248 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

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, for such n,

l ≤ inf{ak : k ≥ n} ≤ sup{ak : k ≥ n}

and this shows, since l is arbitrary that

lim infn→∞

an = lim supn→∞

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. This proves the lemma.The next theorem, known as Fatou’s lemma is another important theorem which justi-

fies the use of the Lebesgue integral.

Theorem 11.3.18 (Fatou’s lemma) Let fn be a nonnegative measurable function with val-ues in [0,∞]. Let g(ω) = liminfn→∞ fn(ω). Then g is measurable and∫

gdµ ≤ lim infn→∞

∫fndµ .

In other words, ∫ (lim inf

n→∞fn

)dµ ≤ lim inf

n→∞

∫fndµ

Proof: Let gn(ω) = inf{ fk(ω) : k ≥ n}. Then

g−1n ([a,∞]) = ∩∞

k=n f−1k ([a,∞]) ∈F .

Thus gn is measurable by Lemma 11.1.6 on Page 225. Also g(ω) = limn→∞ gn(ω) so g ismeasurable because it is the pointwise limit of measurable functions. Now the functions