10.13. RADON NIKODYM THEOREM 303

Proposition 10.13.12 This Lebesgue decomposition is unique. If f , f̂ both work inTheorem 10.13.7, then f = f̂ µ a.e. This function f ∈ L1 (Ω),

∫Ω

f dµ < ∞.

Proof: Say ν ||+ν⊥= ν̂ ||+ ν̂⊥. Then ν || (E)− ν̂ || (E) = ν⊥ (E)− ν̂⊥ (E) . If µ (E) = 0,then the left side is also 0 and so ν⊥ (E)− ν̂⊥ (E) = 0. But then for any E,

ν⊥ (E)− ν̂⊥ (E) =∫

Ehdµ (10.19)

for h a function in L1 (Ω) . This is because ν⊥− ν̂⊥ is a signed measure λ ≪ µ and Corol-lary 10.13.11. From the above, if S, Ŝ are the exceptional sets of µ measure zero,

ν⊥ (E)− ν̂⊥ (E) = ν⊥(E ∪

(S∪ Ŝ

))− ν̂⊥

(E ∪

(S∪ Ŝ

))=∫

E∪(S∪Ŝ)hdµ = 0 (10.20)

because µ(S∪ Ŝ

)= 0 and so the right side of 10.19 must be 0 after all, so ν⊥ (E) = ν̂⊥ (E).

It follows that ν || = ν̂ || also. Now in Theorem 10.13.7, if you have two f , f̂ which work,then

ν || (E) =∫

Ef dµ =

∫E

f̂ dµ = v̂|| (E) (10.21)

and so, f = f̂ a.e. because you can apply this equation to En ≡[

f − f̂ > 1/n]

and concludethat

0 =∫

En

f − f̂ dµ ≥ 1n

µ (En) = 0 (10.22)

so µ([

f − f̂ > 0])

= ∪mµ (En) = 0. Similarly µ([

f̂ − f > 0])

= 0. ■This unique decomposition of a measure ν into the sum of two measures, one absolutely

continuous with respect to µ and the other supported on a set of µ measure zero is calledthe Lebesgue decomposition.

Definition 10.13.13 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 10.13.14 Suppose µ,ν are both σ finite measures defined on (Ω,F ). Thena similar conclusion to the above theorem can be obtained.

ν (E) =∫

Ef dµ +ν (E ∩S) , µ (S) = 0 (10.23)

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 disjoint. Let Fk be the measurable subsets of Ωk, equivalently the intersections withΩk with sets of F . Now let νk (E) ≡ ν (E ∩Ωk) , similar for µk. By Theorem 10.13.7,there exists Sk ⊆ Ωk, and fk as described there, unique up to sets of µ measure 0. Thusµk (Sk) = 0 and νk (E) =

∫E∩Ωk

fkdµk + νk (E ∩Sk) Now let f (ω) ≡ fk (ω) for ω ∈ Ωk.Thus

ν (E ∩Ωk) = ν (E ∩ (Ωk \Sk))+ν (E ∩Sk) =∫

E∩Ωk

f dµ +ν (E ∩Sk) (10.24)

Summing over all k,ν (E) = ν(E ∩SC

)+ν (E ∩S) =

∫E f dµ +ν (E ∩S) . In particular, if

ν≪ µ, then ν (E ∩S) = 0 and so ν (E) =∫

E f dµ. The last claim is obvious from 10.23. ■