32.10. INTEGRALS OF REAL VALUED FUNCTIONS 649

Next note that if a is real and a ≥ 0,(a f )+ = a f+,(a f )− = a f− and if a < 0,(a f )+ =−a f−,(a f )− =−a f+. This follows from a simple computation involving the definition off+, f−. Therefore, if a < 0,∫

a f dµ ≡∫

(a f )+ dµ −∫

(a f )− dµ =∫

(−a) f−dµ −∫

(−a) f+dµ

By Theorem 32.9.1,

=−a(∫

f−dµ −∫

f+dµ

)= a

(∫f+dµ −

∫f−dµ

)≡ a

∫f dµ

The case where a ≥ 0 is easier.Now that we understand how to integrate real valued functions, it is time for another

great convergence theorem, the dominated convergence theorem.

Theorem 32.10.5 (Dominated Convergence theorem) Let fn ∈ L1(Ω) and suppose

f (ω) = limn→∞

fn(ω),

and there exists a measurable function g, with values in [0,∞],2 such that

| fn(ω)| ≤ g(ω) and∫

g(ω)dµ < ∞.

Then f ∈ L1 (Ω) and

0 = limn→∞

∫| fn − f |dµ = lim

n→∞

∣∣∣∣∫ f dµ −∫

fndµ

∣∣∣∣Proof: f is measurable by Corollary 32.2.7. Since | f | ≤ g, it follows that

f ∈ L1(Ω) and | f − fn| ≤ 2g.

By Fatou’s lemma (Theorem 32.8.2),∫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 inf

n→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

∣∣∣∣∫ f dµ −∫

fndµ

∣∣∣∣≥ 0.

This proves the theorem by Lemma 3.3.17 because the limsup and liminf are equal.2Note that, since g is allowed to have the value ∞, it is not known that g ∈ L1 (Ω) .