300 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL

Now choose δ small enough that Cβ φ (δ )< 12 and then subtract the first term on the right

in the above from both sides. It follows from the properties of F again that∫Ω

F ( f )dµ ≤ 2CβCδ−1

∫Ω

F (g)dµ. (10.15)

This establishes the inequality in the case where f is bounded.In general, let fn = min( f ,n) . For n≤ λ , the inequality

µ ([ f > βλ ]∩ [g≤ δλ ])≤ φ (δ )µ ([ f > λ ])

holds with f replaced with fn because both sides equal 0 thanks to β > 1. If n > λ , then[ f > λ ] = [ fn > λ ] and so the inequality still holds because in this case,

µ ([ fn > βλ ]∩ [g≤ δλ ]) ≤ µ ([ f > βλ ]∩ [g≤ δλ ])

≤ φ (δ )µ ([ f > λ ]) = φ (δ )µ ([ fn > λ ])

Therefore, 10.14 is valid with f replaced with fn. Now pass to the limit in∫

ΩF ( fn)dµ ≤

2CβCδ−1∫

ΩF (gn)dµ as n→ ∞ and use the monotone convergence theorem. ■

10.13 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 10.13.1 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 10.13.2 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 10.13.3 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 10.13.4 Let λ be a signed measure on (Ω,F ). If E ∈F with 0 < λ (E), thenE has 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, Theorem2.8.4, there is a maximal chain C . Then ∪C is a set consisting of disjoint measurable setsF ∈F such that λ (F)< 0. Since each set in ∪C has measure strictly less than 0, it follows