982 CHAPTER 27. ORLITZ SPACES

Thus f is measurable and fnm (x)→ f (x) a.e. as m→ ∞.For l > k, ∣∣∣∣ fnk − fnl

∣∣∣∣A <

12k−2

and so

1≥∫

Ω

A

∣∣ fnl (x)− fnk (x)∣∣(

12k−2

)dµ.

By Fatou’s lemma, let l→ ∞ and obtain

1≥∫

Ω

A

∣∣ f (x)− fnk (x)∣∣(

12k−2

)dµ

and so(

f − fnk

)2k−2 ∈ KA (Ω) and so f − fnk ∈ LA (Ω) , fnk ∈ LA (Ω) . Since LA (Ω) is a

vector space, this shows f ∈ LA (Ω) . Also

∣∣∣∣ f − fnk

∣∣∣∣A ≤

12k−2 ,

showing that fnk → f in LA (Ω) . Since a subsequence converges in LA (Ω) , it follows theoriginal Cauchy sequence also converges to f in LA (Ω). This proves the theorem.

Next consider the space, EA (Ω) which will be a subspace of the Orlitz class, KA (Ω)just as LA (Ω) is a vector space containing the Orlitz class.

Definition 27.1.9 Let S denote the set of simple functions, s, such that

µ ({x : s(x) ̸= 0})< ∞.

Then defineEA (Ω)≡ the closure in LA (Ω) of S.

Proposition 27.1.10 EA (Ω)⊆ KA (Ω)⊆ LA (Ω) and they are all equal if (A,Ω) is ∆ regu-lar.

Proof: First note that S⊆ KA (Ω)∩EA (Ω) . Let f ∈ EA (Ω) . Then by the definition ofEA (Ω) , there exists sn ∈ S such that

||sn− f ||A→ 0.

Therefore, for n large enough,

||sn− f ||A <12

and so ∫Ω

A

(| f − sn|( 1

2

) )dµ =

∫Ω

A(|2 f −2sn|)dµ < ∞.