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

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 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 .

Proof of the Hahn decomposition theorem

Preparation: Assume that does not take the value . 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 not allowed. 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
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.