For every such that , one has , i.e., is a positive set for .
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 .
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).
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 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