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Measure and Integration [Pdf] Lecture 2. Measure and Integration There are several ways of presenting the definition of integration with respect to a measure. We will follow, more or less, the approach of Rudin's `Real and Complex Analysis' - this is probably the fastest route. Conventions concerning ±∞. In measure theory it is often essential to use 1 and −∞. We shall view these as two objects and define the following relationship with real numbers (the set of all real numbers will be denoted by R) : −∞ < c < 1 for every c 2 R c + 1 = 1 if −∞ < c ≤ 1, and c + (−∞) = −∞ if −∞ ≤ c < 1 0 · 1 = 0 ±∞; if 0 < c ≤ 1; c(±∞) = ∓∞; if −∞ ≤ c < 0; Note that 1−1 is not defined! The convention 0·1 = 0 might seem like an arbitrary choice, but it is the really the right one : it will be useful in integration and there it will appear as the natural choice. These operations have the usual properties on [0; 1] : commutative, associative and distributive. Throughout our discussion (X; B; µ) will be a measure space. This means that X is a set, B is a collection of subsets of X forming a σ−algebra, and µ is a measure. Recall that to say that B is a σ−algebra means that : (S1) ; 2 B; (S2) if A 2 B then the complement Ac 2 B; 1 (S3) if A1; A2; A3; ::: is a countable collection of sets with each Ai 2 B then [i=1Ai 2 B. Sets in B will be called B−measurable sets, or simply measurable sets. Observations. (i) The intersection of a countable collection A1; A2; ::: of measurable 1 c c sets is also measurable (because \i=1Ai = ([iAi ) is measurable by (S2) and (S3).) m (ii) If A1; :::; Am 2 B then [i=1Ai 2 B; this follows from (S1) and (S3) by taking An = ; m for all n bigger. Using the complements we obtain also that \i=1Ai 2 B. (iii) Notice also that if A; B 2 B then A n B, being equal to A \ Bc, is also in B. To say that µ is a measure on (X; B) means that µ is a mapping B ! [0; 1] such that : (M1) µ(;) = 0 (M2) if A1; A2; ::: is a countable collection of mutually disjoint sets in B (i.e. each Ai 2 B and Ai \ Aj = ; for every i 6= j) then: 1 1 µ([i=1Ai) = µ(Ai): Xi=1 Taking Ai = ; for i bigger than some m, we see (from (M1) and (M2)) that µ is finitely-additive : 1 m m µ([i=1Ai) = µ(Ai) Xi=1 provided, of course, that the Ai are disjoint measurable sets. (Comment : It might seem that finite additivity implies the condition (M1): we have µ(;) = µ(; [ ;) = µ(;) + µ(;), \and so" µ(;) = 0. What is wrong with this argument? Under what additional condition on µ would finite additivity imply (M1) ?) Exercise. If A and B are measurable sets with A ⊂ B, show that µ(A) ≤ µ(B). (Hint : write B as the union of A and B n A; check that B n A (which is B \ Ac) is measurable and use additivity.) The simplest meaningful example of a measure is counting measure: µ(A) = number of elements in A (taken to be 1 if A is infinite) 2.1. Definition. A function f : X ! [−∞; 1] is measurable with respect to the σ−algebra B if the set f −1[a; b] (i.e. the set fx 2 X : f(x) 2 [a; b]g ) is in B, for every a; b 2 [−∞; 1]. Note that, in particular, if f is measurable then (taking b = a) every set of the type f −1(a) is measurable (a 2 [−∞; 1]). 2.2. Terminology. (i) If A ⊂ X, then the indicator function of A is the function 1A : X ! R, defined by : 1; if x 2 A; 1A(x) = 0; otherwise. (ii) A simple function is a function s : X ! R whose range consists of a finite set of points (i.e. the image s(X) is a finite subset of R). If a1; :::; an are all the distinct values of the simple function s then (verify that) n s = ai1Ai Xi=1 n where Ai = fx : s(x) = aig. Notice (verify) that the Ai are disjoint sets and [i=1Ai = X. Notice also that a simple function can be written in different ways in the form c cj1Cj . For instance, 1X is also equal to 1A + 1A , for any A ⊂ X. The represen- tationX described above in terms of the distinct values of s is, of course, unique; which is why we use it in the definition (2.4) of s dµ below. Z 2.3. Exercise. Verify that s is measurable if and only if each Ai is. 2.4. Definition. (Integrating simple functions) If s : X ! [0; 1) is a measurable simple function of the form n s = ai1Ai Xi=1 2 −1 where a1; :::; an are all the distinct values of s (and Ai = s (ai)), then n s dµ = aiµ(Ai): Z Xi=1 Note that a term of the form 0 · 1 might occur in the sum on the right side of the above equation; recall, for this purpose, that 0 · 1 = 0. 2.5. Question. What is the relevance of Problem 2.3 to Definition 2.4 ? m 2.6. Note. If s = cj1Cj is a non-negative measurable simple function then a fact Xj=1 m that is not immediately obvious from Definition 2.4 is that s dµ = cjµ(Cj) (provided Z Xj=1 this sum makes sense). Definition 2.4 says that this is true only when the ci are all the distinct values of s. 2.7. Proposition. If s and t are nonnegative measurable simple functions then so is s + t and (s + t) dµ = s dµ + t dµ. Z Z Z If c 2 [0; 1) is a constant then : cs dµ = c s dµ Z Z n m Proof. Write s = ai1Ai and t = bj1Bj , where a1; :::; an are all the distinct Xi=1 Xj=1 values of s and b1; :::; bm are all the distinct values of t. The function s + t takes the value ai + bj on the set Ai \ Bj (because on this set s takes the value ai and t takes the value bj). Moreover, (verify that) the sets Ai \ Bj are disjoint and cover all of X (i.e. 0 0 (Ai \Bj)\(Ai0 \Bj0 ) = ; unless i = i and j = j ; and [i;j(Ai \Bj) = X.) This is because the Ai's are disjoint and cover X and the Bj's are disjoint and cover X. Therefore, s + t = (ai + bj)1Ai\Bj : 1≤i≤n;X1≤j≤m (To see this, consider any point x 2 X. It must belong to exactly one of the sets Ai \ Bj; say x 2 A1 \ B2. Then the function defined by the sum on the right hand side of the above equation takes the value a1 + b2 at x. But this also the value of s(x) + t(x).) By Problem 2.3 we see that s + t is measurable. It is also clear that s + t is simple and non-negative. We would like to conclude, by Definition 2.4, that it follows that : (s + t) dµ = (ai + bj)µ(Ai \ Bj) (2:1) Z 1≤i≤n;X1≤j≤m 3 However, there is no guarantee that the values ai + bj are distinct and therefore Definition 2.4 cannot be applied directly. Fortunately, equation (2.1) is still true because the sets Ai \ Bj are disjoint (this is the content of Lemma 2.8 below). So we accept equation (2.1). The right side of equation (2.1) can be split up into the sum of two sums : n m m n (ai + bj)µ(Ai \ Bj) = ai µ(Ai \ Bj) + bj µ(Ai \ Bj) : 1≤i≤n;X1≤j≤m Xi=1 Xj=1 Xj=1 Xi=1 Now the sets Ai \ Bj are disjoint and, for fixed i and variable j, they cover Ai ( for [j (Ai \ Bj) = Ai \ [j Bj = Ai \ X = Ai:) Therefore : m µ(Ai \ Bj) = µ(Ai): Xj=1 Similarly, n µ(Ai \ Bj) = µ(Bj): Xi=1 Putting all this together we obtain : n m (s + t) dµ = aiµ(Ai) + bjµ(Bj) Z Xi=1 Xj=1 and we recognize the the two sums on the right hand side to be s dµ and t dµ. Z Z Thus : (s + t) dµ = s dµ + t dµ. Z Z Z That cs dµ = c s dµ, for c any constant in [0; 1), is verified easily by writing out Z Z s as a linear combination of indicator functions and using Definition 2.4. 2.8. Lemma. If C1; :::; Cn are disjoint measurable sets and c1; :::; cn non-negative real numbers then n n ( ci1C ) dµ = ciµ(Ci): Z i Xi=1 Xi=1 n Proof. We shall assume that [i=1Ci = X; for if this is not so then we can always n throw in another set Cn+1 = X n Ci, and use cn+1 = 0, and this would not alter the i[=1 4 value of either side of the above equation (note : here we need to use 0 · 1 = 0 in case n µ(Cn+1) = 1.) Let s = ci1Ci , and let a1; :::am be all the distinct values of s. Note that Xi=1 s is a non-negative measurable simple function. Moreover, (verify that) c1; :::; cn constitute all the values of s (i.e.
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