Hahn decomposition theorem

In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space and any signed measure defined on the -algebra , there exist two -measurable sets, and , of such that:

  1. and .
  2. For every such that , one has , i.e., is a positive set for .
  3. For every such that , one has , i.e., is a negative set for .

Moreover, this decomposition is essentially unique, meaning that for any other pair of -measurable subsets of fulfilling the three conditions above, the symmetric differences and are -null sets in the strong sense that every -measurable subset of them has zero measure. The pair is then called a Hahn decomposition of the signed measure .

Jordan measure decomposition

edit

A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure   defined on   has a unique decomposition into a difference   of two positive measures,   and  , at least one of which is finite, such that   for every  -measurable subset   and   for every  -measurable subset  , for any Hahn decomposition   of  . We call   and   the positive and negative part of  , respectively. The pair   is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of  . The two measures can be defined as

 

for every   and any Hahn decomposition   of  .

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   of a finite signed measure  , one has

 

for any   in  . Furthermore, if   for a pair   of finite non-negative measures on  , then

 

The last expression means that the Jordan decomposition is the minimal decomposition of   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

edit

Preparation: Assume that   does not take the value   (otherwise decompose according to  ). As mentioned above, a negative set is a set   such that   for every  -measurable subset  .

Claim: Suppose that   satisfies  . Then there is a negative set   such that  .

Proof of the claim: Define  . Inductively assume for   that   has been constructed. Let

 

denote the supremum of   over all the  -measurable subsets   of  . This supremum might a priori be infinite. As the empty set   is a possible candidate for   in the definition of  , and as  , we have  . By the definition of  , there then exists a  -measurable subset   satisfying

 

Set   to finish the induction step. Finally, define

 

As the sets   are disjoint subsets of  , it follows from the sigma additivity of the signed measure   that

 

This shows that  . Assume   were not a negative set. This means that there would exist a  -measurable subset   that satisfies  . Then   for every  , so the series on the right would have to diverge to  , implying that  , which is a contradiction, since  . Therefore,   must be a negative set.

Construction of the decomposition: Set  . Inductively, given  , define

 

as the infimum of   over all the  -measurable subsets   of  . This infimum might a priori be  . As   is a possible candidate for   in the definition of  , and as  , we have  . Hence, there exists a  -measurable subset   such that

 

By the claim above, there is a negative set   such that  . Set   to finish the induction step. Finally, define

 

As the sets   are disjoint, we have for every  -measurable subset   that

 

by the sigma additivity of  . In particular, this shows that   is a negative set. Next, define  . If   were not a positive set, there would exist a  -measurable subset   with  . Then   for all   and[clarification needed]

 

which is not allowed for  . Therefore,   is a positive set.

Proof of the uniqueness statement: Suppose that   is another Hahn decomposition of  . Then   is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to  . As

 

this completes the proof. Q.E.D.

References

edit
  • 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].
edit