[PDF] Math 432 - Real Analysis II In class we proved that





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Math 432 - Real Analysis II

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A MONOTONE FUNCTION IS INTEGRABLE Theorem Let f be a monotone function on [a;b] then f is integrable on [a;b] Proof We will prove it for monotonically decreasing functions The proof for increasing functions is similar First note that if f is monotonically decreasing then f(b) • f(x) • f(a) for all x 2 [a;b] so f is bounded on [a;b]

What is a definite integral off?

    The integral offon [a, b] is a real number whose geometrical interpretation is thesigned area under the graphy=f(x) fora?x?b. This number is also calledRb the de?nite integral off. By integratingfover aninterval [a, x] with varying rightend-point, we get a function ofx, called the inde?nite integral off.

Is F integrable on [a1a2],[a2a3],[an1an] andz an?

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Are integrable functions bounded?

    The product of integrable functions is also integrable, as is the quotient pro-vided it remains bounded.

Math 432 - Real Analysis II

Solutions to Homework due March 11

Question 1.Letf(x) =kbe a constant function fork2R.

1. Show thatfis integrable over any [a;b] by using Cauchy's"Pcondition for integrability.

2. Show that

Z b a kdx=k(ba):

Solution 2.

(a) Sincef(x) is the constantkfunction, then clearlyM(f;S) =k=m(f;S) for anyS[a;b]. Thus, for any partitionP=fa=t0< t1<< tn=bgof [a;b], we have that

L(f;P) =nX

k=1m(f;[tk1;tk])(tktk1) =nX k=1k(tktk1) =k(ba): An identical conversation shows thatU(f;P) =k(ba) for any partition. Thus for any" >0, there exists a partition (in fact, any partition) so thatU(f;P)L(f;P) = 0< ". Thus,fis integrable. (b) SinceU(f;P) =k(ba) =L(f;P) for any partitionP, then the Darboux integrals are also equal to k(ba). Therefore,Zb a

kdx=U(k) =L(k) =k(ba):Question 2.In class, we proved that iffis integrable on [a;b], thenjfjis also integrable. Show that the

converse is not true by nding a functionfthat is not integrable on [a;b] but thatjfjis integrable on [a;b].

Solution 2.Consider the function

f(x) =1 ifx2Q

1 ifx62Q:

A computation similar to one in a previous HW shows thatfis not integrable. However,jfjis the constant

function 1, which by Question 1 is integrable.Question 3. (a) Letx;y2S. Show thatjf(x)j jf(y)j jf(x)f(y)j. (b) Letx;y2S. Show thatjf(x)f(y)j M(f;S)m(f;S). (c) Use (a) and (b) to show thatM(jfj;S)m(jfj;S)M(f;S)m(f;S):Hint: To do this, show that for any" >0,

M(jfj;S)m(jfj;S)M(f;S)m(f;S) +":

Solution 3.

1 (a) Letx;y2S. Using the triangle inequality, we have that jf(x)j=jf(x)f(y) +f(y)j jf(x)f(y)j+jf(y)j; which gives us thatjf(x)j jf(y)j jf(x)f(y)j: (b) By denition, ifx;y2S, then the following inequalities are true: m(f;S)f(x)M(f;S) m(f;S)f(y)M(f;S):

The second inequality is identical to

M(f;S) f(y)m(f;S):

Adding this one to the very rst string of inequalities, we get that [M(f;S)m(f;S)]f(x)f(y)M(f;S)m(f;S):

This is identical to our desired statement

jf(x)f(y)j M(f;S)m(f;S): (c) We will show thatM(jfj;S)m(jfj;S)M(f;S)m(f;S) by showing thatM(jfj;S)m(jfj;S)

M(f;S)m(f;S) +"for every" >0. So, let" >0. Since

M(jfj;S) = inffjf(x)jjx2Sg;

then by the approximation theorem, there exists anx02Ssuch thatjf(x0)j> M(jfj;S)"=2. Similarly, by the approximation theorem, there exists ay02Ssuch thatjf(y0)j< m(jfj;S) +"=2:Putting these together, we get that jf(x0)j jf(y0)j> M(jfj;S)m(jfj;S)"=2"=2 =M(jfj;S)m(jfj;S)":

Using the previous parts, we get that

M(jfj;S)m(jfj;S)" jf(x0)j jf(y0)j jf(x0)f(y0)j< M(jfj;S)m(jfj;S):

Therefore, for any" >0,

M(jfj;S)m(jfj;S)M(f;S)m(f;S) +":

Since this is true for every" >0, we have that

M(jfj;S)m(jfj;S)M(f;S)m(f;S):Question 4.Letfandgbe integrable functions on [a;b]. (a) Show that 4fg= (f+g)2(fg)2. (b) Use (a) to show thatfgis also integrable on [a;b].

Solution 4.

2 (a) Beginning with the left-hand side, we get that (f+g)2(fg)2=f2+ 2fg+g2(f22fg+g2) = 4fg: (b) Sincefandgare integrable on [a;b], thenf+gandfgare integrable. Since squares of integrable functions are integrable, then (f+g)2and (fg)2are integrable. Thus, by (a), 4fgis integrable and fgis integrable, as desired.Question 5.Consider the functionfon [0;1] given by f(x) =xifx2Q

0 ifx62Q

(a) LetP=f0 =t0<< tn= 1gbe any partition of [0;1]. Show that

U(f;P) =U(x;P):

(b) Compute the Upper and Lower Darboux sums,U(f) andL(f), and use this to decide iffis integrable.

Solution 5.

(a) For any subinterval [tk1;tk], there are innitely many rationals and irrationals. In particular,

M(f;[tk1;tk]) = supfxjx2Q;x2[tk1;tk]g=tk:

Notice that this is exactly the same values asM(x;[tk1;tk]). Thus,U(f;P) =U(x;P). (b) SinceU(x;P) =U(f;P) for any partitionP, then

U(f) = inffU(f;P)jPis a partition of [a;b]g=

inffU(x;P)jPis a partition of [a;b]g=U(x) = 1=2:

ForL(f), since there are innitely many irrationals in every subinterval [tk1;tk] and thusM(f;[tk1;tk]) =

0:Thus, for any partition,L(f;P) = 0. So,L(f) = 0. SinceU(f)6=L(f), thenfis not integrable.Question 6.Letfbe a bounded function on [a;b]. Suppose that there exists a sequence of partitionsPn

on [a;b] such that limn!1U(f;Pn) = limn!1L(f;Pn):

Show thatfis integrable and that

Z b a f dx= limn!1U(f;Pn) = limn!1L(f;Pn): Solution 6.Since limn!1U(f;Pn) = limn!1L(f;Pn);then lim n!1U(f;Pn)L(f;Pn) = 0: 3 Let" >0. Then, there exists anNsuch that for alln > N, jU(f;Pn)L(f;Pn)0j=U(f;Pn)L(f;Pn)< ": Thus, for the partitionPN+1, the Cauchy integrability is satised. So,fis integrable.

Next, we show thatRb

af dx= limn!1U(f;Pn) =I. Since this limit is also equal to limn!1L(f;Pn) andL(f;Pk)U(f;Pn) for allkandn, we must have that limn!1U(f;Pn)U(f;PN) for eachn. Let " >0. Then there exists anNsuch that for alln > N,jU(f;Pn)Ij=U(f;Pn)I < ":Thus, we have thatZb a f(x)dx=U(f)U(f;Pn)< I+"= limn!1U(f;Pn) +":

UsingL0s, we similarly get that

Z b a f dx >limn!1L(f;Pn)"= limn!1U(f;Pn)":

Thus, we get that

Zb a

f dx= limn!1U(f;Pn) = limn!1L(f;Pn):Question 7.Letfandgbe bounded functions on [a;b]. In what follows, we will show that maxff;ggand

minff;ggare integrable if we know thatfandgare individually integrable. Dene these functions as minff;gg(x) = maxff(x);g(x)g; and similarly for minff;gg. (a) Leta;b2R. Show that minfa;bg=12 (a+b)12 jabj: (b) Use (a) to show that iffandgare integrable, then minff;ggis also integrable. (c) Find an expression similar to one in (a) for maxfa;bg. Prove that your expression is correct. (d) Use (c) to show that iffandgare integrable, then maxff;ggis integrable.

Solution 7.

(a) We proceed with cases. Assume thatab. Then, minfa;bg=a. Sinceab, thenjabj=ba.

Thus,12

(a+b)12 jabj=12 (a+b)12 (ba) =a= minfa;bg: In the other case, ifba, then minfa;bg=b. Sinceba, thenjabj=ab. Thus, 12 (a+b)12 jabj=12 (a+b)12 (ab) =b= minfa;bg: (b) Sincefandgare integrable, thenf+gandfgare integrable. Furthermore, sincefgis integrable, thenjfgjis integrable. Since minff;gg=12 (f+g)12 jfgj is integrable. 4 (c) We conjecture that maxfa;bg=12 (a+b) +12 jabj: Using proofs similar to in (a), this statement is true. (d) Using an argument similar to that in (b), we get that maxff;ggis integrable.5quotesdbs_dbs20.pdfusesText_26
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