Math 432 - Real Analysis II PDF

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Chapter 5. Integration §1. The Riemann Integral Let a and b be two

g(x)dx. Theorem 2.2. Let f and g be integrable functions on [a b]. Then fg is an integrable function on [a

Math 3162 Homework Assignment 7 Grad Problem Solutions

20-Mar-2019 (b)Prove that if f is integrable on [a b]

MAT127B HW Solution 02/26 Chutong Wu 7.4.3 Decide which of the

5 Let f and g be integrable functions on [a b]. (a) Show that if P is any partition of [a

Properties of Riemann-integrable functions Underlying properties of

The preceding property asserts that µ is an additive set function: µ([a b]) = µ([a

MATH 104 HOMEWORK #12

Then by the Exercise 33.7b above. (f +g)2 and (f −g)2 are integrable

Integrability on R

Properties of Integrable Functions. Theorem 5.7 (Linear Property). If fg are integrable on [a

FUNDAMENTALS OF REAL ANALYSIS by Do˘gan C¸ömez III

then f = g almost everywhere and g is Riemann Integrable. = Lebesgue ... If f is Riemann-integrable then f is Lebesgue- integrable and. ∫. [a

Tutorial Sheet: Definite and Improper Integrals 1. Using the definition

(b) If fg are integrable then so is fg and f g.

Math 432 - Real Analysis II

(d) Use (c) to show that if f and g are integrable then max{f

Chapter 5. Integration §1. The Riemann Integral Let a and b be two

A bounded function f on [a b] is said to be (Riemann) integrable if L(f) g(x)dx. Theorem 2.2. Let f and g be integrable functions on [a

MATH 104 HOMEWORK #12

7(b) Show that if f is integrable on [a b]

Chapter 11: The Riemann Integral

Integrability is a less restrictive condition on a function than example if f

Properties of Riemann-integrable functions Underlying properties of

Function additivity. If fg are Riemann-integrable on [a

MAT127B HW Solution 02/26 Chutong Wu 7.4.3 Decide which of the

5 Let f and g be integrable functions on [a b]. (a) Show that if P is any partition of [a

RIEMANN INTEGRATION ON A MULTIDIMENSIONAL

LEMMA: f is integrable iff for every ? there exists a partition P such that If f g are integrable then also the product fg

MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS 7

D. 7.5. Simple consequences. Theorem 7.10. (a) If f and g are integrable on [a b] then f +g is

4.4 Integration of measurable functions.

From the definition f is integrable if and only if

72 Riemann Integrable Functions - Texas Tech University

If f;g: [a;b] !R and both f and gare Riemann integrable then fgis Riemann integrable Proof Apply the Composition theorem The function h(x) = x2 is continuous on any nite interval Then h f= h(f) = f2 and h g= h(g) = g2 are Riemann integrable Also (f+g)2 is Riemann integrable (why?) Therefore fg= 1 2 [(f+ g)2 f2 g2] is Riemann integrable

real analysis - About the Riemann integrability of composite functions

INTEGRABLE FUNCTIONS Then g is a monotonic function from [ab] to R?0 Hence by theorem 7 6 g is integrable on [ab] and Z b a g = Ab a(g) Now let {Pn} be a sequence of partitions of [ab] such that {µ(Pn)} ? 0 and let {Sn} be a sequence such that for each n in Z+ S n is a sample for Pn Then {X (gPnSn)} ? Ab a(g) (8 5) If Pn

Products of Riemann Integrable Functions

ProofofCorollary2: The integrability of f and fn both follow directly from Theorem 1 with ?(y) = y and ?(y) = yn respectively If f and g are bounded and integrable then so is f +g Hence by Theorem 1 with ?(y) = y2 we have that f 2 g2 and (f +g) = f2+g 2+2fg are all integrable The integrability of fg = 1 2 (f+g)2?f ?g2 now

Properties of Riemann Integrable Functions - CoAS

(f +g) = ? b a f +? b a g: Proof f +g is integrable on [a; b]; because for any partition P U(f +g;P)?L(f +g;P) ? U(f;P)?L(f;P)+U(g;P)?L(g;P); and the right side can be made arbitrarily small To prove additivity choose P so that U(f;P)? " 2 < ? b a f < L(f;P)+ " 2 and U(g;P)? " 2 < ? b a g < L(g;P)+ " 2: Then ? b a (f +g

Properties of the Riemann Integral

To prove that fg is integrable when f and g are simply note that fg= (1=2) (f +g)2f2g2!: Property (6) and the linearity of the integral then imply fg is integrable (8) If there exists m;M such that 00 such that m jf(x)j M for all x2[a;b] Note that 1 f(x) 1 f(y) = f(y)f(x) f

Searches related to if f and g are integrable then fg is integrable filetype:pdf

Theorem 1 If f and g are integrable on [a;b] then the product fg is integrable on [a;b] Proof First notice that fg = 1 2 (f +g)2 f2 g2: So it is enough to prove that if f is integrable then f2 is integrable If f is integrable then jfj is integrable Also f2 = jfj2 Let M = supjfj over [a;b] Suppose that P is a partition of [a;b] and I

Is the composite function f g integrable?

For the composite function f ? g, He presented three cases: 1) both f and g are Riemann integrable; 2) f is continuous and g is Riemann integrable; 3) f is Riemann integrable and g is continuous. For case 1 there is a counterexample using Riemann function. For case 2 the proof of the integrability is straight forward.

Is F integrable?

Since f ? is a probability density function, it is trivially integrable, so by the dominated convergence theorem, ? S g n + d ? ? 0 as n ? ?. But ? R g n d ? = 0 so ? R g n + d ? = ? R g n ? d ?.

What makes a function integrable?

Assuming we're talking about Riemann integrals, in order for a function to be Riemann integrable, every sequence of Riemann sums must converge to the same limit. If you sometimes get different limits depending on the partitions of the interval you take, then the function is not integrable.

Is F[A;B]!R Riemann integrable?

A bounded function f: [a;b] !R is Riemann integrable i fis continuous almost everywhere on [a;b]. Theorem 12 (Composition Theorem). Let f : [a;b] !R be Riemann integrable and f([a;b]) [c;d].

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_dbs21.pdfusesText_27

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[PDF] Math 432 - Real Analysis II

Is the composite function f g integrable?

Is F integrable?

What makes a function integrable?

Is F[A;B]!R Riemann integrable?

Math 432 - Real Analysis II

Solutions to Homework due March 11

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

2. Show that

Solution 2.

L(f;P) =nX

Solution 2.Consider the function

1 ifx62Q:

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

Solution 3.

The second inequality is identical to

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

This is identical to our desired statement

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

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

Using the previous parts, we get that

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

Solution 4.

0 ifx62Q

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

Solution 5.

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

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

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

Show thatfis integrable and that

Next, we show thatRb

UsingL0s, we similarly get that

Thus, we get that

Solution 7.

Thus,12