20.1. RADON NIKODYM THEOREM 601

Also, for E ∈S ,

λ (E) =∞

∑n=1

λ (E ∩Sn) =∞

∑n=1

∫XE∩Sn(x) fn(x)dµ

=∞

∑n=1

∫XE∩Sn(x) f (x)dµ

By the monotone convergence theorem

∞

∑n=1

∫XE∩Sn(x) f (x)dµ = lim

N→∞

N

∑n=1

∫XE∩Sn(x) f (x)dµ

= limN→∞

∫ N

∑n=1

XE∩Sn(x) f (x)dµ

=∫ ∞

∑n=1

XE∩Sn(x) f (x)dµ =∫

Ef dµ .

This proves the existence part of the corollary.To see f is unique, suppose f1 and f2 both work and consider for n ∈ N

Ek ≡[

f1− f2 >1k

].

Then0 = λ (Ek ∩Sn)−λ (Ek ∩Sn) =

∫Ek∩Sn

f1(x)− f2(x)dµ .

Hence µ(Ek ∩Sn) = 0 for all n so

µ(Ek) = limn→∞

µ(E ∩Sn) = 0.

Hence µ([ f1− f2 > 0])≤∑∞k=1 µ (Ek) = 0. Therefore, λ ([ f1− f2 > 0]) = 0 also. Similarly

(µ +λ )([ f1− f2 < 0]) = 0.

This version of the Radon Nikodym theorem will suffice for most applications, butmore general versions are available. To see one of these, one can read the treatment inHewitt and Stromberg [64]. This involves the notion of decomposable measure spaces, ageneralization of σ finite.

Not surprisingly, there is a simple generalization of the Lebesgue decomposition partof Theorem 20.1.2.

Corollary 20.1.4 Let (Ω,S ) be a set with a σ algebra of sets. Suppose λ and µ are twomeasures defined on the sets of S and suppose there exists a sequence of disjoint sets ofS , {Ωi}∞

i=1 such that λ (Ωi) ,µ (Ωi) < ∞. Then there is a set of µ measure zero, N andmeasures λ⊥ and λ || such that

λ⊥+λ || = λ , λ ||≪ µ,λ⊥ (E) = λ (E ∩N) = λ⊥ (E ∩N) .