170 CHAPTER 7. THE ABSTRACT LEBESGUE INTEGRAL

Since ε is arbitrary, the two must be equal and they both must equal a. Next supposelimn→∞ an = ∞. Then if l ∈ R, there exists N such that for n≥ N, l ≤ an and therefore, forsuch n,

l ≤ inf{ak : k ≥ n} ≤ sup{ak : k ≥ n}and this shows, since l is arbitrary that liminfn→∞ an = limsupn→∞ an = ∞. The case for−∞ is similar.

Conversely, suppose liminfn→∞ an = limsupn→∞ an = a. Suppose first that a∈R. Then,letting ε > 0 be given, there exists N such that if n≥ N,

sup{ak : k ≥ n}− inf{ak : k ≥ n}< ε

therefore, if k,m > N, and ak > am,

|ak−am|= ak−am ≤ sup{ak : k ≥ n}− inf{ak : k ≥ n}< ε

showing that {an} is a Cauchy sequence. Therefore, it converges to a ∈ R, and as in thefirst part, the liminf and limsup both equal a. If liminfn→∞ an = limsupn→∞ an = ∞, thengiven l ∈ R, there exists N such that for n ≥ N, infn>N an > l. Therefore, limn→∞ an = ∞.The case for −∞ is similar. ■

Here is the dominated convergence theorem.

Theorem 7.8.2 (Dominated Convergence theorem) Let fn ∈ L1(Ω) and suppose

f (ω) = limn→∞

fn(ω),

and there exists a measurable function g, with values in [0,∞],1 such that

| fn(ω)| ≤ g(ω) and∫

g(ω)dµ < ∞.

Then f ∈ L1 (Ω) and

0 = limn→∞

∫| fn− f |dµ = lim

n→∞

∣∣∣∣∫ f dµ−∫

fndµ

∣∣∣∣Proof: f is measurable by Corollary 6.1.4 applied to real and imaginary parts. Since

| f | ≤ g, it follows thatf ∈ L1(Ω) and | f − fn| ≤ 2g.

By Fatou’s lemma (Theorem 7.5.1),∫2gdµ ≤ lim inf

n→∞

∫2g−| f − fn|dµ

=∫

2gdµ− lim supn→∞

∫| f − fn|dµ.

Subtracting∫

2gdµ , 0≤− limsupn→∞

∫| f − fn|dµ. Hence

0 ≥ lim supn→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

∣∣∣∣∫ f dµ−∫

fndµ

∣∣∣∣≥ 0.

This proves the theorem by Lemma 7.8.1 because the limsup and liminf are equal. ■

1Note that, since g is allowed to have the value ∞, it is not known that g ∈ L1 (Ω) .