4 Integration 4.1 Integration of non-negative simple functions

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1 4 Integration 4.1 Integration of non-negative simple functions Throughout we are in a measure space (X, F, µ). Definition Let s be a non-negative F-measurable simple function so that s a i χ Ai, with disjoint F-measurable sets A i, N A i X and a i 0. For any F define the integral of f over to be I (s) a i µ(a i ), with the convention that if a i 0 and µ(a i ) + then 0 (+ ) 0. (So the area under s 0 on R is zero.) xample 13 Consider ([0, 1], L, µ). Define { 1 if x rational f(x) 0 if x irrational. This is a simple function with A 1 Q [0, 1] L and A 0 the set of irrationals in [0, 1] which, as the complement of A 1, is in L. Thus f is measurable and I [0,1] (f) 1µ(Q [0, 1]) + 0µ(Q c [0, 1]) 0, since the Lebesgue measure of a countable set is zero. Lemma 4.1 If are in F and n1 n then lim µ( n) µ(). n (We say that we have an increasing sequence of sets.) If there exists an n such that µ( n ) + then n implies µ() + and the result follows. So assume that µ( n ) < + for all n 1. Then 1

2 1 ( n \ n 1 ) n2 is a disjoint union. Note that n 1 n implies n ( n \ n 1 ) n 1, a disjoint union. So µ( n ) µ( n \ n 1 ) + µ( n 1 ). Because the measures are finite we can rearrange as µ( n \ n 1 ) µ( n ) µ( n 1 ). So µ() µ( 1 ) + µ( n \ n 1 ) n2 µ( 1 ) + lim N lim N µ( N). (µ( n ) µ( n 1 )) n1 (by definition of infinite sum) Theorem 4.2 Let s and t be two simple non-negative F-measurable functions on (X, F, µ) and, F F. Then (i) I (cs) ci (s) for all c R, (ii) I (s + t) I (s) + I (t), (iii)if s t on then I (s) I (t), (iv) If F then I F (s) I (s), (v) If and k1 k then lim k I k (s) I (s). (s of all parts will be omitted from lectures and left to students. the idea is to write out the simple functions for both s and t in terms of common sets C ij as in the proof of Lemma 3.7.) As in Lemma 3.7 write and s t a i χ Ai b j χ Bj j1 a i χ Cij j1 b j χ Cij j1 2

3 with C ij A i B j F. *(i) Note that cs M ca iχ Ai and so I (cs) c ca i µ(a i ) a i µ(a i ) ci (s). *(ii) Then s + t M N j1 (a i + b j )χ Cij. So I (s + t) (a i + b j )µ(c ij ) j1 a i µ(c ij ) + j1 ( N ) a i µ (C ij ) + j1 a i µ(a i ) + I (s) + I (t). b j µ(c ij ) j1 ( M ) b j µ (C ij ) j1 b j µ(b j ) *(iii) Given any 1 i M, 1 j N for which C ij φ we have for any x C ij that a i s(x) t(x) b j so j1 I (s) a i µ(c ij ) j1 j1 I (t). *(iv) By monotonicity of µ we have b j µ(c ij ) 3

4 I F (s) a i µ(a i F ) a i µ(a i ) I (s). *(v) From Lemma 4.1 we know that if we have and k1 k then lim k µ( k ) µ(). Thus lim I k (s) lim k k a i µ(a i k ) a i lim k µ(a i k ) a i µ(a i ) by Lemma 4.1, I (s). 4.2 Integration of non-negative measurable functions. Definition If f : X R + is a non-negative F-measurable function, F, then the integral of f over is fdµ sup {I (s) : s a simple F-measurable function, 0 s f}. Of course, if X we need only that f is defined on some domain containing. Let I(f, ) denote the set {I (s) : s a simple F-measurable function, 0 s f} so the integral equals sup I(f, ). Note The integral exists for all non-negative F-measurable functions though it might be infinite. 4

5 If fdµ we say the integral is defined. If fdµ < we say that f is µ integrable or summable on. Proposition 4.3 For a non-negative, F-measurable simple function, t, we have tdµ I (t). Given any simple F-measurable function, 0 s t we have I (s) I (t) by Theorem 4.2(iii). So I (t) is an upper bound for I(t, ) for which tdµ is the least of all upper bounds. Hence tdµ I (t). Also, tdµ I (s) for all simple F-measurable function, 0 s t, and so is greater than I (s) for any particular s, namely s t. Hence tdµ I (t). Thus tdµ I (t). xample 14 If f k, a constant, then fdµ I (f) kµ(). Theorem 4.4 Throughout, all sets are in F and all functions are nonnegative and F-measurable. (i) For all c 0, cfdµ c fdµ, (15) (ii) If 0 g h on then gdµ hdµ, (iii) If 1 2 and f 0 then fdµ fdµ. 1 2 (i) If c 0 then the right hand side of (15) is 0 as is the left hand side by xample 14. Assume c > 0. If 0 s cf is a simple F-measurable function then so is 0 1s f. c Thus ( ) 1 fdµ I c s 1 c I (s) by Theorem 4.2(i). Hence c fdµ is an upper bound for I(cf, ) for which cfdµ is the least upper bound. Thus c fdµ cfdµ. 5

6 Starting with the observation that if 0 s f is a simple F-measurable function then so is 0 cs cf we obtain (cf) dµ I (cs) by the definition of ci (s) by Theorem 4.2(i). Hence 1 (cf)dµ is an upper bound for I(f, ) for which c fdµ is the least upper bound. Hence 1 (cf)dµ fdµ, that is, cfdµ c fdµ. c Combining both inequalities gives our result. (ii) Let 0 s g be a simple, F-measurable function. Then since g h we trivially have 0 s h in which case I (s) hdµ by the definition of integral. Thus hdµ is an upper bound for I(g, ). As in (i) we get hdµ gdµ. (iii) Let 0 s f be a simple, F-measurable function. Then I 1 (s) I 2 (s) by Theorem 4.2(iii) fdµ 2 by the definition of So 2 fdµ is an upper bound for I(f, 1 ) and so is greater than the least of all upper bounds. Hence 2 fdµ 1 fdµ. Lemma 4.5 Assume F, f 0 is F-measurable and fdµ <. Set A {x : f(x) + }. Then A F and µ(a) 0. Since f is F-measurable then f 1 ({ }) F and so A f 1 ({ }) F. Define { n if x A s n (x) 0 if x / A. Since A F we deduce that s n is an F-measurable simple function. Also s n f and so nµ(a) I (s n ) by definition of I fdµ by definition of < by assumption. 6 2.

7 True for all n 1 means that µ(a) 0. Lemma 4.6 If f is F-measurable and non-negative on F and µ() 0 then fdµ 0. Let 0 s f be a simple, F-measurable function. So s N n1 a nχ An for some a n 0, A n F. Then I (s) N n1 a nµ(a n ). But µ is monotone which means that µ(a n ) µ() 0 for all n and so I (s) 0 for all such simple functions. Hence I(f, ) {0} and so fdµ sup I(f, ) 0. Lemma 4.7 If g 0 and gdµ 0 then µ{x : g(x) > 0} 0. Let A {x : g(x) > 0} and A n {x : g(x) > 1 }. Then n the sets A n {x : g(x) > 1 } F satisfy A n 1 A 2 A 3... with A n1 A n. By lemma 4.1 µ(a) lim n µ(a n ). Using { 1 if x A s n (x) n n 0 otherwise, so s n g on A n we have 1 µ(a n n) I An (s n ) A n gdµ gdµ by the definition of A n Thereom 4.4(iii) 0 by assumption. So µ(a n ) 0 for all n and hence µ(a) 0. Definition If a property P holds on all points in \ A for some set A with µ(a) 0 we say that P holds almost everywhere (µ) on, written as a.e.(µ) on. (*It might be that P holds on some of the points of A or that the set of points on which P does not hold is non-measurable. This is immaterial. But if µ is a complete measure, such as the Lebesgue-Steiltje s measure µ F, then the situation is simpler. Assume that a property P holds a.e.(µ) on. The definition says that the set of points, D say, on which P does not hold can be covered by a set of measure zero, i.e. there exists A : D A and µ(a) 0. Yet if µ is complete then D will be measurable of measure zero. In this section we are not assuming that µ is complete.) 7

8 So, for example, Lemma 4.7 can be restated as Lemma 4.8 If g 0 and gdµ 0 then g 0 a.e.(µ) on. We can extend Theorem 4.4(ii) as follows. Theorem 4.9 If g, h : X R + are F-measurable functions and g h a.e.(µ) then gdµ hdµ. By assumption there exists a set D, of measure zero, such that for all x \ D we have g(x) h(x). Let 0 s g be a simple, F-measurable function, written as s a i χ Ai, with N A i. The problem here is that we may well not have s h. Define s (x) { s(x) if x / D 0 if x D a i χ Ai D c which is still a simple, F-measurable function. Then for x \ D we have s (x) s(x) g(x) h(x), while for x D we have s (x) 0 h(x). Thus s (x) h(x) for all x. Note that A i (A i D c ) (A i D), a disjoint union in which case µ(a i ) µ(a i D c )+µ(a i D) µ(a i ). But A i D D and so µ (A i D) µ(d) 0. Thus µ(a i ) µ(a i D c ). Hence I (s ) a i µ(a i D c ) a i µ(a i ) I (s). 8

9 So I (s) I (s ) hdµ by the definition of integral. Thus hdµ is an upper bound for I(g, ) while gdµ is the least of all upper bounds for I(g, ). Hence hdµ gdµ. Corollary 4.10 If g, h : X R + are F-measurable with g h a.e (µ) on then gdµ hdµ. By assumption there exists a set D of measure zero such that for all x \ D we have g(x) h(x). In particular, for these x we have g(x) h(x) and h(x) g(x). So g h a.e. (µ) on and h g a.e. (µ) on. Hence the result follows from two applications of Theorem 4.9. So, a function may have its values altered on a set of measure zero without altering the value of its integral. In particular, by Lemma 4.5 we may assume that a non-negative integrable function is finite valued. xample 15 (c.f. xample 13) On ([0, 1], L, µ) the function { 1 if x is rational f(x) 0 if x irrational is 0 a.e.(µ) on [0, 1]. So [0,1] fdµ [0,1] 0dµ 0. 9