Hahn decomposition theorem

Short description: Measurability theorem

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

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

Moreover, this decomposition is essentially unique, meaning that for any other pair $(P^{\prime },N^{\prime })$ of $\mathrm{\Sigma }$-measurable subsets of $X$ fulfilling the three conditions above, the symmetric differences $P\mathrm{△}{P}^{\prime }$ and $N\mathrm{△}{N}^{\prime }$ are $\mu$-null sets in the strong sense that every $\mathrm{\Sigma }$-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 $\mathrm{\Sigma }$ 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 $\mathrm{\Sigma }$-measurable subset $E\subseteq N$ and ${\mu }^{-}(E)=0$ for every $\mathrm{\Sigma }$-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

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

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

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

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

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

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 $-\mathrm{\infty }$ (otherwise decompose according to $-\mu$). As mentioned above, a negative set is a set $A\in \mathrm{\Sigma }$ such that $\mu (B)\le 0$ for every $\mathrm{\Sigma }$-measurable subset $B\subseteq A$.

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

Proof of the claim: Define $A_0:=D$. Inductively assume for $n\in {\mathbb{\mathbb{N}}}_0$ that $A_n\subseteq D$ has been constructed. Let

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\mathrm{\Sigma }$-measurable subset $D\subseteq P$ with $\mu (D)<0$. Then $s_n\le \mu (D)$ for all $n\in {\mathbb{\mathbb{N}}}_0$ and

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

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

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

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

this completes the proof. Q.E.D.

References

• 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].

External links

• Hahn decomposition theorem at PlanetMath.
• Hazewinkel, Michiel, ed. (2001), "Hahn decomposition", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046140 
• Hazewinkel, Michiel, ed. (2001), "Jordan decomposition (of a signed measure)", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/h046140