14.11. INTEGRALS OF REAL VALUED FUNCTIONS 339

Definition 14.11.3 Let f ∈ L1 (Ω,µ). Define∫

f dµ ≡∫

f+dµ−∫

f−dµ.

Proposition 14.11.4 The definition of∫

f dµ is well defined and if a,b are real num-bers ∫

(a f +bg)dµ = a∫

f dµ +b∫

gdµ

Proof: First of all, it is well defined because f+, f− are both no larger than | f |. There-fore,

∫f+dµ,

∫f−dµ are both real numbers. Next, why is the integral linear. First consider

the sum. ∫( f +g)dµ ≡

∫( f +g)+ dµ−

∫( f +g)− dµ

Now ( f +g)+− ( f +g)− = f +g = f+− f−+g+−g−. By Lemma 14.11.2 and Theorem14.10.1∫

( f +g)dµ ≡∫

( f +g)+ dµ−∫

( f +g)− dµ =∫

( f++g+)dµ−∫

( f−+g−)dµ

=∫

f+dµ−∫

f−dµ +∫

g+dµ−∫

g−dµ ≡∫

f dµ +∫

gdµ

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 14.10.1,

=−a(∫

f−dµ−∫

f+dµ

)= a

(∫f+dµ−

∫f−dµ

)≡ a

∫f dµ

The case where a≥ 0 is easier.Note how attractive this is. If you have a measurable function f and it is absolutely

integrable, then it is integrable. This is just like the situation with series.Now that we understand how to integrate real valued functions, it is time for another

great convergence theorem, the dominated convergence theorem.

Theorem 14.11.5 (Dominated Convergence theorem) Let fn ∈ L1(Ω) and supposef (ω) = limn→∞ fn(ω), and there exists a measurable function g, with values in [0,∞],1

such that| fn(ω)| ≤ g(ω) and

∫g(ω)dµ < ∞.

Then f ∈ L1 (Ω) and 0 = limn→∞

∫| fn− f |dµ = limn→∞ |

∫f dµ−

∫fndµ|

Proof: f is measurable by Corollary 14.5.4. Since | f | ≤ g, it follows that

f ∈ L1(Ω) and | f − fn| ≤ 2g.

1Note that, since g is allowed to have the value ∞, it is not known that g ∈ L1 (Ω) .