332 CHAPTER 14. THE LEBESGUE INTEGRAL

Note that a simple function achieves each value on a measurable set.

Theorem 14.5.6 Let f ≥ 0 be measurable. Then there exists a sequence of nonneg-ative simple functions {sn} satisfying

0≤ sn(ω) (14.3)

· · · sn(ω)≤ sn+1(ω) · · ·

f (ω) = limn→∞

sn(ω) for all ω ∈Ω. (14.4)

If f is bounded, the convergence is actually uniform. Conversely, if f is nonnegative and isthe pointwise limit of such simple functions, then f is measurable.

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

tn(ω) =2n

∑k=0

knX f−1([ k

n ,k+1

n ))(ω)+2nXI(ω).

Then tn(ω)≤ f (ω) for all ω and limn→∞ tn(ω) = f (ω) for all ω . This is because tn (ω) =2n for ω ∈ I and if f (ω) ∈ [0, 2n+1

n ), then

0≤ f (ω)− tn (ω)≤ 1n. (14.5)

Thus whenever ω /∈ I, the above inequality will hold for all n large enough. Let

s1 = t1, s2 = max(t1, t2) , s3 = max(t1, t2, t3) , · · · .

Then the sequence {sn} satisfies 14.3-14.4. Also each sn has finitely many values and ismeasurable. To see this, note that

s−1n ((a,∞]) = ∪n

k=1t−1k ((a,∞]) ∈F

To verify the last claim, note that in this case the term 2nXI(ω) is not present and forn large enough, 2n/n is larger than all values of f . Therefore, for all n large enough, 14.5holds for all ω . Thus the convergence is uniform.

Now consider the converse assertion. Why is f measurable if it is the pointwise limitof an increasing sequence simple functions?

f−1 ((a,∞]) = ∪∞n=1s−1

n ((a,∞])

because ω ∈ f−1 ((a,∞]) if and only if ω ∈ s−1n ((a,∞]) for all n sufficiently large.

Observation 14.5.7 If f : Ω→R then the above definition of measurability holds withno change. In this case, f never achieves the value ∞. This is actually the case of mostinterest.

Corollary 14.5.8 If f : Ω→ (−∞,∞) is measurable, then there exists a sequence ofsimple functions {sn (ω)} such that |sn (ω)| ≤ | f (ω)| and sn (ω)→ f (ω).