Lebesgue-Radon-Nikodym Theorem

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1 Lebesgue-Radon-Nikodym Theorem Matt Rosenzweig 1 Lebesgue-Radon-Nikodym Theorem In what follows, (, A) will denote a measurable space. We begin with a review of signed measures. 1.1 Signed Measures Definition 1. A signed measure ν on a σ-algebra A is a set function such that 1. ν is extended-valued in the sense that < ν() for all A. If { j } are disjoint subsets of A, then ν j = ν( j ) Note that in order for condition () to hold, the sum ν( j ) must be independent of any rearrangement. If ν( j) is finite, then it follows from a theorem of Riemann that the sum converges absolutely. I have included a proof of this result for the interested reader. Lemma. (Riemann Rearrangement Lemma) Let a n be a series of real numbers which converges, but not absolutely. Suppose α β. Then there exists a rearrangement a n with partial sums s n such that Proof. For n N, set lim inf s n = α, lim sup s n = β p n = a n + a n, q n = a n a n Then p n q n = a n, p n + q n = a n, p n 0, q n 0. I claim that the series p n, q n diverge. If either pn or p n converges, then since a n converges, it follows that a n converges, which contradicts our hypothesis that a n does not converge absolutely. Let P 1, P, denote the nonnegative terms of a n, in the order in which they occur, and let Q 1, Q, denote the absolute values of the negative terms of a n also in their original order. Since N n=1 P n = MN n=1 p n and N n=1 Q n = M N n=1 q n for all N N, both P n, Q n diverge. We will construct sequences (m n ) n=1, (k n ) n=1, such that P P m1 Q 1 Q k1 + P m P m Q k1+1 Q k + satisfies the conclusion of the lemma. Choose real sequences (α n ) n=1, (β n ) n=1 such that α n α, β n β. Let m 1, k 1 be the minimal positive integers such that and Let m, k be the minimal positive integers such that P P m1 > β 1, P P m1 Q 1 Q k1 < α 1 P P m1 Q 1 Q k1 + P m P m > β, and P P m1 Q 1 Q k1 + P m P m Q k1+1 Q k < α 1

2 We continue in this fashion. This selection process is possible since P n, Q n diverge. Let x n, y n denote the partial sums whose last terms are P mn and Q kn, respectively. Then x n b n P mn, y n α n Q kn Since a n is convergent, P mn, Q kn 0 as n. We conclude that x n β, y n α. Since the subsequential limits of the rearrangement series a n are bounded from above by β and bounded from below by α (this is evident from the squeeze theorem), it follows immediately that lim sup s n = β, lim inf s n = α Corollary 3. If every rearrangement of real-valued series n=1 a n converges, then n=1 a n <. 1. Total Variation Given a signed measure ν on a measure space (, A), one might ask if it is always possible to find a positive measure µ which dominates ν in the following sense: ν() µ(), A Moreover, one might ask if there is a smallest such µ in the sense that if µ is any other positive measure which dominates ν, then µ() µ () for all A. It turns out the answer is yes. First, we need to define the notion of the total variation of a measure, analogous to the variation of a measurable function. Definition 4. Define a set function ν : A R, called the total variation of ν, by ν () = sup ν( j ) where the supremum is taken over all countable partitions of. Lemma 5. The total variation ν of a signed measure ν is a positive measure which satisfies ν nu. Proof. The only axiom of measures which ν does not obviously satisfy is σ-additivity. Let { j } be a countable collection of disjoint sets in A, and set = j. For each j, let α j R such that α j < v ( j ). It follows from the definition of supremum and of v that, for each j, there exists a countable collection of disjoint sets {F i,j } i=1 such that j = i=1 F i,j and Since {F i,j } i, is a partition of, we have α j ν(f i,j ) i=1 α j ν(f i,j ) ν () i=1 Taking the supremum over all α j which satisfy α j < v ( j ) yields the inequality ν ( j ) ν () For the reverse inequality, let {F k } be any other partition of. For k fixed, {F k j } is a partition of F k. By the σ-additivity of ν, ν(f k ) = ν(f k j )

3 If ν(f k j ) =, then ν(f k j ) =. It follows from the definition of total variation that ν ( j) =, and the inequality ν(f k ) ν () holds trivially. If ν(f k j ) <, then we can interchange the order of summation to obtaion ν(f k ) = ν(f k j ) = ν(f k j ) ν ( j ), since {F k j } is a partition of j for each fixed j. Since {F k } obtain the reverse inequality ν () ν ( j ), which completes the proof. was an arbitrary partition of, we Analogous to the decomposition of a function f into a difference f = f + f of two nonnegative functions, we can write a signed measure as the difference of two positive measures. Definition 6. For a signed measure ν, we define the positive variation and negative variation of ν by ν + = 1 ( ν + ν) and ν = 1 ( ν ν) If A is such that ν() =, then ν () := 0. It follows from the preceding proposition that ν +, ν are both (positive) measures, and moreover, ν = ν + ν and ν = ν + + ν We say that the signed measure ν is σ-finite if the measure ν is σ-finite. 1.3 Mutual Singularity and Absolute Continuity Definition 7. Two signed measures ν and µ on a measure space (, A) are mutually singular if there are disjoint subsets A, B A so that ν() = ν(a ) and µ() = µ(b ), A If ν, µ are mutually singular, we write ν µ. If ν is a signed measure and ν is a positive measure on A, we say that ν is absolutely continuous with respect to µ if A, µ() = 0 ν() = 0 If ν is absolutely continuous with respect to µ, we write ν µ. Lemma 8. If ν and µ are mutually singular, and ν is also absolutely continuous with respect to µ, then ν vanishes identically. Proof. Let A, B A be disjoint subsets such that ν() = ν(a ) and µ() = µ(b ), A For any A, µ(a ) = 0. By absolute continuity, ν(a ) = ν() = 0. Since A was arbitrary, we conclude that ν = 0. Lemma 9. Let µ be a positive measure and ν be a signed measure. 1. If for every ɛ > 0, there exists δ > 0 such that A, µ() < δ ν() < ɛ, then ν µ. 3

4 . If ν is a finite measure, then the converse holds. Proof. We first prove (1). Let A be such that µ() = 0. It follows from our hypothesis that ν() < ɛ for every ɛ > 0, from which we conclude that ν() = 0. We prove () by contradiction. Suppose there exists ɛ > 0 such that for all δ > 0, there exists δ A with µ( δ ) < δ and ν( δ ) ɛ. For each n N choose n A with µ( n ) 1 and ν( n n ) ɛ. Then ( µ lim sup ) n = µ n=1 k n for all N N. We conclude that µ(lim sup n ) = 0. Since ν µ by hypothesis, we have ν(lim sup n ) = 0. But this is a contradiction since ν(lim sup k k=n n ) = lim sup ν( n ) ɛ 1 k 1.4 Lebesgue-Radon-Nikodym Theorem There are several proofs of the (Lebesgue-)Radon-Nikodym theorem, but I am fond of the one given below because it uses the theory of Hilbert spaces, which is quite elegant. The exposition closely follows that of Stein and Shakarchi in Real Analysis: Measure, Integration, and Hilbert Spaces. Theorem 10. Suppose µ is a σ-finite positive measure on the measure space (, A) and ν is a σ-finite signed measure on A. Then there exist unique signed meaures ν a and ν s on mathcala such that ν a µ, ν s µ, and ν = ν s + ν a. Furthermore, the measure ν a is given by ν a () = f(x)µ(dx), A for some extended µ-integrable function f. Proof. We first consider the case where both µ and ν are positive and finite. Set ρ = µ + nu. We define a functional l : L (, ρ) C by l(ψ) = ψ(x)ν(dx) Clearly, l is linear. I claim that l is bounded. Indeed, since ν, µ are both positive, ( l(ψ) ψ(x) ν(dx) ψ(x) ρ(dx) = ψ(x)1 (x) ρ(dx) (ρ()) 1 ) 1 ψ(x) ρ(dx), where the last inequality follow from the Cauchy-Schwarz inequality. Since L (, ρ) is a Hilbert space, the Riesz representation theorem tells us that there exists a unique (up to a.e. equivalence) g L (, ρ) such that ψ(x)ν(dx) = ψ(x)g(x)ρ(dx), ψ L (, ρ) If A with ρ() > 0, when we set ψ = χ and recall that ν ρ, we obtain 0 1 g(x)ρ(dx) = ν() ρ() ρ() ρ() ρ() = 1, I claim that 0 g(x) 1 ρ-a.e. Indeed, 0 g(x)ρ(dx) for all sets A implies that 0 g(x)ρ(dx) 1 { n ρ g < 1 } { ρ g < 1 } = 0, n N n n g< 1 n Taking the intersection yields ρ {g < 0} = 0. By the same argument, 0 (1 g(x))ρ(dx) for all A implies that g(x) 1 ρ-a.e. Thus, we may assume that 0 g(x) 1 for all x, and we have ψ(1 g)dν = ψgdµ 4

5 Consider the two sets and define two measures ν a and ν s on A by A = {x : 0 g(x) < 1} and B = {x : g(x) = 1} ν a () = ν(a ) and ν s () = ν(b ), A The diligent inclined reader can verify that ν a, ν s are indeed measures. I claim that ν s µ. It is tautological that A, B are disjoint and ν s () = 0 for all measurable subsets A, so we need only show that µ() = 0 for all measurable subsets B. Indeed, taking ψ = 1 in the identity ψgdµ = ψ(1 g)dν yields µ() = 1 gdµ = 1 (1 g)dµ = 1 0dµ = 0 I now claim that ν a µ. Let A be such that µ() = 0. Then 0 = gdµ = (1 g)dν Since 1 g 0, we conclude that (1 g)1 = 0 a.e., which imples that ν a () = ν( A) = 0. I now claim that dv a = fdµ. Let A, and set ψ = ( n k=0 gk )1. Then (1 g n+1 )dν = n+1 gdµ If x B, then (1 g n+1 )(x) = 0, and if x A, then (1 g n+1 )(x) 1, n. In other words, lim 1 g n+1 = 1 A. Since 1 g n+1 and our measure space is finite, the dominated convergence theorem implies that lim (1 g n+1 )dν = 1 A dν = ν(a ) = ν A () Observe that { n+1 g(x) lim g k 1 g(x) x A (x) = x B Set f = g 1 g. From the monotone convergence theorem we conclude that n+1 v a () = lim g k dµ = Furthermore, f L 1 (, µ) since fdµ = ν a() ν() <. We now consider the case where ν, µ are σ-finite positive measures. It follows from the definition of σ-finite that we can find pairwise disjoint sets j A such that = j and µ( j ), ν( j ) < for all j. We define positive, finite measures on A by fdµ µ j () = µ( j ) and ν j () = ν( j ), A For each j, we write ν j = ν j,a + ν j,s, where ν j,s µ j and ν j,a = f j dµ j. Defining f = j f j, ν s = j ν j,s, ν a = j ν j,a completes the argument. If ν is signed, then we apply the preceding argument separately to the positive and negative variations of ν. To see the uniqueness of the decomposition, suppose we also have ν = ν a + ν s, where ν a µ and ν s µ. Then ν a ν a = ν s ν s 5

6 Clearly, the LHS is absolutely continuous with respect to µ. I claim that the RHS is singular with respect to µ. Indeed, let A B, A B be paritions guaranteed in the definition of singular measures. Then (ν s ν s ) () = 0 for all measurable subset A A, and for any \ (A A ), µ() = µ( B \ B ) + µ( B \ B) + µ( B B ) = = 0 We conclude that ν a ν a = 0 = ν s ν s. 6