20.9. SEQUENTIAL COMPACTNESS IN L1 637

Proof: Take { fn} a sequence in K and let A ,S1,Ω1 be as in Lemma 20.9.2. Thus Ais a countable algebra and by assumption, for each E ∈A ,{∫

Efndµ

}is a bounded sequence and so there exists a convergent subsequence. Therefore, from aCantor diagonalization argument, there exists a subsequence, denoted by {gn} such that{∫

Egndµ

}converges for every E ∈A .

Let

M ≡{

E ∈S1 = σ (A ) such that limn→∞

∫E

gndµ exists}.

Then it has been shown that A ⊆M . Suppose Ek ↑ E where Ek ∈M . Then letting ε > 0be given, the assumption shows that for k large enough,∣∣∣∣∫E\Ek

gndµ

∣∣∣∣< ε

for all gn. Therefore, picking such a k,∣∣∣∣∫Egndµ−

∫E

gmdµ

∣∣∣∣≤ 2ε +

∣∣∣∣∫Ek

gndµ−∫

Ek

gmdµ

∣∣∣∣< 3ε

provided m,n are large enough. Therefore, {∫

E gndµ} is a Cauchy sequence and so itconverges.

In the case that Ek ↓ E use the assumption to conclude there exists a k large enough that∣∣∣∣∫Ek\Egndµ

∣∣∣∣< ε

for all gn. Then∣∣∣∣∫Egndµ−

∫E

gmdµ

∣∣∣∣ =

∣∣∣∣∫Ek

gndµ−∫

Ek

gmdµ

∣∣∣∣+

∣∣∣∣∫Ek\Egndµ

∣∣∣∣+ ∣∣∣∣∫Ek\Egmdµ

∣∣∣∣≤

∣∣∣∣∫Ek

gndµ−∫

Ek

gmdµ

∣∣∣∣+2ε < 3ε

provided m,n large enough. Again {∫

E gndµ} is a Cauchy sequence. This shows M is amonotone class and so by the monotone class theorem, Theorem 12.10.5 on Page 320 itfollows M = S 1 ≡ σ (A ).