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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 (Xkl'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
(Xkl’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< cf4dg 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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 Spring '09
 BRUBAKER

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