162 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRAL

and so both sides of 7.1 are equal to ∞. Thus assume for each i > 0,µ (Ei) < ∞. Then itfollows from Lemma 7.2.1 and Lemma 7.1.2,∫

0µ ([s > λ ])dλ =

∫ ap

0µ ([s > λ ])dλ =

p

∑k=1

∫ ak

ak−1

µ ([s > λ ])dλ

=p

∑k=1

(ak−ak−1)p

∑i=k

µ (Ei) =p

∑i=1

µ (Ei)i

∑k=1

(ak−ak−1) =p

∑i=1

aiµ (Ei) ■

Note that this is the same result as in Problem 29 on Page 157 but here there is no questionabout the definition of the integral of a simple function being well defined.

Lemma 7.2.3 If a,b≥ 0 and if s and t are nonnegative simple functions, then∫as+btdµ = a

∫sdµ +b

∫tdµ .

Proof: Let s(ω) = ∑ni=1 α iXAi(ω), t(ω) = ∑

mi=1 β jXB j(ω) where α i are the distinct

values of s and the β j are the distinct values of t. Clearly as+ bt is a nonnegative simplefunction because it has finitely many values on measurable sets. In fact,

(as+bt)(ω) =m

∑j=1

n

∑i=1

(aα i +bβ j)XAi∩B j(ω)

where the sets Ai∩B j are disjoint and measurable. By Lemma 7.2.2,∫as+btdµ =

m

∑j=1

n

∑i=1

(aα i +bβ j)µ(Ai∩B j)

=n

∑i=1

am

∑j=1

α iµ(Ai∩B j)+bm

∑j=1

n

∑i=1

β jµ(Ai∩B j)

= an

∑i=1

α iµ(Ai)+bm

∑j=1

β jµ(B j) = a∫

sdµ +b∫

tdµ . ■

7.3 The Monotone Convergence TheoremThe following is called the monotone convergence theorem. This theorem and relatedconvergence theorems are the reason for using the Lebesgue integral. If limn→∞ fn (ω) =f (ω) and fn is increasing in n, then clearly f is also measurable because of Corollary 6.1.4.Also

f−1 ((a,∞]) = ∪∞k=1 f−1

k ((a,∞]) ∈F

For a different approach to this, see Problem 29 on Page 157.

Theorem 7.3.1 (Monotone Convergence theorem) Suppose that the function f hasall values in [0,∞] and suppose { fn} is a sequence of nonnegative measurable functionshaving values in [0,∞] and satisfying

limn→∞

fn(ω) = f (ω) for each ω.

· · · fn(ω)≤ fn+1(ω) · · ·Then f is measurable and ∫

f dµ = limn→∞

∫fndµ.

162 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRALand so both sides of 7.1 are equal to co. Thus assume for each i > 0, (E;) < ce. Then itfollows from Lemma 7.2.1 and Lemma 7.1.2,[outs>apar=["us>anar=¥ [™ w(s> ayaaka=1 7 %-1vnP P(ax —ayx_1) ) UE, = Yu (BE) Y, (ae —ay-1) =P ait (Ei)k=1 i=lk=1 i=k i=1Note that this is the same result as in Problem 29 on Page 157 but here there is no questionabout the definition of the integral of a simple function being well defined.Lemma 7.2.3 If a,b > 0 and if s and t are nonnegative simple functions, thenostordy =a [sdu+b [ ray.Proof: Let s(@) = V7) & 2a,(@), t(@) = V2) B ; %8;(@) where @; are the distinctvalues of s and the B j are the distinct values of t. Clearly as + bt is a nonnegative simplefunction because it has finitely many values on measurable sets. In fact,Ms:Ms(as + bt)(@) = (aa; + bB ;) Zain; (@)1li=1jwhere the sets A; Bj are disjoint and measurable. By Lemma 7.2.2,mnastord = y( aa; + bB ;)U(AiNB;)j=1i=1_ yay ain Ai nB)+bY YB wlAinB,)i=1 j=l j=li=l= a aA; +b YB ju(B)) =a | sdu+b [tdy. ai=l j=lun.7.3 The Monotone Convergence TheoremThe following is called the monotone convergence theorem. This theorem and relatedconvergence theorems are the reason for using the Lebesgue integral. If lim... f,(@) =f (@) and f, is increasing in n, then clearly f is also measurable because of Corollary 6.1.4.Alsof' ((a,°]) = Ugg | ((a,e]) © FFor a different approach to this, see Problem 29 on Page 157.Theorem 7.3.1 (Monotone Convergence theorem) Suppose that the function f hasall values in |0,°°] and suppose {f,} is a sequence of nonnegative measurable functionshaving values in [0,°] and satisfyinglim fn(@) = f(@) for each o.*fn(@) < fn4i(@) >Then f is measurable and[fa = jim | fra.