244 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

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 11.3.19 is measurable because it is a countable unionof measurable sets. This proves the claim since[

ω : d ( fn (ω) , f (ω))≥ 1m

]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 11.1.5 on Page 224 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=1

Ek(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.