178 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRAL

By Egoroff’s theorem, there exists a set, F with µ (F) < δ 1 such that fn convergesuniformly to f on FC. Therefore, there exists N such that if n > N, then∫

FC| f − fn|dµ <

ε

3.

It follows that for n > N,∫Ω

| f − fn|dµ ≤∫

FC| f − fn|dµ +

∫F| f |dµ +

∫F| fn|dµ

<ε

3+

ε

3+

ε

3= ε,

which verifies 7.10. ■

7.11 Radon Nikodym TheoremLet µ,ν be two finite measures on the measurable space (Ω,F ) and let α ≥ 0. Let λ ≡ν−αµ . Then it is clear that if {Ei}∞

i=1 are disjoint sets of F , then λ (∪iEi) = ∑∞i=1 λ (Ei)

and that the series converges. The next proposition is fairly obvious.

Proposition 7.11.1 Let (Ω,F ,λ ) be a measure space and let λ : F → [0,∞) be ameasure. Then λ is a finite measure.

Proof: Since λ (Ω)< ∞ this is a finite measure. ■

Definition 7.11.2 Let (Ω,F ) be a measurable space and let λ : F →R satisfy: If{Ei}∞

i=1 are disjoint sets of F , then λ (∪iEi) = ∑∞i=1 λ (Ei) and the series converges. Such

a real valued function is called a signed measure. In this context, a set E ∈F is calledpositive if whenever F is a measurable subset of E, it follows λ (F) ≥ 0. A negative set isdefined similarly. Note that this requires λ (Ω) ∈ R.

Lemma 7.11.3 The countable union of disjoint positive sets is positive.

Proof: Let Ei be positive and consider E ≡ ∪∞i=1Ei. If A ⊆ E with A measurable, then

A∩Ei ⊆ Ei and so λ (A∩Ei)≥ 0. Hence λ (A) = ∑i λ (A∩Ei)≥ 0. ■

Lemma 7.11.4 Let λ be a signed measure on (Ω,F ). If E ∈F with 0 < λ (E), then Ehas a measurable subset which is positive.

Proof: If every measurable subset F of E has λ (F) ≥ 0, then E is positive and weare done. Otherwise there exists measurable F ⊆ E with λ (F)< 0. Let the elements of Fconsist of sets of disjoint sets of measurable subsets of E each of which has measure lessthan 0. Partially order F by set inclusion. By the Hausdorff maximal theorem, there is amaximal chain C . Then ∪C is a set consisting of disjoint measurable sets F ∈F suchthat λ (F)< 0. Since each set in ∪C has measure strictly less than 0, it follows that ∪C isa countable set, {Fi}∞

i=1 . Otherwise, there would exist an infinite subset of ∪C with eachset having measure less than − 1

n for some n ∈ N so λ would not be real valued. LettingF = ∪iFi, then E \F has no measurable subsets S for which λ (S) < 0 since, if it did, Cwould not have been maximal. Thus E \F is positive. ■

A major result is the following, called a Hahn decomposition.