Ν(∪Jej) < ∞ Ej. Ν(E) = ∫E F |Ν| : M → |Ν|(E) = Sup |Ν(Ej)|, Ej. Ν ≤ |Ν|. |Ν|(E)

10. Radon-Nikodym derivatives Given a signed measure ν, we define the total variation ν : R as follows: Signed measures | | M → ∞ Let (X, ) be a measurable space. A signed measure ν (E) = sup ν(E ) , M | | | j | is a map ν from to ( , ] with the property that j=1 M −∞ ∞ ∑ if E1,E2,... are disjoint elements of , then where we sup over all ways of decomposing into a M E countable disjoint union of measurable sets E . ∞ j ν( E ) = ν(E ). ∪j j j j=1 Proposition 10.1. The total variation ν is a pos- ∑ | | Notice that this implies that if ν( E ) < , then the itive measure satisfying ∪j j ∞ sum on the RHS is absolutely convergent, for otherwise ν ν . ≤ | | it would not be independent of the ordering of the Ej. Sometimes we refer to (unsigned) measures as positive Proof: We need to show that measures to make the distinction clear. ∞ ∞ ν (E) ν (E ) and ν (E) ν (E ) An example of a signed measure is | | ≤ | | j | | ≥ | | j j=1 j=1 ∑ ∑ whenever E is written as a countable disjoint union of ν(E) = f E measurable sets . ∫ Ej where (X, , µ) is a measure space and f is a fixed To prove , we choose numbers αj < ν (Ej). Then, M ≥ | | real-valued function such that f , the negative part of we can find a partition Ej = iFi,j into measurable − ∪ f, is integrable. (This ensures that ν can never take the sets such that value , which is not allowed.) In fact, we shall soon α ν(F ) . −∞ j ≤ | i,j | prove that this is the only possibility. j ∑

Then F is a partition of E, so we get Notice that the finite signed measures on a measurable ∪i,j i,j α ν(F ) ν (E). space (X, ) form a vector space, denoted M(X). j ≤ | i,j | ≤ | | M j i,j ∑ ∑ Theorem 10.2. M(X) is a complete normed space Taking the sup over all possible α proves . j ≥ under the norm To prove , we take a partition of E into measurable ≤ ν M(X) = ν (X). sets Fk. Then we have ∥ ∥ | |

ν(F ) = ν(F E ) Proof: It is straightforward to show that M(X) is a norm. Suppose | k | k ∩ j ∥·∥ k k j that ν is a Cauchy sequence in M(X). Then for each E , ∑ ∑ ∑ j ∈ M νn(E) νm(E) 0 as m, n , so limn νn(E) exists for ν(Fk Ej) ν (Ej). | − | → → ∞ ≤ | ∩ | ≤ | | each E. Define ν(E) to be limn νn(E). We need to show that ν is k,j j ∑ ∑ a finite signed measure. Since this is true for each way of partitioning E, we find that ν (E) ν (E ) | | ≤ | | j j ∑ as required. The statement ν ν is obvious. ≤ | | □ We can then write any signed measure as the difference of two positive measures, by writing ν + ν ν ν ν = | | + − | | = ν+ + ν . 2 2 − We say that ν is σ-finite if ν is, and then ν+ and ν | | − automatically are as well. To show countable additivity, suppose that E = E is a disjoint Taking m to infinity, we find that ∪i i union of measurable sets. M ν (E ) ν(E ) ϵ for n N(ϵ). | n i − i | ≤ ≥ i=1 ∑ Lemma 10.3. For ϵ > 0 there exists an N(ϵ) so Since this is true for all M, we get ∞ ∞ (10.1) νn(Ei) ν(Ei) ϵ for n N(ϵ). νn(Ei) νm(Ei) < ϵ for n, m N(ϵ). | − | ≤ ≥ | − | ≥ i=1 i=1 ∑ Now we can∑ compute Proof: To begin, choose so that for all M(ϵ) νn νm < ϵ/5 ν(E) ν(Ei) = lim νn(E) ν(Ei) ∥| | − | |∥ | − | n | − | n, m M(ϵ). Then choose I(ϵ) so that νM(ϵ) i I(ϵ)Ei < ϵ/5. i i ≥ | | ∪ ≥ ∑ ∑ Then νm i I(ϵ)Ei < 2ϵ/5 for all m M(ϵ). Finally, choose = lim (νn(Ei) ν(Ei)) ≥ ( ) n | − | | | ∪ ≥ i N(ϵ) M(ϵ) large enough so that νn(Ei) νm(Ei) < ∑ ≥ ( ) i

Certainly then, for every integer M,

M ν (E ) ν (E ) < ϵ for n, m N(ϵ). | n i − m i | ≥ i=1 ∑

Absolute continuity Theorem 10.5 (Radon-Nikodym). Let µ be a σ-finite positive measure on the measur- Definition 10.4. able space (X, ) and ν a σ-finite signed measure. M (1) We say that a signed measure µ is supported on a Then we can write ν = νa + νs where νa is abso-

set A if µ(E) = µ(E A) for all E . lutely continuous w.r.t. µ, and νs and µ are mu- ∩ ∈ M (2) Two signed measures µ and ν are mutually singular tually singular. Moreover, there exists an extended if they are supported on disjoint subsets. This is µ-integrable function f such that denoted µ ν. ⊥ (3) If ν is a signed measure and µ a positive measure, νa(E) = fdµ. ∫E we say that ν is absolutely continuous w.r.t. µ if A function is extended µ-integrable if its negative µ(E) = 0 = ν(E) = 0. • ⇒ part is integrable.

If ν is a finite measure then this last condition is Proof: We first prove when and are both positive and finite | | µ ν equivalent to the assertion that for each ϵ > 0 there measures. Once we have done that, the general case is then not exists δ > 0 such that difficult. We use Hilbert space ideas. Consider the Hilbert space L2(X, ρ) µ(E) < δ = ν (E) < ϵ, ⇒ | | where ρ = µ + ν. Consider the map while in general this is a strictly stronger assertion. L2(X, ρ) ψ l(ψ) = ψ dν. ∋ 7→ Example. Lebesgue measure, delta measures, and ∫ E f on Rn. This is a bounded linear functional, since 7→ E Exercise. Give an example where ν is not finite, ∫ 1/2 | | l(ψ) ψ dν ψ dρ ρ(X) ψ 2 and the first assertion does not imply the second. | | ≤ | | ≤ | | ≤ ∥ ∥L ∫ ∫ 1 using Cauchy-Schwarz. Therefore l is inner product with some element It is tempting to try ψ = (1 g)− , which would then give 2 − g of L (X, ρ): 1 1 ν(E) = dν = (1 g)− (1 g) dν = (1 g)− gdµ E E − − E − (10.2) ψ dν = ψg dρ for all ψ L2(X, ρ). ∫ ∫ ∫ ∈ and the desired conclusion. ∫ ∫ 1 2 However, this is not allowed since (1 g)− / L (X, ρ) necessarily. For any measurable set E, with ρ(E) > 0, set ψ = 1E. Then we − ∈ find that Instead, we approximate, setting 2 n ψ = (1 + g + g + . . . g )1E A ν(E) = 1E dν = 1Eg dρ, ∩ ∫ ∫ which is bounded and therefore in L2. We obtain so n+1 n+1 1 g 0 1 g dρ ρ(E), (1 g ) dν = g − dµ. E E A − E A 1 g ≤ ≤ ∫ ∩ ∫ ∩ − ∫ Since g < 1 on A, 1 gn+1 1 pointwise, so by MCT we get which implies that g 1 a.e. w.r.t. ρ. By changing g on a set of − ↑ ≤ g ρ-measure zero, we can assume that g 1 everywhere. νa(E) = ν(E A) = dν = dµ. ≤ ∩ E A E A 1 g Now we define A to be the set where g < 1 and B to be the set ∫ ∩ ∫ ∩ − This shows that νa is absolutely continuous and we may take f = where g = 1. Putting ψ = 1B, we find that 1 g(1 g)− , which (by putting E = X) the above equation shows is − integrable w.r.t. . ν(B) = 1B dν = 1Bg dρ = 1B dρ = ν(B) + µ(B). µ ∫ ∫ ∫ To prove for σ-finite, positive measures µ, ν, we write X as the Therefore, µ(B) = 0. Since µ is a positive measure, this means that disjoint union of a countable family Ej of sets of finite measure. Let µ is supported in Bc = A. So define µj, νj be the restrictions of µ, ν to Ej. Then we can decompose νj ν (E) = ν(E A), ν (E) = ν(E B). as ν + ν as above. Setting ν = ν and ν = ν a ∩ s ∩ j,a j,s a j j,a s j j,s we satisfy the conditions of the theorem. To treat the case of We have just shown that νs and µ are mutually singular. Now we ∑ ∑ a signed measure, we treat the positive and negative parts of ν show that νa is absolutely continuous w.r.t. µ. separately. First we reformulate Equation (10.2) as □

ψ(1 g) dν = ψg dµ. − ∫ ∫