11.4. THE SPACE L1 255

Proof: f is measurable by Theorem 11.1.8. Since | f | ≤ g, it follows that

f ∈ L1(Ω) and | f − fn| ≤ 2g.

By Fatou’s lemma (Theorem 11.3.18),∫2gdµ ≤ lim inf

n→∞

∫2g−| f − fn|dµ

=∫

2gdµ− lim supn→∞

∫| f − fn|dµ.

Subtracting∫

2gdµ ,

0≤− lim supn→∞

∫| f − fn|dµ.

Hence

0 ≥ lim supn→∞

(∫| f − fn|dµ)≥ lim sup

n→∞

∣∣∣∣∫ f dµ−∫

fndµ

∣∣∣∣≥ lim inf

n→∞

∣∣∣∣∫ f dµ−∫

fndµ

∣∣∣∣≥ 0.

This proves the theorem by Lemma 11.3.17 on Page 247 because the limsup and liminfare equal.

Corollary 11.4.10 Suppose fn ∈ L1 (Ω) and f (ω) = limn→∞ fn (ω) . Suppose also thereexist measurable functions, gn, g with values in [0,∞] such that

limn→∞

∫gndµ =

∫gdµ,

gn (ω)→ g(ω) µ a.e. and both∫

gndµ and∫

gdµ are finite. Also suppose | fn (ω)| ≤gn (ω) . Then

limn→∞

∫| f − fn|dµ = 0.

Proof: It is just like the above. This time g+gn−| f − fn| ≥ 0 and so by Fatou’s lemma,∫2gdµ− lim sup

n→∞

∫| f − fn|dµ =

lim infn→∞

∫(gn +g)− lim sup

n→∞

∫| f − fn|dµ

= lim infn→∞

∫((gn +g)−| f − fn|)dµ ≥

∫2gdµ

and so − limsupn→∞

∫| f − fn|dµ ≥ 0.

Definition 11.4.11 Let E be a measurable subset of Ω.∫E

f dµ ≡∫

f XEdµ.