11.3. THE ABSTRACT LEBESGUE INTEGRAL 243

and continue in this manner obtaining disjoint measurable sets, {Ak}nk=1 such that for ω ∈

Ak the best approximation to f (ω) from Dn is xk. Then

fn (ω)≡n

∑k=1

xkXAk (ω) .

Notemin

k≤n+1d (xk, f (ω))≤min

k≤nd (xk, f (ω))

and so this verifies 11.3.17. It remains to verify 11.3.18.Let ε > 0 be given and pick ω ∈Ω. Then there exists xn ∈D such that d (xn, f (ω))< ε .

It follows from the construction that d ( fn (ω) , f (ω))≤ d (xn, f (ω))< ε. This proves thefirst half.

Now suppose the existence of the sequence of simple functions as described above.Each fn is a measurable function because f−1

n (U) = ∪{Ak : xk ∈U}. Therefore, the con-clusion that f is measurable follows from Theorem 11.1.9 on Page 227.

In the context of this more general notion of measurable function having values in ametric space, here is a version of Egoroff’s theorem.

Theorem 11.3.11 (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

](11.3.19)

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.