228 CHAPTER 9. BASIC FUNCTION SPACES

Then2−k

µ (Bk)≤∫

Bk

∣∣ fnk+1 (ω)− fnk (ω)∣∣p dµ < 4−k

and so µ (Bk) < 2−k. Now if fnk (ω) fails to be a Cauchy sequence, then ω ∈ Bk for in-finitely many k. In other words, ω ∈ ∩∞

n=1∪k≥n Bk ≡ B. This measurable set B has measurezero because

µ (B)≤ µ (∪k≥nBk)≤∞

∑k=n

µ (Bk)<1

2n−1 for every n ∈ N

Therefore, for ω /∈ B,{

fnk

}∞

k=1 is a Cauchy sequence. Let g(ω)≡ 0 on B and let g(ω)≡limk→∞ fnk (ω) if ω /∈ B. Why is g ∈ Lp and why does fn converge to g in Lp? g is the limitof the measurable functions fnkXBC and so it is measurable. By Fatou’s lemma,∫

Ω

|g(ω)|p dµ ≤ lim infk→∞

∫Ω

∣∣ fnk (ω)∣∣p dµ ≤M

and so g ∈ Lp (Ω) . Now by construction,∥∥ fnk − fnk+1

∥∥p < 4−k/p therefore,(∫

Ω

∣∣ fnk − fnk+m

∣∣p dµ

)1/p

≡∥∥ fnk − fnk+m

∥∥p ≤

m−1

∑j=0

∥∥∥ fnk+ j − fnk+ j̄+1

∥∥∥p

≤∞

∑j=k

(4−1/p

) j=

(4−1/p

)k−1

1−4−1/p

Now use Fatou’s lemma to obtain, as m→ ∞,(∫Ω

∣∣ fnk −g∣∣p dµ

)1/p

≤(4−1/p

)k−1

1−4−1/p

The expression on the right converges to 0 as k → ∞ and so limk→∞

∥∥ fnk −g∥∥

p = 0. Itfollows from Theorem 2.3.3 (If the sequence is Cauchy then if a subsequence converges,so does the original sequence.) applied to this normed linear space that

limn→∞∥ fn−g∥p = 0.■

What about L∞ (Ω) , the case conjugate to p = 1? How is the norm defined for L∞ (Ω)?What does it mean to be in L∞ (Ω)?

Definition 9.3.8 A function f is in L∞ (Ω) if it is measurable and if there is someconstant M such that off a set of measure zero, | f (ω)| ≤M. Then ∥ f∥

∞is defined to be the

infimum of all such constants M. Such a function is said to be “essentially bounded”.

Obviously L∞ (Ω) is a vector space. The next task is to verify that ∥·∥∞

is a norm underthe convention that any two functions which are equal off a set of measure zero are thesame.

Proposition 9.3.9 Let f ∈ L∞ (Ω). Then

µ ({ω : | f (ω)|> ∥ f∥∞}) = 0

and if λ < ∥ f∥∞, then µ ({ω : | f (ω)|> λ})> 0.