|
|
| Line 1: |
Line 1: |
| In [[mathematics]], the '''Hahn decomposition theorem''', named after the [[Austria]]n [[mathematician]] [[Hans Hahn (mathematician)|Hans Hahn]], states that given a [[sigma-algebra|measurable space]] (''X'',Σ) and a [[signed measure]] ''μ'' defined on the σ-algebra Σ, there exist two measurable sets ''P'' and ''N'' in Σ such that:
| | The name of the writer is Jayson. Mississippi is exactly where his home is. To perform lacross is some thing he would by no means give up. I am an invoicing officer and I'll be promoted soon.<br><br>Feel free to visit my blog ... email psychic readings ([http://conniecolin.com/xe/community/24580 just click the up coming web site]) |
| | |
| #''P'' ∪ ''N'' = ''X'' and ''P'' ∩ ''N'' = ∅.
| |
| #For each ''E'' in Σ such that ''E'' ⊆ ''P'' one has ''μ''(''E'') ≥ 0; that is, ''P'' is a [[positive and negative sets|positive set]] for ''μ''.
| |
| #For each ''E'' in Σ such that ''E'' ⊆ ''N'' one has ''μ''(''E'') ≤ 0; that is, ''N'' is a negative set for ''μ''.
| |
| | |
| Moreover, this decomposition is essentially unique, in the sense that for any other pair (''P''<nowiki>'</nowiki>, ''N''<nowiki>'</nowiki>) of measurable sets fulfilling the above three conditions, the [[symmetric difference]]s ''P'' Δ ''P''<nowiki>'</nowiki> and ''N'' Δ ''N''<nowiki>'</nowiki> are ''μ''-[[null set]]s in the strong sense that every measurable subset of them has zero measure. The pair (''P'',''N'') is called a ''Hahn decomposition'' of the signed measure ''μ''.
| |
| | |
| ==Jordan measure decomposition==
| |
| A consequence of the Hahn decomposition theorem is the ''Jordan decomposition theorem'', which states that every signed measure ''μ'' has a ''unique'' decomposition into a difference
| |
| μ = μ<sup>+</sup> − μ<sup>–</sup>
| |
| of two positive measures ''μ''<sup>+</sup> and ''μ''<sup>–</sup>, at least one of which is finite,
| |
| such that μ<sup>+</sup>(E) = 0 if E ⊆ N and μ<sup>−</sup>(E) = 0 if E ⊆ P for any Hahn decomposition (P,N) of μ. ''μ''<sup>+</sup> and ''μ''<sup>–</sup> are called the ''positive'' and ''negative part'' of ''μ'', respectively.
| |
| The pair (''μ''<sup>+</sup>, ''μ''<sup>–</sup>) is called a ''Jordan decomposition'' (or sometimes ''Hahn–Jordan decomposition'') of ''μ''.
| |
| The two measures can be defined as
| |
| :<math>\mu^+(E):=\mu(E\cap P)\,</math>
| |
| and
| |
| :<math>\mu^-(E):=-\mu(E\cap N)\,</math>
| |
| for every ''E'' in Σ and any Hahn decomposition (P,N) of μ.
| |
| | |
| 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 (μ<sup>+</sup>, μ<sup>−</sup>) of a finite signed measure μ,
| |
| :<math>
| |
| \mu^+(E) = \sup_{B\in\Sigma, B\subset E} \mu(B)
| |
| </math>
| |
| and
| |
| :<math>
| |
| \mu^-(E) = -\inf_{B\in\Sigma, B\subset E} \mu(B)
| |
| </math>
| |
| for any E in Σ. Also, if μ = ν<sup>+</sup> − ν<sup>–</sup> for
| |
| a pair of finite non-negative measures (ν<sup>+</sup>, ν<sup>–</sup>), then
| |
| :<math>
| |
| \nu^+ \geq \mu^+ \text{ and } \nu^- \geq \mu^- .
| |
| </math>
| |
| The last expression means that the Jordan decomposition is the ''minimal'' decomposition of μ 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 [http://arxiv.org/abs/1206.5449 Fischer (2012)].
| |
| | |
| ==Proof of the Hahn decomposition theorem==
| |
| '''Preparation:''' Assume that ''μ'' does not take the value −∞ (otherwise decompose according to −''μ''). As mentioned above, a negative set is a set ''A'' in Σ such that ''μ''(''B'') ≤ 0 for every ''B'' in Σ which is a subset of ''A''.
| |
| | |
| '''Claim:''' Suppose that a set ''D'' in Σ satisfies ''μ''(''D'') ≤ 0. Then there is a negative set ''A'' ⊆ ''D'' such that ''μ''(''A'') ≤ ''μ''(''D'').
| |
| | |
| '''Proof of the claim:''' Define ''A''<sub>0</sub> = ''D''. [[mathematical induction|Inductively]] assume for a natural number ''n'' that ''A<sub>n</sub>'' ⊆ ''D'' has been constructed. Let
| |
| | |
| :<math>t_n=\sup\{\mu(B): B\in\Sigma,\, B\subset A_n\}</math>
| |
| | |
| denote the [[supremum]] of ''μ''(''B'') for all the measurable subsets ''B'' of ''A<sub>n</sub>''. This supremum might a priori be infinite. Since the empty set ∅ is a possible ''B'' in the definition of ''t<sub>n</sub>'' and ''μ''(∅) = 0, we have ''t<sub>n</sub>'' ≥ 0. By definition of ''t<sub>n</sub>'' there exists a ''B<sub>n</sub>'' ⊆ ''A<sub>n</sub>'' in Σ satisfying
| |
| | |
| :<math>\mu(B_n)\ge \min\{1,t_n/2\}.</math>
| |
| | |
| Set ''A''<sub>''n''+1</sub> = ''A<sub>n</sub>'' \ ''B<sub>n</sub>'' to finish the induction step. Define
| |
| | |
| :<math>A=D\setminus\bigcup_{n=0}^\infty B_n.</math>
| |
| | |
| Since the sets (''B<sub>n</sub>'')<sub>''n''≥0</sub> are disjoint subsets of ''D'', it follows from the [[sigma additivity]] of the signed measure ''μ'' that
| |
| | |
| :<math>\mu(A)=\mu(D)-\sum_{n=0}^\infty\mu(B_n)\le\mu(D)-\sum_{n=0}^\infty\min\{1,t_n/2\}.</math>
| |
| | |
| This shows that ''μ''(''A'') ≤ ''μ''(''D''). Assume ''A'' were not a negative set. That means there exists a ''B'' in Σ which is a subset of ''A'' and satisfies ''μ''(''B'') > 0. Then ''t<sub>n</sub>'' ≥ ''μ''(''B'') for every ''n'', hence the [[series (mathematics)|series]] on the right has to diverge to +∞, which means ''μ''(''A'') = –∞, which is not allowed. Therefore, ''A'' must be a negative set.
| |
| | |
| '''Construction of the decomposition:''' Set ''N''<sub>0</sub> = ∅. Inductively, given ''N<sub>n</sub>'', define
| |
| | |
| :<math>s_n:=\inf\{\mu(D):D\in\Sigma,\, D\subset X\setminus N_n\}.</math>
| |
| | |
| as the [[infimum]] of ''μ''(''D'') for all the measurable subsets ''D'' of ''X'' \ ''N<sub>n</sub>''. This infimum might a priori be –∞.
| |
| Since the empty set is a possible ''D'' and ''μ''(∅) = 0, we have ''s<sub>n</sub>'' ≤ 0. Hence there exists a ''D<sub>n</sub>'' in Σ with ''D<sub>n</sub>'' ⊆ ''X'' \ ''N<sub>n</sub>'' and
| |
| | |
| :<math>\mu(D_n)\le \max\{s_n/2, -1\}\le 0.</math>
| |
| | |
| By the claim above, there is a negative set ''A<sub>n</sub>'' ⊆ ''D<sub>n</sub>'' such that ''μ''(''A<sub>n</sub>'') ≤ ''μ''(''D<sub>n</sub>''). Define ''N''<sub>''n''+1</sub> = ''N<sub>n</sub>'' ∪ ''A<sub>n</sub>''
| |
| to finish the induction step.
| |
| | |
| Define
| |
| | |
| :<math>N=\bigcup_{n=0}^\infty A_n.</math>
| |
| | |
| Since the sets (''A<sub>n</sub>'')<sub>''n''≥0</sub> are disjoint, we have for every ''B'' ⊆ ''N'' in Σ that
| |
| | |
| :<math>\mu(B)=\sum_{n=0}^\infty\mu(B\cap A_n)</math>
| |
| | |
| by the sigma additivity of ''μ''. In particular, this shows that ''N'' is a negative set. Define ''P'' = ''X'' \ ''N''. If ''P'' were not a positive set, there exists a ''D'' ⊆ ''P'' in Σ with ''μ''(''D'') < 0. Then ''s<sub>n</sub>'' ≤ ''μ''(''D'') for all ''n'' and
| |
| | |
| :<math>\mu(N)=\sum_{n=0}^\infty\mu(A_n)\le\sum_{n=0}^\infty\max\{s_n/2, -1\}=-\infty,</math>
| |
| | |
| which is not allowed for ''μ''. Therefore, ''P'' is a positive set.
| |
| | |
| '''Proof of the uniqueness statement:'''
| |
| Suppose that <math>(N',P')</math> is another Hahn decomposition of <math>X</math>. Then <math>P\cap N'</math> is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to <math>N\cap P'</math>. Since
| |
| | |
| :<math>P\,\triangle\,P'=N\,\triangle\,N'=(P\cap N')\cup(N\cap P'),</math>
| |
| | |
| this completes the proof. [[Q.E.D.]]
| |
| | |
| ==References==
| |
| | |
| * {{cite book
| |
| | last = Billingsley
| |
| | first = Patrick
| |
| | title = Probability and Measure -- Third Edition
| |
| | series = Wiley Series in Probability and Mathematical Statistics
| |
| |publisher = John Wiley & Sons
| |
| | location = New York
| |
| | year = 1995
| |
| | isbn = 0-471-00710-2
| |
| }}
| |
| * {{cite arXiv |last=Fischer |first=Tom |eprint=1206.5449 |class=math.ST |title=Existence, uniqueness, and minimality of the Jordan measure decomposition |year=2012 }}
| |
| | |
| ==External links==
| |
| * [http://planetmath.org/?op=getobj&from=objects&id=4014 Hahn decomposition theorem] at [[PlanetMath]].
| |
| * {{springer|title=Hahn decomposition|id=p/h046140}}
| |
| * [http://www.encyclopediaofmath.org/index.php/Jordan_decomposition_(of_a_signed_measure) Jordan decomposition of a signed measure] at [http://www.encyclopediaofmath.org/ Encyclopedia of Mathematics]
| |
| | |
| [[Category:Theorems in measure theory]]
| |
| [[Category:Articles containing proofs]]
| |