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Measure Theory and Integration CHAPTER 1 Measure Theory and Integration One fundamental result of measure theory which we will encounter in this Chapter is that not every set can be measured. More specifically, if we wish to define the measure, or size of subsets of the real line in such a way that the measure of an interval is the length of the interval then, either we find that this measure does not extend all subsets of the real line or some desirable property of the measure must be sacrificed. An example of a desirable property that might need to be sacrificed is this: the (extended) measure of the union of two disjoint subsets might not be the sum of the measures. The Banach-Tarski paradox dramatically illustrates the difficulty of measuring general sets; it is stated in Section ???????? without proof. It is with this in mind that our discussion of measure and integration begins, in this Section, by making precise what constitutes a measurable set and a measurable function and then an integrable function. 1. Algebras of Sets Let Ω be a set and (Ω) denote the power set of Ω which is the set of all subsets of Ω. TheP reader is referred to Chapter 0, Subsection 1 if any notations or set theoretic notions are unfamiliar. Notation: N0 is the set of nonnegative integers; Q is the set of rational numbers; Z is the set of all integers. Suppose Ω and Ω are two sets and f : Ω Ω is a function with 1 2 1 → 2 domain Ω1 and range in Ω2 (but f need not be onto). For any B Ω2, denote by ⊆ 1 f − (B)= x Ω : f(x) B . { ∈ 1 ∈ } the inverse image of the set B under f. Of course f need not be 1 one-to-one (injective) and so f − need not exist as a function. Some of the elementary properties of the inverse image are worthy of note: If B,B′ Ω then ⊆ 2 1 1 1 f − (B B′) = f − (B) f − (B′) 1 ∩ 1 ∩ 1 f − (B B′) = f − (B) f − (B′) 1∪ c 1 c∪ f − (B ) = f − (B) 25 26 1.MEASURETHEORYANDINTEGRATION The verification is asked for in the Exercises at the end of the Section. Definition: If A Ω then define the characteristic function χ on ⊆ A Ω by χA(x)=1if x A and χA(x)=0if x Ω A. ∈ 1 ∈ − Exercise: Determine χA− (B) for an arbitrary B R. There are 4 cases, if = A = Ω. ⊆ ∅ Exercise: (De Morgan’s Laws) If Aα : α is a collection of subsets of a set Ω then the operation {of complementation∈ I} has the properties. C ( α Aα) = α (Aα) C ∪ ∈I ∩ ∈I C ( α Aα) = α (Aα) C ∩ ∈I ∪ ∈I C If = N is countable then we can write n NAn = n NBn where I c c ∪ c∈ ∪ c∈ B1 = A1, B2 = A2 A1, ... Bn = An A1 A2 ... An 1. The Bn are then disjoint. ∩ ∩ ∩ ∩ ∩ − Definition: Let An, n N be a countable collection of sets in Ω. The limit supremum of A ∈ is { n} lim sup An = n N k n Ak n ∩ ∈ ∪ ≥ whereas the limit infimum of A is { n} lim inf An = n N k n Ak. n ∪ ∈ ∩ ≥ Intuitively lim supn An is all points that are in infinitely many of the An whereas liminfn An is all those points that are in all except possibly finitely many of the An. If A is a decreasing sequence of sets, A A A ... then n 1 ⊇ 2 ⊇ 3 ⊇ lim sup An = n NAn = liminf An n ∩ ∈ n Definition: A collection of subsets of Ω is an algebra (field) if F a. Ω . b. If A∈ F then Ac . ∈ F ∈ F c. If Aα : α and if is finite then α . { ∈I}⊆F I ∪ ∈I ∈ F Condition c in the definition could equivalently be replaced by either of the following Conditions. c′. If A, B then A B . ∈ F ∪ ∈ F c′′. If A, B then A B . ∈ F ∩ ∈ F because Condition c′ is equivalent to c; and Condition c′′ in conjunction with Condition b is equivalent to Conditions c′ and b, as can be seen by observing that (A B)= A B. Definition: AC collection∩ Cof subsets∪C of Ω is a σ-algebra (or σ-field) if is an algebra and F F 1. ALGEBRAS OF SETS 27 d. If A : α and if is countable, or finite, then { α ∈I}⊆F I α Aα . ∪ ∈I ∈ F Of course condition d implies condition c in the definition of an algebra. Conditions b and d together are equivalent to b and d′ where d′. If A : α and if is countable, or finite, then { α ∈I}⊆F I α Aα . ∩ ∈I ∈ F Example 1.1. (i) The power set (Ω) of a set Ω is a σ- algebra. P (ii) , Ω is a σ-algebra. (iii) {∅, A,} Ac, Ω is a σ-algebra, if A is any subset of Ω. {∅ } Example (i) is the largest σ-algebra of subsets of Ω; Example (ii) is the smallest and Example (iii) is the smallest σ-algebra that contains the set A. Example 1.2. Consider the set of all real intervals of the form [a,b) along with all intervals of the form [a, ) and ( ,b) where a and b are real numbers. Define to be the∞ set of all−∞ finite unions of disjoint intervals of this form alongF with the null set and R itself. We claim that is an algebra. Certainly . If A then we must check that AcFis in and that is obvious∅ if ∈A F= or A∈= FR and so F ∅ we may suppose A = 1 j n[aj,bj) where a1 <b1 <a2 <b2 < ∪ ≤ ≤ c −∞ ≤ a3 <...<bn . Then A is of the same form: indeed if a1 > ≤ ∞ c −∞ and bn < then A = ( ,a1) [b1,a2) ... [bn 1,an) [bn, ). ∞c −∞ ∪ ∪ ∪ − ∪ ∞ Therefore A ; the cases where a1 = or b1 = or both are handled similarly.∈ F Therefore is closed under−∞ taking complements.∞ It remains to check thatF is closed under finite unions and it suffices to show that if A, B F then A B . If A = or ∈ F ∪ ∈ F ∅ A = R then this is obvious and so we suppose A = 1 j n[aj,bj) where ∪ ≤ ≤ a1 < b1 < a2 < b2 < a3 <...<bn and further it suffices to−∞ consider ≤ the cases that B = [c,d) where≤c<d ∞ or B = ( ,d) or B = [c, ) or B = R. If A B = then there is nothing−∞ to show ∞ ∩ ∅ but if B meets any subinterval [aj,bj) of A or shares an endpoint then B [aj,bj) is again an interval of the same type, that is half open and half∪ closed. (Alternatively one argues that there are four cases: c A or not and d A or not.) Therefore A B is in . ∈ ∈ ∪ F Example 1.3. Sometimes it is more convenient to work with consisting of all finite disjoint intervals of the form (a,b] (as opposedF to [a,b) above) and ( ,b] and (a, ). Again in this case is an algebra but not a σ-algebra.−∞ ∞ F 28 1.MEASURETHEORYANDINTEGRATION Proposition 1.4. Let α : α be a non empty collection of σ-algebras (resp. algebras) on{F a given∈ set I}Ω indexed by a set . Then I the intersection α α is also a σ-algebra (resp. algebra). ∩ ∈I F Proof. We need only check that α α satisfies the three prop- erties of σ-algebras The set Ω is certainly∩ ∈I inF all σ-algebras and hence c in α α. If A α α then A is in every α. Finally if (An)n N ∩ ∈I F ∈ ∩ ∈I F F ∈ is a sequence of sets in α α then n NAn is in every α and this completes the proof in the∩ case∈I F of σ-algebras∪ ∈ and the proofF for algebras is analogous. Suppose now that (Ω) is any collection of subsets of a set Ω. Then there is a smallestS ⊆Pσ-algebra which contains which we shall denote this σ( ). To see this observe that the set of allSσ-algebras that contain is nonS empty because it contains (Ω) and so the intersection of all σ-algebrasS that contain is a σ-algebra,P by the Proposition and it certainly contains and soS it must be the minimal such σ-algebra. Definition: If S is a σ-algebra then a subset with the property that = σF( ) is said to be a system of generatorsS ⊆ F of . Consider nowF the caseS that Ω = Rn, where n 1. Similar consider-F n ≥ ations apply if Ω is any topological space. On R x =(x1,...,xn), we 2 1/2 ∋ define the norm x =( 1 j n xj ) and the corresponding (usual) topology where| a| base for≤ the≤ | neighborhoods| of a point x Rn is n ∈ Vr(x) : r > 0 where Vr(x) = y R : x y < r is an open ball{ of radius r}centered at x. Then{ ∈ a set U |is said− | to be} open in Rn, if, for every x U there is r > 0 so that Vr(x) U. The set of all open sets is denoted∈ and we define the Borel⊆subsets of Rn to be = σ( ), that is theO smallest σ-algebra that contains . We shall sometimesB O write (Rn) for when wishing to emphasizeO that it is the Borel σ-algebra onB Rn. InB general, if Ω is a topological space and is the set of all open subsets of Ω then the Borel σ-algebra on ΩO is σ( ).
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