{\mathrm{\infty }}max\left(\frac{s_n}{2},-1\right)=-\mathrm{\infty },}$

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

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

${\displaystyle P\mathrm{△}{P}^{\prime }=N\mathrm{△}{N}^{\prime }=(P\cap N^{\prime })\cup (N\cap P^{\prime }),}$

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 atPlanetMath.
• "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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