328 CHAPTER 12. THE CONSTRUCTION OF MEASURES

Theorem 12.11.2 Let (Ω,F ,µ) be a complete measure space and let f ≤ g≤ h be func-tions having values in [0,∞] . Suppose also that f (ω) = h(ω) a.e. ω and that f and h aremeasurable. Then g is also measurable. If

(Ω,F ,µ

)is the completion of a σ finite mea-

sure space (Ω,F ,µ) as described above in Theorem 12.11.1 then if f is measurable withrespect to F having values in [0,∞] , it follows there exists g measurable with respect to F, g≤ f , and a set N ∈F with µ (N) = 0 and g = f on NC. There also exists h measurablewith respect to F such that h ≥ f , and a set of measure zero, M ∈F such that f = h onMC.

Proof: Let α ∈ R.[ f > α]⊆ [g > α]⊆ [h > α]

Thus[g > α] = [ f > α]∪ ([g > α]\ [ f > α])

and [g > α]\ [ f > α] is a measurable set because it is a subset of the set of measure zero,

[h > α]\ [ f > α] .

Now consider the last assertion. By Theorem 11.3.9 on Page 241 there exists an in-creasing sequence of nonnegative simple functions, {sn} measurable with respect to Fwhich converges pointwise to f . Letting

sn (ω) =mn

∑k=1

cnkXEn

k(ω) (12.11.54)

be one of these simple functions, it follows from Theorem 12.11.1 there exist sets, Fnk ∈F

such that Fnk ⊆ En

k and µ(Fn

k

)= µ

(En

k

). Then let

tn (ω)≡mn

∑k=1

cnkXFn

k(ω) .

Thus tn = sn off a set of measure zero, Nn ∈F , tn ≤ sn. Let N′ ≡ ∪nNn. Then by Theorem12.11.1 again, there exists N ∈F such that N ⊇ N′ and µ (N) = 0. Consider the simplefunctions,

s′n (ω)≡ tn (ω)XNC (ω) .

It is an increasing sequence so let g(ω) = limn→∞ sn′ (ω) . It follows g is mesurable withrespect to F and equals f off N.

Finally, to obtain the function, h ≥ f , in 12.11.54 use Theorem 12.11.1 to obtain theexistence of Fn

k ∈F such that Fnk ⊇ En

k and µ(Fn

k

)= µ

(En

k

). Then let

tn (ω)≡mn

∑k=1

cnkXFn

k(ω) .

Thus tn = sn off a set of measure zero, Mn ∈F , tn ≥ sn, and tn is measurable with respectto F . Then define

s′n = maxk≤n

tn.