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HAHN DECOMPOSITION AND RADON-NIKODYM THEOREM WITH A PARAMETER D.NOVIKOV 5 0 Abstract. The paper contains a simple proof of the classical Hahn decomposition theorem for 0 chargesand,asacorollary,anexplicitmeasurableinparameterconstructionofaRadon-Nikodym 2 derivativeofonemeasurebyanother. n a J 4 1. Introduction 1 This text appeared as an answer to a question of E.B.Dynkin, and was initially meant to appear ] A as an appendix to [1]. It contains a straightforward construction of a Hahn decomposition for a charge on a measurable space. As an application, we prove measurability by parameter of the C Radon-Nikodym derivative of one measure by another. . h A required in [1] slightly weaker result was later proved by a much shorter and less elementary t arguments. a m I would like to thank E.B.Dynkin for asking the question and for transforming the initial version of the text to a readable one. [ 1 1.1. Our goal is to prove v 5 Theorem 1.1. Suppose that P(dν) is a probability measure on a Luzin measurable space (M,M), 1 and let P(µ,C), µ∈M, C ∈M be a probability measure in C and a M-measurable function in µ. 2 If, for every µ∈M, the measure P(µ,·) is absolutely continuous with respect to P, then there exists 1 a M×M-measurable version of the Radon-Nikodym derivative P(µ,dν). 0 P(dν) 5 Every function f : M × M → R ∪ {0} can be characterized by a countable family of sets 0 + / Sr ={f ≤r} where r ≥0 are rationals. Namely, h t (1.1) f(µ,ν)=inf{r:(µ,ν)∈Sr}. a m This way we establish a 1-1 correspondence between f and families S with the properties: r v: 1.1.A. Sr ⊂Sr′ for any r <r′, 1.1.B. S =M×M i r r X A function f is M×M-measurable if and only if all S belong to M. S r r a 1.2. First, we prove Lemma 1.1. Assume that the family S ∈ M satisfies the conditions 1.1.A–1.1.B and that f is r defined by (1.1). Put S (µ)=S ∩({µ}×M). If r r P(µ,dν)≤rP(dν) on S (µ), r (1.2) P(µ,dν)>rP(dν) on M\S (µ), r then f is a version of P(µ,dν). P(dν) [We write P′(dν) ≥ P′′(dν) on B if P′(A) ≥ P′′(A) for all A ∈ M such that A ⊂ B. The condition(1.2)means that S (µ),M\S (µ) is a Hahn decompositionof the chargeP(µ,·)−rP(·).] r r Proof. We need to demonstrate that, for every A ∈ M, P(µ,A) = fP(dν). Fix δ > 0 and A introduce a function R (1.3) f (µ,ν)=mδ on {mδ ≤f <(m+1)δ},m=0,1,2,.... δ 1 2 D.NOVIKOV By (1.2), P(µ,dν)≤(m+1)δP(dν) on {f <(m+1)δ}, (1.4) P(µ,dν)>mδP(dν) on {f ≥mδ}. By (1.3) and (1.4), (1.5) P(µ,dν)>f P(dν)≥m/(m+1)P(µ,dν) on {mδ ≤f <(m+1)δ}. δ Since m/(m+1) is monotone increasing, (1.6) f P(dν)≥(n+1)/(n+2)P(µ,dν) on {f >(n+1)δ}. δ We have f P(dν)=I +J where A δ δ δ R (1.7) I = f 1 P(dν)≤nδ δ δ fδ≤nδ ZA and J = f 1 P(dν). Since {f >nδ}⊃{f >(n+1)δ}, we have, by (1.6), δ A δ fδ>nδ δ R (1.8) J ≥ f 1 P(dν)≥(n+1)/(n+2)P(µ,A∩{f >(n+1)δ}) δ δ f>(n+1)δ ZA Since P(µ,dν)≤(n+1)δP(dν) on {f ≤(n+1)δ}, we have (1.9) P(µ,A∩{f >(n+1)δ})≥P(µ,A)−P(µ,{f ≤(n+1)δ})≥P(µ,A)−(n+1)δ, and we get (1.10) J ≥(n+1)/(n+2)(P(µ,A)−(n+1)δ). δ Let δ → 0.Then f ↑ f. By (1.7), I → 0 and therefore J → fP(dν). It follows from (1.10) δ δ δ A that fP(dν)≥(n+1)/(n+2)P(µ,A). By tending n to ∞ we get fP(dν)≥P(µ,A). On the A R A other hand, by (1.5), fP(dν)≤P(µ,A). (cid:3) R A R R 2. Construction of family S r 2.1. On Hahn decomposition. Assume that λ is a charge of bounded variation on a measurable space (X,A). Hahn decomposition corresponding to λ is a representation of X as a disjoint union of two sets X and X , such that λ(A) ≥ 0 for all measurable A ⊂ X , and λ(A) ≥ 0 for all − + + measurable A⊂ X . We construct such a decomposition in a way to be used for constructing sets − S . r Theorem 2.1. Let β = inf λ(A) and let E ⊂ A be such that λ(E ) < β +ǫ , where ǫ > 0 A∈A k k k k and ǫ <∞. Put k X = E . P m k k≥m \ The pair (2.1) X = X , X =X \X − m + − m≥1 [ is a Hahn decomposition corresponding to λ. Proof. 1◦. Note that, if A ⊂ E , then β ≤≪ {E \A} =≪ (E )− ≪ (A) < β +ε − ≪ (A) and k k k k therefore ≪ (A) < ε . Hence, if A ⊂ X , then ≪ (A) ≤ 0 which implies that ≪ (A) ≤ 0 for all k m subsets A of X . − 2◦. First, note that (2.2) λ(A ∩A )≤λ(A )+λ(A )−β for all A ,A ∈A. 1 2 1 2 1 2 This follows from the identities: ≪(A )=λ(A \A )+λ(A ∩A ), 1 1 2 1 2 ≪(A )=λ(A \A )+λ(A ∩A ), 2 2 1 1 2 λ(A \A )+λ(A ∩A )+λ(A \A )=λ(A ∪A )≥β 1 2 1 2 2 1 1 2 HAHN DECOMPOSITION AND RADON-NIKODYM THEOREM WITH A PARAMETER 3 Since ≪ (E ) ≤ β +ε , (2.2) implies, by induction in ℓ, that λ(E +···+E ) ≤ β +ε + k k m m+ℓ m ···+ε and therefore λ(X )≤β+εm where εm = ǫ . m+ℓ m k≥m k Now, suppose that λ(A)=−δ <0 for some A⊂X , and take m such that εm <δ. Then, since + P A∩X =∅ and, therefore, A∩X =∅, we get − m λ(A∪X )=λ(A)+λ(X )<β m m which contradicts the definition of β. (cid:3) 2.2. Measurable decomposition related to charge depending on parameter. Theorem 2.2. Let λ(µ,·), µ ∈ M, be a family of charges on a Luzin measurable space (M,M). Assume that λ(µ,A) is a M-measurable function of µ ∈ M for any A ∈ M. Then there exists a set X ∈ M×M, such that its sections X (µ) = X ∩({µ}×M) define Hahn decomposition − − − X (µ),M\X (µ) of the charge λ(µ,cdot). − − Proof. Let {I }⊂M, I =∅, be a countable family of subsets of M with the following properties n 1 C for any A ∈ M, any ǫ > 0 and any µ ∈ M there exists n = n(A,ǫ,µ)i such that λ(µ,A\ I )+λ(µ,I \A)<ǫ n n Finite unions of intervals with rationalendpoints formsuch a family for [0,1]with BorelA-algebra. Therefore such a family exists for any Luzin measurable space. Denote β(µ) = inf λ(µ,I ). Since λ(µ,I ) are M-measurable functions of µ, the β(µ) is also n n n M-measurable. Choose a sequence of positive ǫ such that ǫ <∞. Define k k E = PA ×I , k N,k N N≥1 [ where A is the set of all µ such that N is the smallest number with the property λ(µ,I ) < N,k N β(µ)+ǫ . k Note that for any fixed µ the family of sections {E (µ)} , E (µ) = E ∩({µ}×M), satisfy k k≥1 k k conditions of Theorem 2.1: λ(µ,E (µ))<β(µ)+ǫ(k). k Therefore, if we put X = E , X = X ⊂M×M, m k − m k≥m m≥1 \ [ then the sections X (µ) define Hahn decomposition of λ(µ,·) for any fixed µ. − Let us prove that X ∈ M × M. It is enough to prove that A ∈ M. The set Y = − N,k n,k {µ|λ(µ,I )−β(µ)<ǫ }isM-measurablesincebothλ(µ,I )andβ(µ)areM-measurable. Therefore n k n A =Y \ Y ∈M. N,k N,k n,k n<N [ (cid:3) 2.3. ProofofTheorem1.1. LetXr bethesetsconstructedinTheorem2.2forthechargeP(µ,·)− − rP(·). By construction,they satisfy the condition 1.2. The followingmodification of Xr satisfies in − addition the conditions 1.1.A-1.1.B, and, therefore, defines a M×M-measurable version of Radon- Nikodym derivative. Put (2.3) S =(M×M)\ Xr , S = Xr′ ∪S , 0 r −! r r′≤r − 0 [ [ where r,r′ denote non-negative rational numbers.   Evidently, Sr ∈ M×M. Evidently, Sr′ ⊆ Sr for r′ < r, so the condition 1.1.B holds. Since Xr ⊂S and by definition of S the condition 1.1.B holds as well. − r 0 LetuschecktheconditionsoftheLemma1.1,i.e. thatS (µ),M\S (µ)isaHahndecomposition r r of the charge P(µ,·)−rP(·). 4 D.NOVIKOV Note that (2.4) S (µ)\Xr(µ)=S (µ)∪ Xr′(µ)\Xr(µ) , r − 0 − − r[′≤r(cid:16) (cid:17) andXr(µ) define a HahndecompositionofP(µ,·)−rP(·). Therefore,it is enoughto provethatall − the sets onthe rightside are disjoint fromthe support of P(µ,·)−rP(·). Since P(µ,·) is absolutely continuous with respect to P(·), it is enough to prove that all sets on the right hand side of 2.4 are disjoint from the support of P(·). By definition of S , P(µ,·)−rP(·) is non-negative on S (µ) for any r ≥0. This is possible only 0 0 if P is zero on S (µ). 0 The charge P(µ,·)−rP is non-negative on Xr′(µ)\Xr(µ), and the charge − − P(µ,·)−r′P isnon-positiveonXr′(µ)\Xr(µ). Forr′ <r thisispossibleonlyifP iszeroonXr′(µ)\Xr(µ). (cid:3) − − − − This construction can be repeated almost word-by-wordto prove Theorem 2.3. Suppose {λ ,µ } are two families of measure on a measurable space (Y,A ) and x x Y the parameter x belongs to a measurable space (X,A ). Assume that for each x the measure µ is X x absolutely continuous with respect to λ . x Assumethatλ ,µ aremeasurablefunctionsin x: foranyA∈σ thefunctions λ (A)andµ (A) x x Y x x are σ -measurable. x Assume that there exists a countable family {Sn}n∈N ⊂σY such that for any A∈σY, any x∈X, and any ǫ>0 one can find S =S (A,x,ǫ) such that λ (A∆S )<ǫ and µ (A∆S )<ǫ. n n x n x n Then there exists a σ ×σ -measurable version of the Radon-Nikodym derivative µx. X Y λx References [1] E.B.Dynkin,X-harmonic functions,preprint