204 CHAPTER 9. THE LEBESGUE INTEGRAL

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) =

∑mj=1 ∑

ni=1(aα i + bβ j)XAi∩B j(ω) where the sets Ai ∩B j are disjoint and measurable. By

Lemma 9.2.2, ∫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 . ■

9.3 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

f−1 ((a,∞]) = ∪∞k=1 f−1

k ((a,∞]) ∈F

For a different approach to this, see Problem 22 on Page 199.

Theorem 9.3.1 (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.

Proof: By Lemma 9.1.5 limn→∞

∫fndµ = supn

∫fndµ

= supn

suph>0

∞

∑k=1

µ ([ fn > kh])h = suph>0

supN

supn

N

∑k=1

µ ([ fn > kh])h

= suph>0

supN

N

∑k=1

µ ([ f > kh])h = suph>0

∞

∑k=1

µ ([ f > kh])h =∫

f dµ. ■

Note how it was important to have∫

∞

0 [ f > λ ]dλ in the definition of the integral andnot [ f ≥ λ ]. You need to have [ fn > kh] ↑ [ f > kh] so µ ([ fn > kh])→ µ ([ f > kh]) . Toillustrate what goes wrong without the Lebesgue integral, consider the following example.

Example 9.3.2 Let {rn} denote the rational numbers in [0,1] and let

fn (t)≡{

1 if t /∈ {r1, · · · ,rn}0 otherwise