In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space (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: 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 and 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**** *Sig**ma** has** a uni**qu**e*** decomposition into**** a difference**** of**** two positive**** measures****,**** *mu***^{+} and *mu***^{-},**** at**** least one of**** which is**** finite****,**** such**** that**** **** for every *Sig**ma**-mea**su**ra**bl**e*** subset**** **** and **** for every *Sig**ma**-mea**su**ra**bl**e*** subset****,**** for any Hahn**** decomposition (P,N) of**** *mu***.**** We call \mu^{+} and \mu^{-} the positive and negative part of \mu, respectively. The pair is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of \mu. The two measures can be defined as for every 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 of a finite signed measure \mu, one has for any E in \Sigma. Furthermore, if for a pair of finite non-negative measures on X, then 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 - \infty (otherwise decompose according to - \mu). As mentioned above, a negative set is a set such that for every \Sigma-measurable subset. Claim: Suppose that satisfies. Then there is a negative set such that. Proof of the claim: Define A_{0} := D. Inductively assume for that has been constructed. Let 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, we have. By the definition of t_{n}, there then exists a \Sigma-measurable subset satisfying Set to finish the induction step. Finally, define As the sets are disjoint subsets of D, it follows from the sigma additivity of the signed measure \mu that This shows that. Assume A were not a negative set. This means that there would exist a \Sigma-measurable subset that satisfies \mu(B) > 0. Then for every, so the series on the right would have to diverge to + \infty, implying that , which is a contradiction, since. Therefore, A must be a negative set. Construction of the decomposition: Set. Inductively, given N_{n}, define as the infimum of \mu(D) over all the \Sigma-measurable subsets D of. This infimum might a priori be - \infty. As \varnothing is a possible candidate for D in the definition of s_{n}, and as, we have. Hence, there exists a \Sigma-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 \Sigma-measurable subset that by the sigma additivity of \mu. In particular, this shows that N is a negative set. Next, define. If P were not a positive set, there would exist a \Sigma-measurable subset with \mu(D) < 0. Then for all and 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'. As this completes the proof. Q.E.D.