654 CHAPTER 24. THE BOCHNER INTEGRAL

24.1.1 Eggoroff’s TheoremIn the context of a more general notion of measurable function having values in a metricspace, here is a version of Egoroff’s theorem. Here we introduce a finite measure µ . Noneof the above section had anything to do with a measure.

Theorem 24.1.18 (Egoroff) Let (Ω,F ,µ) be a finite measure space,(µ(Ω) < ∞)and let fn, f be X valued measurable functions where X is a separable metric space andfor all ω /∈ E where µ(E) = 0, fn (ω)→ f (ω). Then for every ε > 0, there exists a set,F ⊇ E, µ(F)< ε, such that fn converges uniformly to f on FC.

Proof: First suppose E = /0 so that convergence is pointwise everywhere. Let

Ekm = {ω ∈Ω : d ( fn (ω) , f (ω))≥ 1/m for some n > k}.

Claim:[ω : d ( fn (ω) , f (ω))≥ 1

m

]is measurable.

Proof of claim: Let {xk}∞

k=1 be a countable dense subset of X and let r denote a positiverational number, Q+. Then

∪k∈N,r∈Q+ f−1n (B(xk,r))∩ f−1

(B(

xk,1m− r))

=

[d ( f , fn)<

1m

](24.2)

Here is why. If ω is in the set on the left, then d ( fn (ω) ,xk)< r and d ( f (ω) ,xk)<1m − r.

Therefore,

d ( f (ω) , fn (ω))< r+1m− r =

1m.

Thus the left side is contained in the right. Now let ω be in the right side. That isd ( fn (ω) , f (ω))< 1

m . Choose 2r < 1m−d ( fn (ω) , f (ω)) and pick xk ∈ B( fn (ω) ,r). Then

d ( f (ω) ,xk)≤ d ( f (ω) , fn (ω))+d ( fn (ω) ,xk)<1m−2r+ r =

1m− r

Thus ω ∈ f−1n (B(xk,r))∩ f−1

(B(xk,

1m − r

))and so ω is in the left side. Thus the two

sets are equal. Now the set on the left in 24.2 is measurable because it is a countableunion of measurable sets. This proves the claim since

[ω : d ( fn (ω) , f (ω))≥ 1

m

]is the

complement of this measurable set.Hence Ekm is measurable because Ekm = ∪∞

n=k+1

[ω : d ( fn (ω) , f (ω))≥ 1

m

]. For fixed

m,∩∞k=1Ekm = /0 because fn (ω) converges to f (ω). Therefore, if ω ∈Ω there exists k such

that if n > k, | fn (ω)− f (ω)| < 1m which means ω /∈ Ekm. Note also that Ekm ⊇ E(k+1)m.

Since µ(E1m)< ∞, Theorem 9.2.4 on Page 242 implies

0 = µ(∩∞k=1Ekm) = lim

k→∞µ(Ekm).

Let k(m) be chosen such that µ(Ek(m)m) < ε2−m and let F =∞⋃

m=1Ek(m)m. Then µ(F) <

ε because

µ (F)≤∞

∑m=1

µ(Ek(m)m

)<

∞

∑m=1

ε2−m = ε

Now let η > 0 be given and pick m0 such that m−10 < η . If ω ∈FC, then ω ∈

∞⋂m=1

ECk(m)m.

Hence ω ∈ ECk(m0)m0

so d ( f (ω) , fn (ω)) < 1/m0 < η for all n > k(m0). This holds for all

ω ∈ FCand so fn converges uniformly to f on FC.