302 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL
Thus k→ f k (ω) is increasing. Let f (ω) ≡ limk→∞ f k (ω). Thus f = 0 on S. Now letE be measurable. Thus µ (E) = µ (E ∩S)+ µ
(E ∩SC
), similar for λ and λ
(E ∩SC
)=
∑n λ(E ∩SC ∩∆k
n). To save space, let Ẽ ≡ E ∩SC. Then using 10.16∫
XẼ f kdµ ≤∞
∑n=0
αkn+1µ
(Ẽ ∩∆
kn
)≤
∞
∑n=0
αknµ
(Ẽ ∩∆
kn
)+
∞
∑n=0
2−kµ
(Ẽ ∩∆
kn
)≤
∞
∑n=0
ν
(Ẽ ∩∆
kn
)+2−k
µ(Ẽ)= ν
(Ẽ)+2−k
µ(Ẽ)≤∫
XẼ f kdµ +2−kµ(Ẽ)
(10.18)
From the monotone convergence theorem it follows ν(Ẽ)=∫
XẼ f dµ =∫
XE f dµ .This proves most of the following theorem which is the Radon Nikodym theorem.
Theorem 10.13.7 Let ν and µ be finite measures defined on a measurable space(Ω,F ). Then there exists a measurable set S with µ (S) = 0 and a nonnegative measurablefunction ω → f (ω) such that ν (E) =
∫E f dµ +ν (E ∩S) .
Proof: Let S be defined in Lemma 10.13.6 so S≡Ω\(∪nNk
n)
and µ (S) = 0. If E ∈F ,and f as described above,
ν (E) = ν(E ∩SC)+ν (E ∩S) =
∫E∩SC
f dµ +ν (E ∩S) =∫
Ef dµ +ν (E ∩S)
Thus if E ⊆ SC, we have ν (E) =∫
E f dµ . ■
Definition 10.13.8 Let µ,ν be finite measures on (Ω,F ). Then ν≪ µ means thatwhenever µ (E) = 0, it follows that ν (E) = 0.
Sometimes people write f = dλ
dµ, in the case ν ≪ µ and this is called the Radon
Nikodym derivative.
Proposition 10.13.9 If ν ,µ are finite measures and ν ≪ µ, then there exists nonneg-ative measurable f such that ν (E) =
∫E f dµ.
Proof: In Theorem 10.13.7, ν (E ∩S) = 0 because µ (S) = 0 and ν ≪ µ . ■
Definition 10.13.10 Let S be in the above theorem. Then
ν || (E)≡ ν(E ∩SC)= ∫
E∩SCf dµ =
∫E
f dµ
while ν⊥ (E)≡ ν (E ∩S) . Thus ν ||≪ µ and ν⊥ is nonzero only on sets which are containedin S which has µ measure 0.
Corollary 10.13.11 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 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 . ■