658 CHAPTER 33. THE LEBESGUE MEASURE AND INTEGRAL IN Rp

of these intervals, denoted as {(ai,bi)}ni=1 , which overlap, such that a ∈ (a1,b1) ,b1 ∈

(a2,b2) , · · · ,b ∈ (an,bn) . Therefore, m([a,b])≤ ∑ni=1 (bi −ai) . It follows

n

∑i=1

(bi −ai)≥ m([a,b])≥n

∑i=1

(bi −ai)− ε ≥ (b−a)− ε

Therefore, since(a− ε

2 ,b+ε

2

)⊇ [a,b], (b−a)+ ε ≥ m([a,b]) ≥ (b−a)− ε Since ε is

arbitrary, (b−a) = m([a,b]) . Consider [a+δ ,b−δ ] . From what was just shown, it fol-lows that m([a+δ ,b−δ ]) = (b−a)− 2δ ≤ m((a,b)) and so, since this holds for everyδ ,(b−a)≤ m((a,b))≤ m([a,b]) = (b−a). This shows 3.)

33.2 One Dimensional Lebesgue MeasureTheorem 33.2.1 Let F denote the σ algebra of Theorem 32.4.4, associated withthe outer measure m in Theorem 33.1.1, on which m is a measure. Then every open intervalis in F . All open sets are in F and all half open and closed intervals are in F .

Proof: The first task is to show (a,b) ∈ F . I need to show that for every S ⊆ R,

m(S)≥ m(S∩ (a,b))+m(

S∩ (a,b)C)

(33.1)

Suppose first S is an open interval, (c,d) . If (c,d) has empty intersection with (a,b) oris contained in (a,b) there is nothing to prove. The above expression reduces to nothingmore than m(S) = m(S). Suppose next that (c,d) ⊇ (a,b) . In this case, the right side ofthe above reduces to

m((a,b))+m((c,a]∪ [b,d))≤ b−a+a− c+d −b

= d − c = m((c,d))

The only other cases are c ≤ a < d ≤ b or a ≤ c < d ≤ b. Consider the first of these cases.Then the right side of 33.1 for S = (c,d) is

m((a,d))+m((c,a]) = d −a+a− c = m((c,d))

The last case is entirely similar. Thus 33.1 holds whenever S is an open interval. Now it isclear 33.1 also holds if m(S) = ∞. Suppose then that m(S)< ∞ and let

S ⊆ ∪∞k=1 (ak,bk)

such that

m(S)+ ε >∞

∑k=1

(bk −ak) =∞

∑k=1

m((ak,bk)) .

Then since m is an outer measure, and using what was just shown,

m(S∩ (a,b))+m(

S∩ (a,b)C)

≤ m(∪∞k=1 (ak,bk)∩ (a,b))+m

(∪∞

k=1 (ak,bk)∩ (a,b)C)

≤∞

∑k=1

m((ak,bk)∩ (a,b))+m((ak,bk)∩ (a,b)C

)≤

∞

∑k=1

m((ak,bk))≤ m(S)+ ε.