220 CHAPTER 9. THE LEBESGUE INTEGRAL

Theorem 9.11.8 Let ν and µ be finite measures defined on a measurable space(Ω,F ) where ν ≪ µ . Then there exists a unique up to a set of measure zero nonnegativemeasurable function ω → f (ω) such that ν (E) =

∫E f dµ.

Proof: If you had f̂ which also works, then consider the set En where f̂ (ω)> f (ω)+1/n. Then 0 =

∫En

(f̂ (ω)− f (ω)

)dµ ≥ 1

n µ (En) . Thus µ (En) = 0 and so also[f̂ − f > 0

]= ∪nEn

is a set of measure 0. Similarly[

f − f̂ > 0]

is a set of measure zero and so f = f̂ for a.e.ω . ■

Sometimes people write f = dλ

dµand dλ

dµis called the Radon Nikodym derivative.

Corollary 9.11.9 In the above situation, let λ be a signed measure and let λ ≪ µ

meaning that if µ (E) = 0⇒ λ (E) = 0. Here assume that µ is a finite measure. Then thereexists a unique up to a set of measure zero h ∈ L1 such that λ (E) =

∫E hdµ .

Proof: Let P∪N be a Hahn decomposition of λ . Let

λ+ (E)≡ λ (E ∩P) , λ− (E)≡−λ (E ∩N) .

Then both λ+ and λ− are absolutely continuous measures and so there are nonnegative h+and h− with λ− (E) =

∫E h−dµ and a similar equation for λ+. Then 0 ≤ −λ (Ω∩N) ≤

λ− (Ω) < ∞, similar for λ+ so both of these measures are necessarily finite. Henceboth h− and h+ are in L1 so h ≡ h+− h− is also in L1 and λ (E) = λ+ (E)− λ− (E) =∫

E (h+−h−)dµ . ■

Definition 9.11.10 A measure space (Ω,F ,µ) is σ finite if there are countablymany measurable sets {Ωn} such that µ is finite on measurable subsets of Ωn.

There is a routine corollary of the above theorem.

Corollary 9.11.11 Suppose µ,ν are both σ finite measures defined on (Ω,F ) withν ≪ µ . Then a similar conclusion to the above theorem can be obtained. ν (E) =

∫E f dµ

for f a nonnegative measurable function. If ν (Ω) < ∞, then f ∈ L1 (Ω). This f is uniqueup to a set of µ measure zero.

Proof: Since both µ,ν are σ finite, there are{

Ω̃k}∞

k=1 such that ν(Ω̃k),µ(Ω̃k)

are

finite. Let Ω0 = /0 and Ωk ≡ Ω̃k \(∪k−1

j=0Ω̃ j

)so that µ,ν are finite on Ωk and the Ωk are dis-

joint. Let Fk be the measurable subsets of Ωk, equivalently the intersections with Ωk withsets of F . Now let νk (E) ≡ ν (E ∩Ωk) , similar for µk. By Theorem 9.11.8, there existsfk as described there, unique up to sets of µ measure 0. Thus νk (E) =

∫E∩Ωk

fkdµk.Nowlet f (ω)≡ fk (ω) for ω ∈Ωk. Thus ν (E ∩Ωk) =

∫E∩Ωk

f dµ . Summing over all k,ν (E) =∫E f dµ. ■

9.12 Iterated IntegralsThis is about what can be said for the σ algebra of product measurable sets. First it isnecessary to define what this means.