212 CHAPTER 9. THE LEBESGUE INTEGRAL

Corollary 9.8.3 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 9.8.4 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).

Another very important observation applies to the case where Ω is also a metric space.In this lemma, spt( f ) denotes the closure of the set on which f is nonzero.

Definition 9.8.5 Let K be a set and let V be an open set containing K. Then thenotation K ≺ f ≺ V means that f (x) = 1 for all x ∈ K and spt( f ) is a compact subset ofV . spt( f ) is defined as the closure of the set where f is not zero. It is called the “support”of f . A function f ∈ Cc (Ω) for Ω a metric space if f is continuous on Ω and spt( f ) iscompact. This Cc (Ω) is called the continuous functions with compact support.

Now that the Lebesgue integral has been presented, it is time to show the way that themeasure of Theorem 8.8.2 represents the functional.

Proposition 9.8.6 Let L be a positive linear functional on Cc (X) for X a metric spaceand let µ be the measure described by Theorem 8.8.2. Then for all f ∈ Cc (X) ,L( f ) =∫

X f dµ where this is the Lebesgue integral just described.

Proof: Let f ∈Cc(X), f real-valued, and suppose f (X)⊆ [a,b]. Choose t0 < a and lett0 < t1 < · · · < tn = b, ti− ti−1 < ε . Let Ei = f−1((ti−1, ti])∩ spt( f ). Note that ∪n

i=1Ei is aclosed set equal to spt( f ). ∪n

i=1Ei = spt( f ). Since X = ∪ni=1 f−1((ti−1, ti]). Let Vi ⊇ Ei,Vi is

open and let Vi satisfy

f (x)< ti + ε for all x ∈Vi, µ(Vi \Ei)< ε/n. (9.9)