Hahn decomposition theorem

Hahn decomposition theorem In mathematics, the Hahn decomposition theorem, named after the Austrian mathematician Hans Hahn, states that for any measurable space {displaystyle (X,Sigma )} and any signed measure {displaystyle mu } defined on the {displaystyle sigma } -algebra {displaystyle Sigma } , there exist two {displaystyle Sigma } -measurable sets, {displaystyle P} and {displaystyle N} , of {displaystyle X} such that: {displaystyle Pcup N=X} and {displaystyle Pcap N=varnothing } . For every {displaystyle Ein Sigma } such that {displaystyle Esubseteq P} , one has {displaystyle mu (E)geq 0} , i.e., {displaystyle P} is a positive set for {displaystyle mu } . For every {displaystyle Ein Sigma } such that {displaystyle Esubseteq N} , one has {displaystyle mu (E)leq 0} , i.e., {displaystyle N} is a negative set for {displaystyle mu } .

Moreover, this decomposition is essentially unique, meaning that for any other pair {displaystyle (P',N')} of {displaystyle Sigma } -measurable subsets of {displaystyle X} fulfilling the three conditions above, the symmetric differences {displaystyle Ptriangle P'} and {displaystyle Ntriangle N'} are {displaystyle mu } -null sets in the strong sense that every {displaystyle Sigma } -measurable subset of them has zero measure. The pair {displaystyle (P,N)} is then called a Hahn decomposition of the signed measure {displaystyle mu } .

Contents 1 Jordan measure decomposition 2 Proof of the Hahn decomposition theorem 3 References 4 External links Jordan measure decomposition A consequence of the Hahn decomposition theorem is the Jordan decomposition theorem, which states that every signed measure {displaystyle mu } defined on {displaystyle Sigma } has a unique decomposition into a difference {displaystyle mu =mu ^{+}-mu ^{-}} of two positive measures, {displaystyle mu ^{+}} and {displaystyle mu ^{-}} , at least one of which is finite, such that {displaystyle {mu ^{+}}(E)=0} for every {displaystyle Sigma } -measurable subset {displaystyle Esubseteq N} and {displaystyle {mu ^{-}}(E)=0} for every {displaystyle Sigma } -measurable subset {displaystyle Esubseteq P} , for any Hahn decomposition {displaystyle (P,N)} of {displaystyle mu } . We call {displaystyle mu ^{+}} and {displaystyle mu ^{-}} the positive and negative part of {displaystyle mu } , respectively. The pair {displaystyle (mu ^{+},mu ^{-})} is called a Jordan decomposition (or sometimes Hahn–Jordan decomposition) of {displaystyle mu } . The two measures can be defined as {displaystyle {mu ^{+}}(E):=mu (Ecap P)qquad {text{and}}qquad {mu ^{-}}(E):=-mu (Ecap N)} for every {displaystyle Ein Sigma } and any Hahn decomposition {displaystyle (P,N)} of {displaystyle 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 {displaystyle (mu ^{+},mu ^{-})} of a finite signed measure {displaystyle mu } , one has {displaystyle {mu ^{+}}(E)=sup _{Bin Sigma ,~Bsubseteq E}mu (B)quad {text{and}}quad {mu ^{-}}(E)=-inf _{Bin Sigma ,~Bsubseteq E}mu (B)} for any {displaystyle E} in {displaystyle Sigma } . Furthermore, if {displaystyle mu =nu ^{+}-nu ^{-}} for a pair {displaystyle (nu ^{+},nu ^{-})} of finite non-negative measures on {displaystyle X} , then {displaystyle nu ^{+}geq mu ^{+}quad {text{and}}quad nu ^{-}geq mu ^{-}.} The last expression means that the Jordan decomposition is the minimal decomposition of {displaystyle 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 {displaystyle mu } does not take the value {displaystyle -infty } (otherwise decompose according to {displaystyle -mu } ). As mentioned above, a negative set is a set {displaystyle Ain Sigma } such that {displaystyle mu (B)leq 0} for every {displaystyle Sigma } -measurable subset {displaystyle Bsubseteq A} .

Claim: Suppose that {displaystyle Din Sigma } satisfies {displaystyle mu (D)leq 0} . Then there is a negative set {displaystyle Asubseteq D} such that {displaystyle mu (A)leq mu (D)} .

Proof of the claim: Define {displaystyle A_{0}:=D} . Inductively assume for {displaystyle nin mathbb {N} _{0}} that {displaystyle A_{n}subseteq D} has been constructed. Let {displaystyle t_{n}:=sup({mu (B)mid Bin Sigma ~{text{and}}~Bsubseteq A_{n}})} denote the supremum of {displaystyle mu (B)} over all the {displaystyle Sigma } -measurable subsets {displaystyle B} of {displaystyle A_{n}} . This supremum might a priori be infinite. As the empty set {displaystyle varnothing } is a possible candidate for {displaystyle B} in the definition of {displaystyle t_{n}} , and as {displaystyle mu (varnothing )=0} , we have {displaystyle t_{n}geq 0} . By the definition of {displaystyle t_{n}} , there then exists a {displaystyle Sigma } -measurable subset {displaystyle B_{n}subseteq A_{n}} satisfying {displaystyle mu (B_{n})geq min !left(1,{frac {t_{n}}{2}}right).} Set {displaystyle A_{n+1}:=A_{n}setminus B_{n}} to finish the induction step. Finally, define {displaystyle A:=D{Bigg backslash }bigcup _{n=0}^{infty }B_{n}.} As the sets {displaystyle (B_{n})_{n=0}^{infty }} are disjoint subsets of {displaystyle D} , it follows from the sigma additivity of the signed measure {displaystyle mu } that [need a justification that does not use subtraction] {displaystyle mu (A)=mu (D)-sum _{n=0}^{infty }mu (B_{n})leq mu (D)-sum _{n=0}^{infty }min !left(1,{frac {t_{n}}{2}}right).} This shows that {displaystyle mu (A)leq mu (D)} . Assume {displaystyle A} were not a negative set. This means that there would exist a {displaystyle Sigma } -measurable subset {displaystyle Bsubseteq A} that satisfies {displaystyle mu (B)>0} . Then {displaystyle t_{n}geq mu (B)} for every {displaystyle nin mathbb {N} _{0}} , so the series on the right would have to diverge to {displaystyle +infty } , implying that {displaystyle mu (A)=-infty } , which is not allowed. Therefore, {displaystyle A} must be a negative set.

Construction of the decomposition: Set {displaystyle N_{0}=varnothing } . Inductively, given {displaystyle N_{n}} , define {displaystyle s_{n}:=inf({mu (D)mid Din Sigma ~{text{and}}~Dsubseteq Xsetminus N_{n}}).} as the infimum of {displaystyle mu (D)} over all the {displaystyle Sigma } -measurable subsets {displaystyle D} of {displaystyle Xsetminus N_{n}} . This infimum might a priori be {displaystyle -infty } . As {displaystyle varnothing } is a possible candidate for {displaystyle D} in the definition of {displaystyle s_{n}} , and as {displaystyle mu (varnothing )=0} , we have {displaystyle s_{n}leq 0} . Hence, there exists a {displaystyle Sigma } -measurable subset {displaystyle D_{n}subseteq Xsetminus N_{n}} such that {displaystyle mu (D_{n})leq max !left({frac {s_{n}}{2}},-1right)leq 0.} By the claim above, there is a negative set {displaystyle A_{n}subseteq D_{n}} such that {displaystyle mu (A_{n})leq mu (D_{n})} . Set {displaystyle N_{n+1}:=N_{n}cup A_{n}} to finish the induction step. Finally, define {displaystyle N:=bigcup _{n=0}^{infty }A_{n}.} As the sets {displaystyle (A_{n})_{n=0}^{infty }} are disjoint, we have for every {displaystyle Sigma } -measurable subset {displaystyle Bsubseteq N} that {displaystyle mu (B)=sum _{n=0}^{infty }mu (Bcap A_{n})} by the sigma additivity of {displaystyle mu } . In particular, this shows that {displaystyle N} is a negative set. Next, define {displaystyle P:=Xsetminus N} . If {displaystyle P} were not a positive set, there would exist a {displaystyle Sigma } -measurable subset {displaystyle Dsubseteq P} with {displaystyle mu (D)<0} . Then {displaystyle s_{n}leq mu (D)} for all {displaystyle nin mathbb {N} _{0}} and {displaystyle mu (N)=sum _{n=0}^{infty }mu (A_{n})leq sum _{n=0}^{infty }max !left({frac {s_{n}}{2}},-1right)=-infty ,} which is not allowed for {displaystyle mu } . Therefore, {displaystyle P} is a positive set. Proof of the uniqueness statement: Suppose that {displaystyle (N',P')} is another Hahn decomposition of {displaystyle X} . Then {displaystyle Pcap N'} is a positive set and also a negative set. Therefore, every measurable subset of it has measure zero. The same applies to {displaystyle Ncap P'} . As {displaystyle Ptriangle P'=Ntriangle N'=(Pcap N')cup (Ncap P'),} this completes the proof. Q.E.D. References 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]. External links Hahn decomposition theorem at PlanetMath. "Hahn decomposition", Encyclopedia of Mathematics, EMS Press, 2001 [1994] Jordan decomposition of a signed measure at Encyclopedia of Mathematics hide vte Measure theory Basic concepts Absolute continuityLebesgue integrationLp spacesMeasureMeasure space Probability spaceMeasurable space/function Sets Almost everywhereBorel setCarathéodory's criterionConvergence in measure -systemEssential range infimum/supremumLocally measurableπ-systemσ-algebraNon-measurable set Vitali setNull setSupportTransverse measure Types of Measures BaireBanachBesovBorelComplexCompleteContent(Logarithmically) ConvexDiscreteFiniteInner(Quasi-) InvariantLocally finiteMaximisingMetric outerOuterPerfectPre-measure(Sub-) ProbabilityProjection-valuedRadonRandomRegular Borel regularInner regularOuter regularSaturatedSet functionσ-finites-finiteSignedSingularSpectralStrictly positiveTightVector Particular measures CountingDiracEulerGaussianHaarHarmonicHausdorffIntensityLebesgueLogarithmicProductPushforwardSpherical measureTangentTrivialYoung Main results Carathéodory's extension theoremConvergence theorems DominatedMonotoneVitaliDecomposition theorems HahnJordanEgorov'sFatou's lemmaFubini'sHölder's inequalityMinkowski inequalityRadon–NikodymRiesz–Markov–Kakutani representation theorem Other results Disintegration theoremLebesgue's density theoremLebesgue differentiation theoremSard's theorem Applications Probability theoryReal analysisSpectral theory Categories: Theorems in measure theory

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