11.3. THE ABSTRACT LEBESGUE INTEGRAL 241

This proves the lemma.Note that

∫s could equal +∞ if µ (Ak) =∞ and ak > 0 for some k, but

∫s is well defined

because s≥ 0. Recall that 0 ·∞ = 0.

Lemma 11.3.8 If a,b≥ 0 and if s and t are nonnegative simple functions, then∫as+bt = a

∫s+b

∫t.

Proof: Let

s(ω) =n

∑i=1

α iXAi(ω), t(ω) =m

∑i=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 isa nonnegative simple function because it is measurable and has finitely many values. Also,

(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. By Lemma 11.3.7,∫as+bt =

m

∑j=1

n

∑i=1

(aα i +bβ j)µ(Ai∩B j)

= an

∑i=1

α iµ(Ai)+bm

∑j=1

β jµ(B j)

= a∫

s+b∫

t.

This proves the lemma.

11.3.4 Simple Functions And Measurable FunctionsThere is a fundamental theorem about the relationship of simple functions to measurablefunctions given in the next theorem.

Theorem 11.3.9 Let f ≥ 0 be measurable. Then there exists a sequence of nonnegativesimple functions {sn} satisfying

0≤ sn(ω) (11.3.14)

· · · sn(ω)≤ sn+1(ω) · · ·

f (ω) = limn→∞

sn(ω) for all ω ∈Ω. (11.3.15)

If f is bounded the convergence is actually uniform.

Proof: Letting I ≡ {ω : f (ω) = ∞} , define

tn(ω) =2n

∑k=0

knX[k/n≤ f<(k+1)/n](ω)+nXI(ω).