Hahn decomposition theorem explained

(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 .

1. 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 .

1. 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 , 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 \qquad (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_ \mu(B) \quad \text \quad (E) = - \inf_ \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 ^{+} and \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 - inf ty (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 N _{0} that A _{n} \subseteq D has been constructed. Lett _{n} : = \sup(\{ \mu(B) \mid B \in \Sigma ~ 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 , (t _{n})/(2) \right).

Set

A _{n} : = A _{n} \setminus B _{n} to finish the induction step. Finally, defineA : = D \ backslash c up ^{inf ty} _{n =0} B _{n} .

As the sets

(B _{n} ) ^{inf ty} _{n =0} are disjoint subsets of D , it follows from the sigma additivity of the signed measure \mu that\mu(D) = \mu(A) + \sum ^{inf ty} _{n =0} \mu(B _{n} ) \geq \mu(A) + \sum ^{inf ty} _{n =0} min \left( 1 , (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 N _{0} , so the series on the right would have to diverge to + inf ty , implying that \mu(D) = + inf ty , 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} , defines _{n} : = inf (\{ \mu(D) \mid D \in \Sigma ~ 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 - inf ty . 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( (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} : = N _{n} \cup A _{n} to finish the induction step. Finally, defineN : = c up ^{inf ty} _{n =0} A _{n} .

As the sets

(A _{n} ) ^{inf ty} _{n =0} are disjoint, we have for every \Sigma -measurable subset B \subseteq N that\mu(B) = \sum ^{inf ty} _{n =0} \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 N _{0} and\mu(N) = \sum ^{inf ty} _{n =0} \mu(A _{n} ) \leq \sum ^{inf ty} _{n =0} max \left( (s _{n})/(2) , - 1 \right) = - inf ty,

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' . AsP \triangle P' = N \triangle N' = (P \cap N') \cup (N \cap P'),

this completes the proof. Q.E.D.

References

• Book: Billingsley , Patrick . Probability and Measure -- Third Edition. Wiley Series in Probability and Mathematical Statistics. John Wiley & Sons. New York. 1995. 0-471-00710-2.
• Fischer . Tom . 1206.5449 . math.ST . Existence, uniqueness, and minimality of the Jordan measure decomposition . 2012 .

External links

• Hahn decomposition theorem at PlanetMath.