Chapter 4

# Chapter 4 - DJL Properties pg integrals In this section we...

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Unformatted text preview: DJL Properties pg integrals In this section, we prove the four basic properties of the integral that we shall need. Theorem. (Properties pf the integral) (1) (Linearitx property.) g: f and g are integrable pp [a,b], then pp is of + dg (here c and d are constants), and furthermore b b b I (cf+dg) = c J f + d J g. a (2) (Additivitx propertx.)' Suppose f ig defined 93 [a,c] and a < b <(:. Then the two integrals 95 the right exist if and only i: the integral 93 the left exists. (3) (Comparison propertx.) ££_ f(x) < g(x) for all x i3 [a,b], then b b L’ f < L. g, provided both integrals exist. (4) (Reflection property.) if f ii integrable pp [a,b], then f(-x) ig integrable pp [-b,-a], and -a b I f(*x) dx = I f(x) dx. -b a , D,2‘ We use the first three of these properties repeatedly. 0 0 I C b Property (4) is used only in deriv1ng the formula for I xp dx. a Let us note that once we make the convention that a a I f = 0' and that L) f = - I f if a < b, then the formula c ’b c I f = I f + I f a a b holds without regard to the requirement that a g b < c. The proof is left as an exercise. Proof. First, one verifies these properties for step functions. This is quite straightforward. Property (3) has already been proved; proper— ties (1) and (2) will be assigned as exercisessand property (4) is proved as follows: Let s be a step function on [a,b] relative to the partition x0....,xn. Let s(x) = 5k for x in (Xk-l'xk)‘ The function u(x) = s(-x) is then a step function relative to the partition -xk,...,—xl,-xo of the interval [-b,-a]. Indeed, if x is in the interval (—xk,—xk_l), then -x is in the interval (Xk-l’xk)’ so that u(x) = s(—x) = sk. D.3 Then by definition, But s(x) dx ‘—-~—. 0‘ II at Mt: s -(x- _); l k k k l and these two expressions are equal. Thus (4) holds for step functions. Step 2. we first prove property (1) in the case where c and d are non—negative. Suppose that f and g are integrable on [a,b]. Choose step functions si and ti such that slﬁfitl and \$2§g\$t2 and b b - b b a I I 8 t - I s < and t - s < -—————-. L 1 a l 2(c+1) a 2 a 2 2(d+1) Then let s = cs1 + ds2 and let t = ct1 + dtz. New 5 and t are step functions, and (since c and d are non—negative) sl< cf4-dg g t . \ Furthermore, by property (1) for step functions, 13.4 b b b i b b b St _fs = [cgtl'tdjt21—[Cj'514’d 82] a a a a a a b b b b = c[ t —Js]+d[ft -—§s] a1 a1 a2 32 ct _9 g S E 3 2(c+l) + (d+1) < ‘2‘ “ ‘2" Hence the integral of of + dg exists by the Riemann condition. Now, by definition of the integral, we have b b b b b b J 51 < I f < I t1 and I 32 <‘J g < I t2, » a we multiply the first set of inequalities by c, and the second by d, and add, obtaining the inequalities: b b b b J.33 b b S s = c'[ 51 + d3 52 5 c g f + d g s c t1 + d.[ t2 f I, t . a a a a a a a a Here we use property (1) for step functions again. Since the expression in the box lies between the integrals of s and» t, by the Riemann condition it must equal the integral of cf + dg . Step 3. To complete the proof of property (1), it suffices to show that b b 8" (—f) = -g f. a V a This is easy. Given i >0, choose step functions 3 and t such that s <.f \$ t on [a,b], and \ D.5 Then -s and -t are step fumtions on [a,b], and -tg -f.<~ —s on [a,b]. Furthermore, b b b b [f—s]-[f—t]=—fs+ft<£. a a a a Here we use property. (1) for step functions. Thus the integral of —f exists, by the Riemann condition. Now by definition of the integral b b b J3<J f_<[t. a a a Multiplying these inequalities by -1, we conclude that b b b b b a Let) =-f t 5 -1"); S if: =§a'(~8). Here we use property (1) for step functions, again. Since the expression in the box lies between the integrals of -t and -s , by the Riemann condition it must equal the integral of -f. Step it Now we prove property (2). We. consider first the "existence" part of the statement. Suppose the integrals b . C f f and I f a b exist. Choose step functions 31 and t1 with 515 f < t1 on [a,b], and choose step functions s2 and t2 with szxéfgt2 on [b,c], such that b b c c I t]. -I 51 < 8/2 and L t2 —I 52 < 9/2. a a b The values of these functions at the partition points do not matter, so we can assume that t1 and t2 are equal at c , and \$1 and 52 are equal at C. Then tl and t2 combine to define a step function t such that f < t on [a,c}, and s and' 52 combine to form a step function 5 such that l s < f on [a,C] . Furthermore, using property (2) for step functions. A: r) n- I 93““ n m II M U: H + 1: t2} - U: 51 " LC 52} (by .the way 5 and t were constructed) < E. .8. , a + a c Hence I ' f exists , by the Riemann condition. a c Conversely, suppose I f exists. Then given a > 0,. a We can choose step functions 5 and t with s < f < t on {3.0}, such that Let s1 and t1 be the restrictions of s and t, respectively, to [a,b], and let 52 and t2 be their res- trictions to [b,c] . As before, ‘using property (2) for step fucntions, U: tl + Lctz] — U: sl + L: 52] < a, [£1ka 4 I: 51] V* U: t.2 ' [be 52] < 8. Since each expression in parentheses is nonnegative, each is b c f and I f exist. 'b Now in either of these cases, we have wermve or less than 5. Hence I ‘ a by definition. Adding, we obtain k W 0 U) H u m n n . c c . ~ Since the expression in the box lies between. 5a s and 5a t, the Riemann condition implies that it equals I: f . Step g: we prove the comparison property (3). D.8 Consider the set of all step functions 3 such that s s f on [a,b]; also consider the set of all step functions .t such that g < t on [a,b]. Because f < g on [a,b], we conclude that s < t on [a,b], whence because (3) holds for step functions. Holding t fixed and letting s vary, we conclude that {I a b s} < L t, for any fixed t 9 9. Now letting t vary, we see that {j a b 'That is, yr) < ‘fm. b Since both f and g are integrable, we have £(f) [ ' b 3(9) = I 9, so our result is proved. a and Step g; Finally, we prove the reflection property (4). D.9 Given a > 0, choose step functions 3 and t b SO . b that s <‘f < t on [a,b] and I t - I s < t. Then s(—x) a a and t(—x) are step functions on [-b,—a], and s(-x) < f(-x) < t(—x) on [~b,-a]. Now -a -a b b [ t(-x) - I s(-X) = J t - I s < e; -b —b here ye use the fact that (4) holds for step functions. Thus -a I f(-x) ~b ,e>cists. , hx_the Riemann condition. Using (4) for step functions again, b —a -a I s - L S(-x) < [_ f(-x) a b b Since the expression in the box lies between the integrals of s and t, it H -a b < J t(-x) z I t, -b a must by the Riemann condition equal the integral of f. D.lO Exercises 1. Prove property (1) for step functions. [Hint: If s and t are step functions, the first thing to do is to choose a partition P that is compatible with both 3 and t. Then show cs + dt is a step function compatible with Pd 2. Prove property (2) for step functions. [Hint: If P1 is a partition of [a,b] and P2 is a partition of [b,c], then Pl U P2 is a partition of [a,c].] 3. We know (2) holds if a < b < c. Show that with our convention, it holds in all cases: a=b, a<c<b, c<a<b, a = c, b < a < c, C < b < a. b = c, b < c < a, 4; Let 1r0 < - ' - < xn be a partition of [a,b}. Let s be a step function on [a,b] such that s(x) = 3k for xk_1 < x < xk. Let h be an increasing function on [a,b]. Suppose we deﬁne [b s dh = i=1 sk- (hag—hum». a (a) Show that this integral is well—deﬁned. (b) Show that this integral satisﬁes the linearity, additivity, and comparison properties. You need to use the fact that h is increasing in order to prove one of these properties; which one? ' [This deﬁnition is actually an important one in mathematics. It leadsb to a generalization of the integral called the Riemann—Stieltjeu i;nteg1;al one deﬁnes I: f dh by using upper and lower integrals, just as before. This integral is important in probability theory.] ' ...
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