27.1. BASIC THEORY 983

Since S⊆ KA (Ω) , ∫Ω

A(2 |sn|)dµ < ∞.

Therefore, 2 f −2sn ∈ KA (Ω) and 2sn ∈ KA (Ω) and so, since KA (Ω) is convex,

2 f −2sn

2+

2sn

2= f ∈ KA (Ω) .

This shows EA (Ω)⊆ KA (Ω) .Next consider the claim these spaces are all equal in the case that (A,Ω) is ∆ regular. It

was already shown in Proposition 27.1.5 that in this case,

KA (Ω) = LA (Ω)

so it remains to show EA (Ω) = KA (Ω). Is every f ∈KA (Ω) the limit in LA (Ω) of functionsfrom S? First suppose µ (Ω) = ∞. Then A(r | f |) ≤ KrA(| f |) and so A(r | f |) ∈ L1 (Ω) forany r. Let ε > 0 be given and let

Ωδ ≡ {x : | f (x)| ≥ δ}

Then by the dominated convergence theorem,

limδ→0+

∫Ω\Ωδ

A(| f |ε

)dµ = 0.

Choose δ such that ∫Ω\Ωδ

A(| f |ε

)dµ <

12

and let sn→ f XΩδpointwise with |sn| ≤

∣∣ f XΩδ

∣∣ . Then sn = 0 on Ω\Ωδ and so

∫Ω

A(| f − sn|

ε

)dµ =

∫Ωδ

A(| f − sn|

ε

)dµ

+∫

Ω\Ωδ

A(| f |ε

)dµ ≤

∫Ωδ

A(| f − sn|

ε

)dµ +

12.

By the dominated convergence theorem,∫Ωδ

A(| f − sn|

ε

)dµ <

12

for all n large enough. Therefore, for such n,∫Ω

A(| f − sn|

ε

)dµ < 1

and so || f − sn||A ≤ ε showing that in this case EA (Ω)⊇ KA (Ω) since ε > 0 is arbitrary.