330 CHAPTER 14. THE LEBESGUE INTEGRAL

Then F ≡ ∪kFk is Fσ because it is still a countable union of closed sets. Recall that thecountable union of countable sets is countable. Also µ (E \F)≤ ∑k µ (Ek \Fk) = 0. How-ever, it may not be clear why ∪kGk would be Gδ . However, the above implies that thereexists an open set Vi ⊇ Ei∩Ak such that µ (Ei∩Ak)+

ε

2k2i+1 > µ (Vi) . Then if Vk ≡∪∞i=1Vi,

µ (Vk)≤∑i

µ (Vi)<∞

∑i=0

(µ (Ei∩Ak)+

ε

2k2i+1

)=

ε

2k +µ (E ∩Ak) , Vk ⊇ E ∩Ak

µ (Vk \ (E ∩Ak))<ε

2k

Then µ(∪∞

k=0Vk \E)≤ ∑

∞k=0 µ (Vk \ (E ∩Ak)) < 2ε . It follows that there exists open Wn

containing E such that µ (Wn \E) < 1/n. These Wn can be assumed decreasing. Thus ifG ≡ ∩nWn,µ (G\E) = 0. Hence G ⊇ E ⊇ F and µ (G\F) = µ (G\E) + µ (E \F) =0. Thus G is closed with respect to complements and countable disjoint unions so fromLemma 14.2.3 it contains σ (I ) = B (R) but G was defined to consist of sets of B (R) soG = B (R).

The first claim 14.1 was just shown. Let l < µ (E) then µ (E ∩ [−n,n]) > l for largeenough n and so there is a closed set K contained in E ∩ [−n,n] such that l < µ (K) also.This shows the first of 14.2. There is nothing to show in the second if µ (E) =∞. So assumeµ (E) is finite. Then letting G be from the first part, G = ∩nWn where Wn is open and theseare decreasing open sets. We can assume µ (W1) < µ (E) + 1 from the argument givenabove to show 14.1. Thus µ (G) = µ (E) = limn→∞ µ (Wn) and so for large n,µ (E)+ ε >µ (Wn). This shows the second part of 14.1.

This shows that all those Lebesgue Stieltjes measures are regular.Next is to define the kind of function which can be integrated. The measure space of

this section dealing with the Lebesgue Stieltjes measures is specific to R but what comesnext is the general notion in an abstract measure space.

14.5 Measurable FunctionsYou can integrate nonnegative measurable functions. All this will be presented in general.Thus the functions are defined on a measure space. I am going to present this in the generalsetting but you can apply it to the measure space just developed consisting of the LebesgueStieltjes measure on R or on the counting measure of Example 14.1.2.

Notation 14.5.1 In whatever context f−1 (S) ≡ {ω : f (ω) ∈ S}. It is called the inverseimage of S and everything in the theory of the Lebesgue integral is formulated in terms ofinverse images. For a real valued f , f−1 (λ ,∞) may sometimes be written as [ f > λ ].

Lemma 14.5.2 Let f : Ω→ (−∞,∞] where F is a σ algebra of subsets of Ω. Thefollowing are equivalent.

f−1((d,∞]) ∈F for all finite d,

f−1((−∞,d)) ∈F for all finite d,

f−1([d,∞]) ∈F for all finite d,

f−1((−∞,d]) ∈F for all finite d,

f−1 ((a,b)) ∈F for all a < b,−∞ < a < b < ∞.