27.1. BASIC THEORY 981

= |α| ||u||A + |β | ||v||A .Now let Ω1 ⊆Ω.

||u||A,Ω1≡ inf

{t > 0 :

∫Ω1

A(|u(x)|

t

)dµ ≤ 1

}≤ inf

{t > 0 :

∫Ω

A(|u(x)|

t

)dµ ≤ 1

}≡ ||u||A,Ω .

This occurs because if t is in the second set, then it is in the first so the infimum of thesecond is no smaller than that of the first. This proves the proposition.

Next it is shown that LA (Ω) is a Banach space.

Theorem 27.1.8 LA (Ω) is a Banach space and every Cauchy sequence has a subsequencewhich also converges pointwise a.e.

Proof: Let { fn} be a Cauchy sequence in LA (Ω) and select a subsequence{

fnk

}such

that ∣∣∣∣ fnk+1 − fnk

∣∣∣∣A ≤ 2−k.

Thus

fnm (x) = fn1 (x)+m−1

∑k=1

fnk+1 (x)− fnk (x) .

Let

gm (x)≡ | fn1 (x)|+m−1

∑k=1

∣∣ fnk+1 (x)− fnk (x)∣∣ .

Then

||gm||A ≤ || fn1 ||A +∞

∑k=1

2−k ≡ K < ∞.

Let

g(x)≡ limm→∞

gm (x)≡ | fn1 (x)|+∞

∑k=1

∣∣ fnk+1 (x)− fnk (x)∣∣ .

Now K > ||gm||A so

1≥∫

A(|gm (x)|

K

)dµ.

By the monotone convergence theorem,

1≥∫

A(|g(x)|

K

)dµ

showing g(x)< ∞ a.e., say for all x /∈ E where E is a measurable set having measure zero.Let

f (x) ≡ XEC (x)

(fn1 (x)+

∑k=1

(fnk+1 (x)− fnk (x)

))= lim

m→∞XEC (x) fnm (x) .

27.1. BASIC THEORY 981= |@| |u|, + 1B] IIv|l4-Now let Q) CQ.IIWlao, = int{r>0: [ a( #2!) au <i}int{r>0: [a (M1) au <i} = Ilsa:This occurs because if t is in the second set, then it is in the first so the infimum of thesecond is no smaller than that of the first. This proves the proposition.Next it is shown that L4 (Q) is a Banach space.IATheorem 27.1.8 L, (Q) is a Banach space and every Cauchy sequence has a subsequencewhich also converges pointwise a.eé.Proof: Let {f,,} be a Cauchy sequence in L4 (Q) and select a subsequence { f;, } suchthat|| Frcs — Snel |, $2.Thusfr (8) = Fin (2) + py Fes (8) — fu (8):Letgn) = lfm (I+ py Mines (8) — fu (2)ThenIsnlla <lMinlly+ 24K <eLet -g(x) = lim 8m (x) = fr, (x i+ py | fru (*) — fy, (X )I-1> [a( 2) aw.By the monotone convergence theorem,1> [a( Ee") aushowing g(x) < co a.e., say for all x ¢ E where E is a measurable set having measure zero.LetNow K > ||gmll4 80f(x) = 24000 (1 694 +E (Fun (X — fay (% )= jim Zc (x) fim (x).