Chapter 33

The Lebesgue Measure and Integral inRp

33.1 An Outer Measure on P (R)It is needed to find a measure which delivers length. Recall P (S) denotes the set of allsubsets of S. To begin with, it is shown there is an outer measure which gives length.

Theorem 33.1.1 There exists a function m : P (R)→ [0,∞] which satisfies the fol-lowing properties.

1. If A ⊆ B, then 0 ≤ m(A)≤ m(B) ,m( /0) = 0.

2. m(∪∞

k=1Ai)≤ ∑

∞i=1 m(Ai)

3. m([a,b]) = b−a = m((a,b)).

Proof: First it is necessary to define the function m. This is contained in the followingdefinition.

Definition 33.1.2 For A ⊆ R,

m(A) = inf

{∞

∑i=1

(bi −ai) : A ⊆ ∪∞i=1 (ai,bi)

}

In words, you look at all coverings of A with open intervals. For each of these opencoverings, you add the lengths of the individual open intervals and you take the infimum ofall such numbers obtained.

Then 1.) is obvious because if a countable collection of open intervals covers B, then italso covers A. Thus the set of numbers obtained for B is smaller than the set of numbers forA. Why is m( /0) = 0? Then /0 ⊆ (a−δ ,a+δ ) and so m( /0) ≤ 2δ for every δ > 0. Lettingδ → 0, it follows that m( /0) = 0.

Consider 2.). If any m(Ai) = ∞, there is nothing to prove. The assertion simply is∞ ≤ ∞. Assume then that m(Ai)< ∞ for all i. Then for each m ∈N there exists a countableset of open intervals, {(am

i ,bmi )}

∞

i=1 whose union contains Am such that

m(Am)+ε

2m >∞

∑i=1

(bmi −am

i ) .

Then using Theorem 6.6.4 on Page 165,

m(∪∞m=1Am)≤ ∑

i,m(bm

i −ami ) =

∞

∑m=1

∞

∑i=1

(bmi −am

i )≤∞

∑m=1

m(Am)+ε

2m =∞

∑m=1

m(Am)+ ε,

and since ε is arbitrary, this establishes 2.).Next consider 3.). By definition, there exists a sequence of open intervals, {(ai,bi)}∞

i=1whose union contains [a,b] such that m([a,b]) + ε ≥ ∑

∞i=1 (bi −ai) . Since [a,b] is com-

pact, finitely many of these intervals also cover [a,b]. It follows there exist finitely many

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