11.5. VITALI CONVERGENCE THEOREM 257

For the last part, is suffices to verify a single function in L1 (Ω) is uniformly integrable.To do so, note that from the dominated convergence theorem,

limR→∞

∫[| f |>R]

| f |dµ = 0.

Let ε > 0 be given and choose R large enough that∫[| f |>R] | f |dµ < ε

2 . Now let µ (E)< ε

2R .Then ∫

E| f |dµ =

∫E∩[| f |≤R]

| f |dµ +∫

E∩[| f |>R]| f |dµ

< Rµ (E)+ε

2<

ε

2+

ε

2= ε.

This proves the lemma.The following theorem is Vitali’s convergence theorem.

Theorem 11.5.3 Let { fn} be a uniformly integrable set of complex valued functions,

µ(Ω)< ∞and fn(x)→ f (x)

a.e. where f is a measurable complex valued function. Then f ∈ L1 (Ω) and

limn→∞

∫Ω

| fn− f |dµ = 0. (11.5.25)

Proof: First it will be shown that f ∈ L1 (Ω). By uniform integrability, there existsδ > 0 such that if µ (E)< δ , then ∫

E| fn|dµ < 1

for all n. By Egoroff’s theorem, there exists a set, E of measure less than δ such that onEC, { fn} converges uniformly. Therefore, for p large enough, and n > p,∫

EC

∣∣ fp− fn∣∣dµ < 1

which implies ∫EC| fn|dµ < 1+

∫Ω

∣∣ fp∣∣dµ.

Then since there are only finitely many functions, fn with n ≤ p, there exists a constant,M1 such that for all n, ∫

EC| fn|dµ < M1.

But also, ∫Ω

| fm|dµ =∫

EC| fm|dµ +

∫E| fm|

≤ M1 +1≡M.