7.8. THE DOMINATED CONVERGENCE THEOREM 171

Corollary 7.8.3 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

n→∞

∫(gn +g)dµ− lim sup

n→∞

∫| f − fn|dµ

= lim infn→∞

∫(gn +g)dµ− lim sup

n→∞

∫| f − fn|dµ

= lim infn→∞

∫((gn +g)−| f − fn|)dµ ≥

∫2gdµ

and so − limsupn→∞

∫| f − fn|dµ ≥ 0. Thus

0 ≥ lim supn→∞

(∫| f − fn|dµ

)≥ lim inf

n→∞

(∫| f − fn|dµ

)≥∣∣∣∣∫ f dµ−

∫fndµ

∣∣∣∣≥ 0. ■

Definition 7.8.4 Let E be a measurable subset of Ω.∫E

f dµ ≡∫

f XEdµ.

If L1(E) is written, the σ algebra is defined as {E ∩A : A ∈ F} and the measure isµ restricted to this smaller σ algebra. Clearly, if f ∈ L1(Ω), then f XE ∈ L1(E) and iff ∈ L1(E), then letting f̃ be the 0 extension of f off of E, it follows f̃ ∈ L1(Ω).

What about something ordinary, the integral of a continuous function?

Theorem 7.8.5 Let f be continuous on [a,b]. Then∫ b

af (x)dx =

∫[a,b]

f dm

where the integral on the left is the usual Riemann integral and the integral on the right isthe Lebesgue integral.

Proof: From Theorems 6.5.1 and 6.8.2 f X[a,b] is Lebesgue measurable. Assume forthe sake of simplicity that f (x)≥ 0. If not, apply what is about to be shown to f+ and f−.Let sn (x) be a step function and let this converge uniformly to f (x) on [a,b] with sn (x) = 0for x /∈ [a,b]. For example, let

sn (x)≡n

∑j=1

f(x j−1

)X[x j−1,x j) (x)