[Undergraduate Texts in Mathematics] Real Mathematical Analysis || Functions of a Real Variable

3 Functions of a Real Variable

1 Differentiation

The function f : (a, b) ---+ IR is differentiable at x if

(1) lim f(t) - f(x) = L t---+x t - x

exists. By definition t this means L is a real number and for each E > 0 there exists a 8 > 0 such that if 0 < It - x I < 8 then the differential quotient above differs from L by < E. The limit L is the derivative of f at x, f' (x). In calculus language, ilx = t - x is the change in the independent variable x, while ilf = f (t) - f (x) is the resulting change in the dependent variable Y = f (x). Differentiability at x means that

. ilf f'(x) = hm -.

Llx---+O ilx

We begin by reviewing the proofs of some standard calculus facts.

t This concept of limit is slightly different from the limit of a sequence. Here t is a continuous parameter that tends to x, whereas for the sequence (an), the parameter n is an integer that grows without bound. A limit definition general enough to include both concepts is discussed in Exercise 26.

C. C. Pugh, Real Mathematical Analysis© Springer Science+Business Media New York 2002

140 Functions of a Real Variable Chapter 3

1 The Rules of Differentiation (a) Differentiability implies continuity. (b) If f and g are differentiable at x then so is f + g, the derivative

being (f + g)'(x) = f'(x) + g'(x).

( c) If f and g are differentiable at x then so is f . g, the derivative being

(f. g)' (x) = f'(x) . g(x) + f(x) . g'(x).

(d) The derivative of a constant is zero, c' = O. (e) If f and g are differentiable at x and g (x) =1= 0 then f j g is differen

tiable at x, the derivative being

(fjg)'(x) = (f'(x) . g(x) - f(x)· g'(x))jg(x)2.

(f) If f is differentiable at x and g is differentiable at y = f (x) then go f is differentiable at x, the derivative being

(g 0 f)'(x) = g'(y) . f'(x).

Proof (a) Continuity in the calculus notation amounts to the assertion that 6.f -+ 0 as 6.x -+ O. This is obvious. If the fraction I)..f j 6.x tends to a finite limit while its denominator tends to zero, then its numerator must also tend to zero.

(b) Since 6.(f + g) = 6.f + 6.g,

6.(f + g) = 6.f + 6.g -+ f'(x) + g'(x). 6.x 6.x 6.x

as 6.x -+ O. (c) Since 6.(f . g) = 6.f . g(x + 6.x) + f(x) . 6.g, continuity of g at x

implies that

_6._(f_. g_) = _6._f g(x + 6.x) + f(x)-6.g -+ f'(x)g(x) + f(x)g'(x), 6.x 6.x 6.x

as 6.x -+ O. (d) If c is a constant then 6.c = 0 and c' = O. (e) Since

g(x)6.f - f(x)6.g 6.(fjg) = g(x + 6.x)g(x) ,

the formula follows when we divide by 6.x and take the limit.

Section 1 Differentiation 141

(f) The shortest proof of the chain rule for y = f (x) is by cancellation:

flg flg fly , , - = -- -+ g (y)f (x). flx fly flx

A slight flaw is present, fly may be zero when flx is not. This is not a big problem. Differentiability of g at y implies that

f"...g , - = g (y) +cr f"...y

where cr = cr(f"...y) -+ 0 as f"...y -+ O. Define cr(O) = O. The formula

f"...g = (g'(y) + cr)f"...y

holds for all small f"...y, including f"...y = O. Continuity of f at x (which is true by (a)) implies that f"...y -+ 0 as f"...x -+ O. Thus

f"...g, f"...y" - = (g (y) + cr)- -+ g (y)f (x) f"...x f"...x

as f"...x -+ O. o

2 Corollary The derivative of a polynomial ao + alx + ... + anxn exists

everywhere and equals al + 2a2x + ... + nanxn- I .

Proof Immediate from the differentiation rules. o A function f : (a, b) -+ lR. that is differentiable at each x E (a, b) is

differentiable.

3 Mean Value Theorem A continuous function f : [a, b] -+ lR. that is

differentiable on the interval (a, b) has the mean value property: there

exists a point e E (a, b) such that

Proof Let

feb) - f(a) = f'(e)(b - a).

s = feb) - f(a)

b-a

be the slope of the secant of the graph of f. See Figure 56. The function ¢ (x) = f (x) - Sx is differentiable and has the same value

bf(a) - af(b) v=-----

b-a


..... ·a~t \\Ue . sec"

• a

Figure 56 The secant line for the graph of J.

1\; ./\. · .. . · .. . · .. . · .. . .. . · . . · .. . · .. . · .. . · .. .

. . \ ~: :~. . .

..... I ............... . e .................... I • .................... .. e .................... I •

a e e b a e b a e b a e b

Figure 57 <p'(0) = O.

at a and b. Differentiability implies continuity. <p is continuous and therefore takes on maximum and minimum values. Since it has the same value at both endpoints, <p has a maximum or a minimum that occurs at a point 0 E (a, b). See Figure 57. Then <p'(0) = 0 (see Exercise 6) and J(b) - J(a) = 1'(O)(b - a).

4 Corollary If J is differentiable and 11'(x)1 :s M Jorall x E (a, b) then J satisfies the global Lipschitz condition: Jor all t, x E (a, b),

IJ(t) - J(x)1 :s Mit - xl.

In particular if l' (x) = 0 Jor all x E (a, b) then J (x) is constant.

Proof IJ(t) - J(x)1 = 11'(O)(t - x) 1 for some 0 between x and t. D

Remark The Mean Value Theorem is the most important theorem in calculus for making estimates.

It is often convenient to deal with two functions simultaneously, and for that we have the following result.

5 Ratio Mean Value Theorem Suppose that the functions J and g are continuous on an interval [a, b] and differentiable on the interval (a, b).


Then there is a () E (a, b) such that

b.fg'«()) = b.gf'«())

where b.f = feb) - f(a) and b.g = g(b) - g(a). (If g(x) == x, the Ratio Mean Value Theorem becomes the ordinary Mean Value Theorem.}

Proof If b.g i= 0, then the theorem states that for some (),

/),.f f'(e)

b.g g'«())

This ratio expression is how to remember the theorem. The whole point here is that f' and g' are evaluated at the same (). The function

(x) = b.f(g(x) - g(a)) - /),.g(f(x) - f(a))

is differentiable and its value at both endpoints a, b is O. Since is continuous it takes on a maximum and a minimum somewhere in the interval [a, b]. Since has equal values at the endpoints of the interval, it must take on a maximum or minimum at some point () E (a, b); i.e., () i= a, b. Then '«()) = 0 and b.fg'«()) = b.gf'«()) as claimed. D

6 L'Hospital's Rule If f and g are differentiable functions defined on an interval (a, b), both of which tend to 0 at b, and if the ratio of their derivatives f' (x) / g' (x) tends to a finite limit L at b then f (x) / g(x) also tends to L at b. (We assume that g(x), g'(x) i= O.}

Rough proof Let x E (a, b) tend to b. Imagine a point t E (a, b) tending to b much faster than x does. It is a kind of "advance guard" for x. Then f(t)/f(x) and g(t)/g(x) are as small as we wish, and by the Ratio Mean Value Theorem, there is a () E (x, t) such that

f(x)

g(x)

f(x) - 0 . f(x) - f(t)

g(x) - 0 g(x) - g(t)

f'«())

glee)

The latter tends to L because () is sandwiched between x and t as they tend to b. The symbol "~" means approximately equal. See Figure 58. D

Complete proof Given E > 0 we must find 8 > 0 such that if Ix - bl < 8 then If(x)/g(x) - LI < E. Since f'(x)/g'(x) tends to L at b there does exist 8 > 0 such that if x E (b - 8, b) then

I f'(x) - LI < ~. g'(x) 2


---- Ilightyear -------- I mile --I inch-

• • • • • a x e b

Figure 58 x and t escort f) toward b.

For each x E (b - 8, b) determine a point t E (b - 8, b) which is so near to b that

g(X)2E If(t)1 + Ig(t)1 < 4(lf(x)1 + Ig(x)1)

Ig(t) I < Ig(x) I 2 .

Since f(t) and get) tend to 0 as t tends to b, and since g(x) t= 0 such a t exists. It depends on x, of course. By this choice of t and the Ratio Mean Value Theorem

1

f(x) - LI = 1 f(x) _ f(x) - f(t) + f(x) - f(t) - LI g(x) g(x) g(x) - get) g(x) - get)

1

g(x)f(t) - f(x)g(t) I If'(f) I < + ---L <E - g(x)(g(x) - get»~ g'ee) ,

which completes the proofthat f(x)/ g(x) ~ L as x ~ b. o

It is clear that L'Hospital's Rule holds equally well as x tends to b or to a. It is also true that it holds when x tends to ±oo or when f and g tend to ±oo. See Exercises 7,8.

From now onfeelfree to use L'Hospital's Rule!

7 Theorem Iff is differentiable on (a, b) then its derivative function l' (x) has the intermediate value property.

Differentiability of f implies continuity of f, and so the Intermediate Value Theorem applies to f and states that f takes on all intermediate values, but this is not what Theorem 7 is about. Not at all. Theorem 7 concerns l' not f· The function l' can well be discontinuous, but nevertheless it too takes on all intermediate values. In a clear abuse of language, functions like l' possessing the intermediate value property are called Darboux continuous, even when they are discontinuous! Darboux was the first to realize how badly discontinuous a derivative function can be. Despite the fact that


f' has the intermediate value property, it can be discontinuous at almost every point of [a, b]. Strangely enough, however, f' can not be discontinuous at every point. If f is differentiable, !' must be continuous at a dense, thick set of points. See Exercise 24 and the next section for the relevant definitions.

Proof Suppose that a < Xl < X2 < band

ex = f'(XI) < Y < f'(X2) = (3.

We must find e E (XI,X2) suchthat!'(e) = Y. Choose a small h, 0 < h < X2 - Xl, and draw the secant segment a(x)

between the points (x, f (x» and (x + h, f (x + h» on the graph of f. Slide x from Xl to X2 - h continuously. This is the sliding secant method. See Figure 59.

cr(x)

a b

x x+h

Figure 59 The sliding secant.

When h is small enough, slopea(xI) ~ f'(XI) and slopea(x2 - h) ~ !'(X2). Thus

slopea(XI) < Y < slopea(x2 - h).

Continuity of f implies that for some x E (Xl, X2 - h), slopea(x) = Y. The Mean Value Theorem then gives a e E (x, x + h) such that f' (e) = Y.

o 8 Corollary The derivative of a differentiable function never has a jump discontinuity.

Proof Near a jump, a function omits intermediate values. o

Pathological examples

Non-jump discontinuities of f' may very well occur. The function


{

2 . 1 f(x) = : SIll~ if x> 0

if x :::: 0

Chapter 3

is differentiable everywhere, even at x = 0, where f' (0) = O. Its derivative function for x > 0 is

, 1 I f (x) = 2x sin - - cos -,

x x

which oscillates more and more rapidly with amplitude approximately 1 as x --+ O. Since f' (x) fr 0 as x --+ 0, f' is discontinuous at x = O. Figure 60 shows why f is differentiable at x = 0 and has f' (0) = O. Although the graph oscillates wildly at 0, it does so between the envelopes y = ±x2, and any curve between these envelopes is tangent to the x-axis at the origin. Study this example, Figure 60.

0.01

0.005 0.5

0 0

-0.005 -0.5

-1 -0.01

0 0.05 0.1 0 0.05 0.1

Figure 60 The graphs of the function y = x 2 sin(l/x) and its envelopes y = ±x2 ; and the graph of its derivative.

A similar but worse example is

{

X3/2 sin ~ if x > 0 g(x) = x

o if x :::: 0

Its derivative at x = 0 is g' (0) = 0, while at x f. 0 its derivative is

, 3 . 1 1 1 g (x) = - JX SIll - - - cos -,

2 x Jx x

which oscillates with increasing frequency and unbounded amplitude as x --+ 0 because 1/ Jx blows up at x = O. See Figure 61.


0.04 ,--------------, 15~------------,

\0 0.Q2

o o

-5 -D.02

-10

-D.04 L-___ ~ ___ ----'

o 0.05 0.1 0.05 0.1

Figure 61 The function y = x 3/ 2 sin(1/x), its envelopes y = ±x3/ 2, and its derivative.

Higher Derivatives

The derivative of 1', if it exists, is the second derivative of I,

(f')'(x) = I"(x) = lim I'(t) - 1'(x). t-4X t - x

Higher derivatives are defined inductively and written I(r) = (f(r-l))'. If I(r)(x) exists then I is rth order differentiable at x. If I(r) (x) exists for each x E (a, b) then I is rth order differentiable. If I(r)(x) exists for all r and all x then I is infinitely differentiable, or smooth. The zero-th derivative of I is I itself, I(O)(x) = I(x).

9 Theorem II I is rth order differentiable and r 2: 1 then I(r-l)(x) is a continuous function olx E (a, b).

Proof Differentiability implies continuity and I(r-l)(x) is differentiable. D

10 Corollary A smoothfunction is continuous. Each derivative 01 a smooth function is smooth and hence continuous.

Proof Obvious from the definition of smoothness and Theorem 9. D

Smoothness Classes

If I is differentiable and its derivative function l' (x) is a continuous function of x, then I is continuously differentiable, and we say that I is of class ct. If I is rth order differentiable and I(r) (x) is a continuous function of x, then I is continuously rth order differentiable, and we say that I is


of class cr. If f is smooth, then by the preceding corollary, it is of class cr for all finite r and we say that f is of class Coo. To round out the notation, we say that a continuous function is of class Co.

Thinking of cr as the set of functions of class cr, we have the regularity hierarchy,

CO ::J C 1 ::J •.. ::J n C' = Coo. rEN

Each inclusion Cr ::J Cr+1 is proper; there exist continuous functions that are not of class C 1, C 1 functions that are not of class C 2 , and so on. For example,

f (x) = Ix I is of class CO but not of class C 1•

f(x) = x Ixl is of class C 1 but not of class C2•

f(x) = Ixl3 is of class C2 but not of class C3

Analytic Functions

A function that can be expressed locally as a convergent power series is analytic. More precisely, the function f : (a, b) -+ lR is analytic if for each x E (a, b), there exists a power series

and a 8 > 0 such that if Ihl < 8 then the series converges and

00

f(x +h) = I:arhr. r=O

The concept of series convergence will be discussed further in Section 3 and Chapter 4. Among other things we show in Section 2 of Chapter 4 that analytic functions are smooth; and if f (x + h) = L arhr then

This gives uniqueness of the power series expression of a function: if two power series express the same function f at x then they have identical coefficients, namely f(r) (x) / r!. See Exercise 4.36 for a stronger type of uniqueness, namely the identity theorem for analytic functions.

We write C{J) for the class of analytic functions.


A Non-analytic Smooth Function

The fact that smooth functions need not be analytic is somewhat surprising; i.e., CW is a proper subset of Coo. A standard example is

I-l/X

e(x) = ~ if x> 0

if x .:::: 0

Its smoothness is left as an exercise in the use of L'Hospital's rule and induction, Exercise 14. At x = 0 the graph of e(x) is infinitely tangent to the x-axis. Every derivative e(r) (0) = O. See Figure 62.

0.4 ,------r-----.---------r----,------,

0.35

0.3

0.25

0.2

0.15

0.1

0.05

oL---"""'-----'--------'------'--------' o 0.2 0.4 0.6 0.8

Figure 62 The graph of e(x) = e-1/ x .

It follows that e(x) is not analytic. For if it were then it could be expressed near x = 0 as a convergent series e(h) = Larhr, and ar = e(r)(O)/rL Thus a r = 0 for each r, and the series converges to zero, whereas e(h) is different from zero when h > O. Although not analytic at x = 0, e(x) is

analytic elsewhere. See also Exercise 4.35.

Taylor Approximation

The rth order Taylor polynomial of an rth order differentiable function f at x is

f "(x) f(r) (x) r f(k) (x) P(h) = f(x) + f'(x)h + __ h 2 + ... + h r = " hk.

2! r! ~ k!


The coefficients f(k)(x)/ k! are constants, the variable is h. Differentiation of P with respect to h at h = 0 gives

P(O) = f(x)

pI(O) = f'ex)

11 Taylor Approximation Theorem Assume that f : (a, b) -+ lR is rth order differentiable at x. Then

(a) P approximates f to order r at x in the sense that the Taylor remainder

R(h) = f(x + h) - P(h)

is rth order flat at h = 0; i.e., R(h)/ hr -+ 0 as h -+ O. (b) The Taylor polynomial is the only polynomial of degree::::: r with this

approximation property. (c) If, in addition, f is (r + 1)st order differentiable on (a, b) then for

some 0 between x and x + h,

f (r+l) (0) R(h) = hr+l.

(r + I)!

Proof (a) The first r derivatives of R(h) exist and equal 0 at h = O. If h > 0 then repeated applications of the Mean Value Theorem give

R(h) = R(h) - 0 = R'(OI)h = (R' (Ol) - O)h = R"(02)e1h

= ... = R(r-l) (Or-l)er-2 ... 01h

where 0 < Or-I < ... < 01 < h. Thus

as h -+ 0.1f h < 0 the same is true with h < 01 < ... < Or-I < O. (b) If Q(h) is a polynomial of degree::::: r, Q =1= P, then Q - P is not

rth order flat at h = 0, so f(x + h) - Q(h) can not be rth order flat either. (c) Fix h > 0 and define

R(h) tr+1 get) = f(x + t) - pet) - hr+1 t

r+1 = R(t) - R(h) hr+1


forO::: t::: h.NotethatsinceP(t)isapolynomialofdegreer,p(r+I)(t) = o for all t, and

Also, g(O) = g'(O) = ... = g(r) (0) = 0, and g(h) = R(h) - R(h) = O. Since g = 0 at 0 and h, the Mean Value Theorem gives a tl E (0, h) such that g'(tl) = O. Since g'(O) and g'(tl) = 0, the Mean Value Theorem gives a t2 E (0, tl) such that gil (t2) = O. Continuing, we get a sequence t1 > t2 > ... > tr+1 > 0 such that g(k) (tk) = O. The (r + l)st equation, g(r+l)(tr+l) = 0, implies that

Thus, () = x + tr+ 1 makes the equation in (C) true. If h < 0 the argument is symmetric. 0

12 Corollary For each r E N, the smooth non-analytic function e(x) satisfies lim e(h)j hr = O.

h--+O

Proof Obvious from the theorem and the fact that e(r) (0) = 0 for all r. 0

The Taylor series at x of a smooth function f is the infinite Taylor polynomial

00 f(r)( ) T(h) = " x hr.

~ r! r=O

In calculus, you compute the Taylor series of functions such as sinx, arctan x, eX, etc. These functions are analytic: their Taylor series converge and express them as power series. In general, however, the Taylor series of a smooth function need not converge to the function, and in fact it may fail to converge at all. The function e (x) is an example of the first phenomenon. Its Taylor series at x = 0 converges, but gives the wrong answer. Examples of divergent and totally divergent Taylor series are indicated in Exercise 4.35.

The convergence of a Taylor series is related to how quickly the rth

derivative grows (in magnitude) as r ~ 00. In Section 6 of Chapter 4 we give necessary and sufficient conditions on the growth rate that determine whether a smooth function is analytic.


Inverse Functions

A strictly monotone continuous function f : (a, b) ---* IR bijects (a, b) onto some interval (c, d) where c = f(a), d = f(b) in the increasing case. It is a homeomorphism (a, b) ---* (c, d) and its inverse function f-I : (c, d) ---*

(a, b) is also a homeomorphism. These facts were proved in Chapter 2. Does differentiability of f imply differentiability of f- I ? If j' i= 0 the

answer is "yes." Keep in mind, however, the function f : x 1-+ x 3 • It shows that differentiability of f- I fails when j' (x) = O. For the inverse function is y 1-+ yl/3, which is non-differentiable at y = o. 13 Inverse Function Theorem in dimension one If f : (a, b) ---* (c, d) is a differentiable surjection and j' (x) is never zero then f is a homeomorphism, and its inverse is differentiable with derivative

_I I 1 (f ) (y) = f'(x)

where y =f(x).

Proof If j' is never zero then by the intermediate value property of derivatives, it is either always positive or always negative. We assume for all x that j' (x) > o. If a < s < t O. Thus fis strictly monotone. Differentiability implies continuity, so f is a homeomorphism (a, b) ---* (c, d). To check differentiability of f- I at y E (c, d), define

Then y = f(x) and ~y = f(x + ~x) - fx = ~f. Thus

~f-I f-I(y + ~y) - f-I(y) ~x 1 1

~y ~y ~y ~y/~x ~f1~x

Since f is a homeomorphism, ~x ---* 0 if and only if ~y ---* 0, so the limit of ~f-I I ~y exists and equals II j'(x). 0

If a homeomorphism f and its inverse are both of class cr, r ::: 1, then f is a Cr diffeomorphism.

14 Corollary If f : (a, b) ---* (c, d) is a homeomorphism of class cr, 1 :5 r :5 00, and f' i= 0 then f is a cr diffeomorphism.


Proof Ifr = 1,theformulaU-1)'(y) = l/J'(x) = 1/J'U-1(y)) implies that U-1)'(y) is continuous, so I is a C 1 diffeomorphism. Induction on r :::: 2 completes the proof. 0

The corollary remains true for analytic functions: the inverse of an analytic function with non-vanishing derivative is analytic. The generalization of the inverse function theorem to higher dimensions is a principal goal of Chapter 5.

A longer but more geometric proof of the one dimensional inverse function theorem can be done in two steps.

(i) A function is differentiable if and only if its graph is differentiable.

(ii) The graph of I-I is the reflection of the graph of I across the diagonal, and is thus differentiable.

See Figure 63.

feb) ~ •.••.•••••.••.•••••••••••• "

y = f(x) ~ .••••.•..•••••.•.••••

f(a)~' ••• ,

, . , . '

' . .

~:: ~ - ~- .... ~ - - - .. - . -. - .~ ... ...... . ~ -..... ~. - -: a f(a) x b f(x) feb)

Figure 63 A picture proof of the inverse function theorem in R


2 Riemann Integration

Let f : [a, b] ---+ lR be given. Intuitively, the integral of f is the area under its graph; i.e., for f 2: 0,

lb f(x) ds = area 1U

where 1U is the undergrapb of f,

1U = {(x, y) : a ~ x ~ band ° ~ y ~ f(x)}.

The precise definition involves approximation. A partition pair consists of two finite sets of points P, T C [a, b]; P = {xo, ... , xn} and T = {ti, ... , tn}, interlaced as

a = Xo ~ ti ~ Xi ~ t2 ~ X2 ~ ... ~ tn ~ Xn = b.

We assume the points xo, ... , Xn are distinct. The Riemann sum corresponding to f, P, T is

n

R(f, P, T) = L f(ti)/),.Xi = f(ti)/),.Xi + f(t2)/),.X2 + ... + f(tn)/),.xn i=i

where /),.Xi = Xi - Xi-i. The Riemann sum R is the area of rectangles which approximate the area under the graph of f. See Figure 64. Think of the points ti as "sample points." We sample the value of the function f at ti.

(tj) . . . . .. . ...

a b

Figure 64 The area ofthe strip is f(ti)/),.Xi.

Section 2 Riemann Integration 155

The mesh of the partition P is the length of the largest subinterval [Xi-I, xd.A partition with large mesh is coarse; one with small mesh is fine. In general, the finer the better. A real number I is the Riemann integral of f over [a, b] if it satisfies the following approximation condition:

V E > 0 38 > 0 such that if P, T is any partition pair then

mesh P < 8 =} 1 R - II < E

where R = R (f, P, T). If such an I exists it is unique, we denote it as

l

b f(x) dx = I = lim R(f, P, T),

a meshP---+O

and we say that f is Riemann integrable with Riemann integral I. See Exercise 26 for a formalization of this limit definition.

15 Theorem If f is Riemann integrable then it is bounded.

Proof Suppose not. Let I = J: f (x) dx. There is some 8 > 0 such that 1 R - II < 1 for all partition pairs P, T with mesh P < 8. Fix such a partition pair P = {xo, ... , x n }, T = {tl, ... , tn }. If f is unbounded on [a, b] then there is also a subinterval [Xio-I, Xio] on which it is unbounded. Choose a new set T' = {t~, ... , t~} with t[ = ti for all i =1= io and choose t[o such that

If(t:o) - f(tio)Ib.Xio > 2.

This is possible since the supremum of {If(t)1 : Xio-I ::: t ::: Xio} is 00. Let R' = R (f, P, T'). Then I R - R'I > 2, contrary to the factthat both Rand R' differ from I by < 1. 0

Let n denote the set of all functions that are Riemann integrable over [a,b],

16 Theorem (Linearity of the Integral) (a) n is a vector space and f f--+ 1: f (x) dx is a linear map n ---+ JR. (b) The constant function h(x) = k is integrable and its integral is

k(b - a).

Proof (a) Riemann sums behave naturally under linear combination:

R(f + cg, P, T) = R(f, P, T) + cR(g, P, T),

and it follows that their limits, as mesh P ---+ 0, give the expected formula

lb f(x) + cg(x) dx = lb f(x) dx + clb

g(x) dx.


(b) Every Riemann sum for the constant function hex) = k is k(b - a), so its integral equals this number too. 0

17 Theorem (Monotonicity of the Integral) If f, g E Rand f .::: g then

lb f(x) dx .::: lb g(x) dx.

Proof For each partition pair P, T, we have R (j, P, T) .::: R (g, P, T). 0

18 Corollary Iff E Rand If I .::: M then II: f(x)dxl.::: M(b - a).

Proof By Theorem 16, the constant functions ±M are integrable. By Theorem 17, -M .::: f(x) .::: M implies that

-M(b - a) .::: lb f(x) dx .::: M(b - a).

Darboux integrability

The lower sum and upper sum of a function f : [a, b] -+ [-M, M] with respect to a partition P of [a, b] are

where

n n

L(j, P) = LmiLlxi and U(j, P) = LMiLlXi i=1

mi = inf{f(t) : Xi-I'::: t .::: Xi},

Mi = sup{f(t) : Xi-I'::: t .::: X;}.

i=1

We assume f is bounded in order to be sure that mi and Mi are real numbers. Clearly

L(j, P) .::: R(j, P, T) .::: U(j, P)

for all partition pairs P, T. See Figure 65. The lower integral and upper integral of f over [a, b] are

L = sup L(j, P) and 7 = inf U(j, P). p p

P ranges over all partitions of [a, b] when we take the supremum and infimum. If the lower and upper integrals of f are equal, L = 7, then f is Darboux integrable and their common value is its Darboux integral.

19 Theorem Riemann integrability is equivalent to Darboux integrability, and when a function is integrable, its three integrals - lower, upper, and Riemann - are equal.

Section 2

lower sum

··7

a

Riemann sum

Riemann Integration

.. ~::

~

: : : :/

b

v:::~~:~:: Vr \... [/: .::::::::: ::::<k/:<::: :::: :::1:::/::: ........................... ~: .. . ......... ·····L···· .............. .

a Xi b upper sum

Figure 65 The lower sum, the Riemann sum, and the upper sum.

157


To prove Theorem 19 it is convenient to refine a partition P by adding more partition points; the partition pi refines P if pI :J P.

Suppose first that pi = P U {w} where w E (Xio-l, Xio). The lower sums for P and P' are the same except that mioAxio in L(f, P) splits into two terms in L (f, Pi). The sum of the two terms is at least as large as mio AXio. For the infimum of f over the intervals [Xio-l, w] and [w, Xio] is at least as large as mio. Similarly, U(f, Pi) ::s U(f, P). See Figure 66.

Repetition continues the pattern and we formalize it as the

Refinement Principle Refining a partition causes the lower sum to increase and the upper sum to decrease.

lower summand

. .......... -

refined lower summand

refined upper summand

Figure 66 Refinement increases L and decreases U.

. . . . . . . .

upper summand

The common refinement p* of two partitions P, pi of [a, b] is

p* = P U p'.

According to the refinement principle

L(f, P) ::s L(f, p*) ::s U(f, p*) ::s U(f, pI).

We conclude: each lower sum is less than or equal to each upper sum; the lower integral is less than or equal to the upper; and thus

A bounded function f : [a, b] -+ lR is Darboux integrable

if and only ifYE > 0 3P such that U(f, P) - L(f, P) < E. (2)

Proof of Theorem 19 Assume that f is Darboux integrable: the lower and upper integrals are equal, say their common value is I. Given E > 0 we must find 8 > 0 such that IR - II < E whenever R = R(f, P, T) is a


Riemann sum with mesh P < o. By Darboux integrability and (2) there is a partition PI of [a, b] such that

E VI - LI < -

2

where VI = V(j, PI), LI = L(j, PI). Fix 0 = E/8Mnl where nl is the number of partition points in PI. Let P be any partition with mesh P < o. (Since 0 « E, think of PI as coarse and P as much finer.) Let p* be the common refinement P U PI. By the refinement principle,

L < L* < V* < V 1_ _ _ I

where L * = L(j, p*) and V* = V(j, p*). Thus,

V* - L* < ~ 2·

Write P = {Xi} and p* = {xll for 0 :::: i :::: n and 0 :::: j :::: n*. The sums V = L Mi .D.Xi and V * = ~ M7 .D.x 7 are identical except for terms with

for some i, j. There are at most n I - 2 such terms and each is of magnitude at most Mo. Thus,

* E V - V < (nl - 2)2Mo < -. 4

Similarly, L * - L < E/4, and so

V - L = (V - V*) + (V* - L *) + (L * - L) < E.

Since I and R both belong to the interval [L, V], we see that IR - II < E.

Therefore I is Riemann integrable. Conversely, assume that I is Riemann integrable with Riemann integral

I. By Theorem 15, I is bounded. Let E > 0 be given. There exists a 0 > 0 such that for all partition pairs P, T with mesh P < 0, I R - I I < E /4 where R = R(j, P, T). Fix any such P and consider L = L(j, P),V = V(j, P). There are choices of intermediate sets T = {td, T' = (tn such that each l(ti) is so close to mi and each 1(1{) is so close to Mi that R - L < E/4 and V - R' < E/4 where R = R(j, P, T) and R' = R(j, P, T'). Since mesh P < 0, we know that IR - II < E/4 and IR' - II < E/4. Thus

V - L = (V - R') + (R' -l) + (l - R) + (R - L) < E.


Since L I, I are fixed numbers that belong to the interval [L, U] oflength E, and E is arbitrary, the E-principle implies that

L = I = I,

which proves that f is Darboux integrable and that its lower, upper, and Riemann integrals are equal. D

According to Theorem 19 and (2) we get

20 Riemann's Integrability Criterion

A bounded function is Riemann integrable if and only if 'VE > 03P such that U(f, P) - L(f, P) < E.

Example Every continuous function f : [a, b] -+ lR is Riemann integrable. (See also Corollary 22 to the Riemann-Lebesgue Theorem, below.) Since [a, b] is compact and f is continuous, f is uniformly continuous. See Theorem 44 in Chapter 2. Let E > 0 be given. Uniform continuity provides a 8 > 0 such that if It - sl < 8 then If(t) - f(s)1 < E/2(b - a). Choose any partition P with mesh P < 8. On each partition interval [Xi-I, Xi], we have M; - mi < E/(b - a). Thus

By Riemann's integrability criterion f is Riemann integrable.

Example A piecewise continuous function is continuous except at a finite number of points. A step function is constant except at a finite number of points where it is discontinuous. Clearly, a step function is a special type of piecewise continuous function. See Figure 67.

The characteristic function (or indicator function) of a set E c R X E, is 1 at points of E and 0 at points of EC

• See Figure 68. A step function is a finite sum of constants times characteristic functions

of intervals. See Figure 67. Bounded piecewise continuous functions are Riemann integrable. See Corollary 23 below. Some characteristic functions are Riemann integrable, others aren't.

Example The characteristic function of Q is not integrable on [a, b] . It is defined as XIQ(x) = 1 when X E Q and XIQ(x) = 0 when X ¢. Q. See Figure 69. Every lower sum L(XIQ' P) is 0 and every upper sum is b - a. By Riemann's criterion, X IQ is not integrable. Note that X IQ is discontinuous at every point, not merely at rational points.


Figure 67 The graphs of a piecewise continuous function and a step function.

Figure 68 The graph of a characteristic function and the region below the graph.

The fact that X IQi fails to be Riemann integrable is actually a failing of Riemann integration theory, for the function X IQi is fairly tame. Its integral ought to exist and it ought to be 0, because the undergraph is just countably many line segments of height 1, and their area ought to be 0.

Example The rational ruler function is Riemann integrable. At each rational number x = p/q, we set f(x) = l/q, while f(x) = ° when x is irrational. See Figure 70. The integral of f is 0. Note that f is discontinuous at every x E Ql and is continuous at every x E Q)C.

Example The Zeno's staircase function Z (x) = 1/2 on the first half of [0, 1], Z(x) = 3/4 on the next quarter of [0, 1], and so on. See Figure 71. It is Riemann integrable and its integral is 2/3. The function has infinitely


Figure 69 The graph of X <Q and the region below its graph.

o. .1 .2 .33.4 .5 .6 .7 .75 .9 I.

Figure 70 The graph of the rational ruler function and the region below its graph.

many discontinuity points, one at each point (2k - I) 12k. In fact, every monotone function is Riemann integrable. t See Corollary 24 below.

t To prove this directly is not hard. See also Corollary 24 below. The key observation to make is that a monotone function is not much different than a continuous function. It has only jump discontinuities,


Figure 71 Zeno's staircase.

These examples raise a natural question:

Exactly which functions are Riemann integrable?

To give an answer to the question, and for many other applications, the following concept is very useful. A set Z C lR is a zero set if for each E > 0 there is a countable covering of Z by open intervals (a;, b;) such that

00

Lb; -a;::::: E.

;=1

The sum of the series is the total length of the covering. Think of zero sets as negligible; if a property holds for all points except those in a zero set then one says that the property holds almost everywhere, abbreviated "a.e."

21 Riemann-Lebesgue Theorem Afunction f : [a, b] --+ lR is Riemann integrable if and only if it is bounded and its set of discontinuity points is a zero set.

The set D of discontinuity points is exactly what its name implies,

D = {x E [a,b]: fisdiscontinuousatthepointx}.

A function whose set of discontinuity points is a zero set is continuous almost everywhere. The Riemann-Lebesgue theorem states that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere.

Examples of zero sets are

and only countably many of them; given any E > 0, there are only finitely many at which the jump is ~ E. See Exercise 2.30.


(a) Any subset of a zero set. (b) Any finite set. (c) Any countable union of zero sets. (d) Any countable set. (e) The middle-thirds Cantor set.

Chapter 3

(a) is clear. For if Zo C Z where Z is a zero set, and if E > 0 is given, then there is some open covering of Z by intervals whose total length is :::: E; but the same collection of intervals covers Zo, which shows that Zo is also a zero set.

(b) Let Z = {Zl, ... , Zn} be a finite set and let E > 0 be given. The intervals (Zi - E/2n, Zi + E/2n), for i = 1, ... , n, cover Z and have total length E. Therefore Z is a zero set. In particular, any single point is a zero set.

(c) This is a typical "E"/2n -argument." Let Zl, Z2, ... be a sequence of zero sets and Z = UZj. We claim that Z is a zero set. Let E > 0 be given. The set Z 1 can be covered by countably many intervals (ail, bil ) with total length I)bi1 - ail) :::: E/2. The set Z2 can be covered by countably many intervals (ai2, bi2 ) with total length 'I)bi2 - ai2) :::: E/4. In general, the set Z j can be covered by countably many intervals (ai), bi}) with total length

00

"(b·· - a··) < ~. ~ I) I) - 2) i=1

Since the countable union of countable sets is countable, the collection of all the intervals (ai), bi}) is a countable covering of Z by open intervals, and the total length of all these intervals is

00

< L:;j j=l

Thus Z is a zero set and (c) is proved. (d) This is implied by (b) and (c).

E E E - + - + - + ... = E. 248

(e) Let E > 0 be given and choose n E N such that 2n j3n < E. The middle-thirds Cantor set C is contained inside 2n closed intervals of length 1 /3 n , say It, ... , I2". Enlarge each closed interval Ii to an open interval (ai, bi) ::J Ii such that bi - ai = E/2n. (Since 1/3n < E/2n, and Ii has length 1/3n , this is possible.) The total length of these 2n intervals (ai, bi) is E. Thus C is a zero set.

In the proof of the Riemann-Lebesgue Theorem, it is useful to focus on the "size" of a discontinuity. A simple expression for this size is the


oscillation of J at x,

oscxCf) = lim sup J(t) - lim inf J(t). t--+x t--+x

Equivalently,

oscx(f) = lim diam J([x - r, x + r]) r--+O

(Of course, r > 0.) It is clear that J is continuous at x if and only if oSCx (f) = O. It is also clear that if I is any interval containing x in its interior then

M[ - m[ :::: oscxCf)

where M[ and m[ are the supremum and infimum of J(t) as t varies in I. See Figure 72.

:x ................. ,. ............ - - .. _.-

Figure 72 The partition intervals Ii with large oscillation have i E fl. These are "bad" intervals.

Proof of the Riemann-Lebesgue Theorem The set D of discontinuity points of J : [a, b] -+ [-M, M] naturally filters itself as the countable union

00

where DK = {x E [a, b] : oscxCf) :::: K}.

and K = 1/ k. According to ( a), (c) above, D is a zero set if and only if each Dl/ k is a zero set.

Assume that J is Riemann integrable and let E, K > 0 be given. By Theorem 19 there is a partition P such that

U - L = L(Mi - mi)L'.ui < EK.


Any partition interval Ii = [Xi-I, xil that contains a point of DK in its interior has Mi - mi ~ K. Since Z)Mi - mi)tJ..xi < EK, there can not be too many such intervals. (This is the key step in the estimates.) More precisely, if we sum over the i's such that Ii contains a point of DK in its interior then

Except for the zero set of points which lie at partition points, DK is contained in finitely many open intervals whose total length is < E. Since E is arbitrary, each DK is a zero set, K = 1, 1/2, 1/3, .... By (c), D is a zero set.

Conversely, assume that the discontinuity set D of f : [a, b] ---+ [-M, M] is a zero set. Let E > 0 be given. By Riemann's integrability criterion, to prove that f is Riemann integrable it suffices to find L = L(f, P) and U = U(f, P) such that U - L < E. Choose K > 0 so that

E K<---

2(b-a)

By (a), DK C D is a zero set, so there is a countable covering of DK by open intervals Jj = (aj, bj ) with total length :s E/4M. Also, for each X E [a, b] \ DK there is an open interval Ix containing X such that

sup{f(t) : t E Ix} - inf{f(t) : t E Ix} < K.

Consider the collection V of open intervals Jj and Ix such that j E N and x E [a, b] \ DK • It is an open covering of [a, b]. Compactness of [a, b] implies that V has a Lebesgue number A > O.

Let P be any partition of [a, b] having mesh P < A. We claim that U(f, P) - L(f, P) < E. Each partition interval Ii is contained wholly in some Ix or wholly in some Jj . (This is what Lebesgue numbers are good for.) Set

J = {i : h is contained in some Jj}.

See Figure 73. For some finite m, JI U ... U Jm contains those partition


small oscillation on fu:;

xo=a

Ih;with iEI D.X; with i EJ

Figure 73 The partition intervals Ii with large oscillation have i E J. These are "bad" intervals.

intervals Ii with i E J. Also, {l, ... , n} = I U J. Then

n

U-L L(Mi - m;)6.xi i=1

< L(Mi - m;)6.xi + L(Mi - mi)6.xi iEJ iO

< L2M6.Xi + LK6.xi iEJ iO

m

< 2MLbj -aj + K(b -a) j=!

E E < -+- E.

2 2

For the total length of the intervals Ii contained in the intervals i j is no greater than L b j - a j . As remarked at the outset, Riemann's integrability criterion then implies that f is integrable. 0

The Riemann-Lebesgue Theorem has many consequences, ten of which we list as corollaries.

22 Corollary Every continuous function is Riemann integrable, and so is every bounded piecewise continuous function.


Proof The discontinuity set of a continuous function is empty, and is therefore a zero set. The discontinuity set of a piecewise continuous function is finite, and is therefore also a zero set. A continuous function defined on a compact interval [a, b] is bounded. The piecewise continuous function was assumed to be bounded. By the Riemann-Lebesgue Theorem, both these functions are Riemann integrable. 0

23 Corollary The characteristic function of S C [a, b] is Riemann integrable if and only if the boundary of S is a zero set.

Proof as is the discontinuity set of Xs. See also Exercise 5.44. 0

24 Corollary Every monotone function is Riemann integrable.

Proof The set of discontinuities of a monotone function f : [a, b] --+ lR is countable and therefore is a zero set. (See Exercise 30 in Chapter 2.) Since f is monotone, its values lie in the interval between f(a) and feb), so f is bounded. By the Riemann-Lebesgue Theorem, f is Riemann integrable.

o

25 Corollary The product of Riemann integrable functions is Riemann integrable.

Proof Let f, g E R be given. They are bounded and their product is bounded. By the Riemann-Lebesgue Theorem their discontinuity sets, DC!) and D(g), are zero sets, and DC!) U D(g) contains the discontinuity set of the product f . g. Since the union of two zero sets is a zero set, the Riemann-Lebesgue Theorem implies that f . g is Riemann integrable. 0

26 Corollary Iff: [a,b] --+ [c,d]isRiemannintegrableand¢: [c,d]--+ lR is continuous, then the composite ¢ 0 f is Riemann integrable.

Proof The discontinuity set of ¢ 0 f is contained in the discontinuity set of f, and therefore is a zero set. Since ¢ is continuous and [c, d] is compact, ¢ 0 f is bounded. By the Riemann-Lebesgue Theorem, ¢ 0 f is Riemann integrable. D

27 Corollary If fER then I fiE R.

Proof The function ¢ : y 1--+ I y I is continuous, so x 1--+ I f (x) I is Riemann integrable according to Corollary 26. D


28 Corollary If a < c < band f : [a, b] -+ lR is Riemann integrable then its restrictions to [a , c], [c, b] are Riemann integrable, and

lb f(x) dx = lc

f(x) dx + lb f(x) dx.

Conversely, Riemann integrability on [a, c] and [c, b] implies Riemann integrability on [a , b].

Proof See Figure 74. The union of the discontinuity sets for the restrictions of f to the subintervals [a, c], [c, b] is the discontinuity set of f. The latter is a zero set if and only if the former two are, and so by the RiemannLebesgue Theorem, f is Riemann integrable if and only if its restrictions to [a , c], [c, b] are.

Let X [a, c] , X [c,b] be the characteristic functions of [a, c], [c, b]. By Corollary 22 they are integrable, and by Corollary 25, so are the products X [a ,c]' f and X [c,b] . f. Since

f = X[a ,c] • f + X(c,b] . f,

the addition formula follows from linearity of the integral, Theorem 16. 0

a c b

Figure 74 Additivity of the integral is equivalent to additivity of area.

29 Corollary If f : [a, b] -+ [0, M] is Riemann integrable and has integral zero then f(x) = 0 at every continuity point x of f. That is, f(x) = 0 almost everywhere.

Proof Suppose not: let Xo be a continuity point of f and assume that f(xo) > O. Then for some 8 > 0 and each x E (xo - 8, Xo + 8), f(x) ~ f(xo)/2. The function

(

f(xo) if x E (xo - 8, Xo + 8) g(x) = 0 2

otherwise


satisfies 0 ::S g(x) ::S f(x) everywhere. See Figure 75. By monotonicity of the integral, Theorem 17,

f(xo)8 = lb g(x) dx ::S lb f(x) dx = 0,

a contradiction. Hence f (x) = 0 at every continuity point. o

~\

~'I>'V

.. ~ ____ ~~ ____ gr_ap_h~g • • /' f\.." _ _ _

a b

Figure 75 The shaded rectangle prevents the integral of f being zero.

Corollary 26 and Exercises 34, 36, 46, 48 deal with the way that Riemann integrability behaves under composition. If fEn and ifJ is continuous then ifJ 0 fEn, although the composition in the other order, f 0 ifJ, may fail to be integrable. Continuity is too weak a hypothesis for such a "change of variable." See Exercise 36. However, we have the following result.

30 Corollary Iff is Riemann integrable and 1/1 is a bijection whose inverse satisfies a Lipschitz condition then f 0 1/1 is Riemann integrable.

Proof More precisely, we assume that f : [a, b] --+ lR is Riemann integrable, 1/1 bijects [c, d] onto [a, b], 1/I(c) = a, 1/I(d) = b, and for some constant K and all s, t E [a, b],

/1/I-I(s) _1/I-1(t)/ ::S K Is - tl.

We then assert that f 0 1/1 is a Riemann integrable function [c, d] --+ R Note that 1/1-1 is a homeomorphism. For it is a continuous bijection whose domain of definition is compact.

Let D be the set of discontinuity points of f. Then D' = 1/I-I(D) is the set of discontinuity points for f 0 1/1. Let E > 0 be given. There is an open covering of D by intervals (ai, bi) whose total length is ::S E / K. The homeomorphic intervals (a;, b;) = 1/1 -I (ai, bi) cover D' and have total length

Lb; - a; ::S L K(bi - ai) ::S E.

Therefore D' is a zero set and by the Riemann Lebesgue Theorem, f 0 1/1 is integrable. 0


31 Corollary If fER and 0/ : [c, d] -+ [a, b] is a C1 diffeomorphism then f 0 0/ is Riemann integrable.

Proof The hypothesis that 0/ is a C1 diffeomorphism means that it is a continuously differentiable bijection whose derivative is nowhere zero. We assume that o/'(t) > 0. Since 0/' is continuous and positive on [c, d] there is a constant K > ° such that for all e E [c, d], 0/' (e) ~ K. The Mean Value Theorem implies that for all u, v E [c, d], there exists a e between u and v

such that o/(u) - o/(v) = o/'(e)(u - v).

Thus, Io/(u) - o/(v)1 ~ K lu - vi.

For any s, t E [a, b], set u = 0/-1 (s), V = 0/-1 (t). Then

Is - tl ~ K io/-1(s) - 0/-1(t)i,

which is a Lipschitz condition on 0/-1 with Lipschitz constant K = K- 1.

By Corollary 30, f 0 0/ is Riemann integrable. D

Versions of the preceding theorem and corollary remain true without the hypotheses that 0/ bijects. The proofs are harder because 0/ can fold infinitely often. See Exercise 39.

In calculus you learn that the derivative of the integral is the integrand. This we now prove.

32 Fundamental Theorem of Calculus If f : [a, b] -+ lR. is Riemann integrable then its indefinite integral

F(x) = l x

f(t) dt

is a continuous function of x. The derivative of F (x) exists and equals f (x) at all points x at which f is continuous.

Proof #1 Obvious from Figure 76.

Proof #2 Since f is Riemann integrable, it is bounded; say I f (x) IsM for all x. By Corollary 28

IF(y) - F(x)1 = 11Y

f(t) dtl s M Iy - xl·

Therefore, F is continuous: given E > 0, choose 8 < E / M, and observe that Iy - xl < 8 implies that IF(y) - F(x)1 < M8 < E. In exactly the same way, if f is continuous at x then


F(x)

a b

Figure 76 Why does this picture give a proof of the Fundamental Theorem of Calculus?

F(x + h) - F(x) IjX+h ----- = - f(t) dt ---+ f(x)

h h x

as h ---+ 0. For if

then

m(x, h) = inf{f(s) : Is - xl s Ihl}

M(x, h) = sup{f(s) : Is - xl s Ihl}

l1 X

+h l1 x

+h

m(x, h) = - m(x, h) dt s - f(t) dt h x h x

l1 x+

h

S h x M(x, h) dt = M(x, h).

When f is continuous at x, m(x, h) and M(x, h) converge to f(x) as h ---+ 0, and so must the integral sandwiched between them,

IjX+h - f(t) dt. h x

(If h < ° then k J:+h f(t) dt is interpreted as -k J:+h f(t) dt.) 0

33 Corollary The derivative of an indefinite Riemann integral exists almost everywhere and equals the integrand almost everywhere.


Proof Assume that f : [a, b) ~ lR is Riemann integrable and F(x) is its indefinite integral. By the Riemann-Lebesgue Theorem, f is continuous almost everywhere, and by the Fundamental Theorem of Calculus, F' (x)

exists and equals f (x) wherever f is continuous. 0

A second version of the Fundamental Theorem of Calculus concerns antiderivatives. If one function is the derivative of another, the second function is an antiderivative of the first.

Note When G is an anti derivative of of g : [a, b) -+ JR, we have

G'(x) = g(x)

for all x E [a, b], not merely for almost all x E [a, b).

34 Corollary Every continuous function has an antiderivative.

Proof Assume that f : [a, b) -+ JR is continuous. By the Fundamental Theorem of Calculus, the indefinite integral F (x) has a derivative everywhere, and F' (x) = f (x) everywhere. 0

Some discontinuous functions have an anti derivative and others don't. Surprisingly, the wildly oscillating function

f(x) = to. rr Slll

x

if x :::: ° if x > °

has an antiderivative, but the jump function

g(x)

does not. See Exercise 42.

--1°1 if x :::: ° if x > °

35 Antiderivative Theorem An antiderivative of a Riemann integrable junction, if it exists, differs from the indefinite integral by a constant.

Proof We assume that f : [a, b) -+ JR is Riemann integrable, that G is an antiderivative of f, and we assert that for all x E [a, b),

G(x) = l x

f(t) dt + C,


where C is a constant. (In fact, C = G(a).) Partition [a, x] as

a = Xo < Xl < ... < Xn = X,

and choose tk E [Xk-l, Xk] such that

Such a tk exists by the Mean Value Theorem applied to the differentiable function G. Telescoping gives

n n

G(X) - G(a) = L G(Xk) - G(Xk-l) = L ICtk)/:!;.Xb k=l k=l

which is a Riemann sum for I on the interval [a, x]. Since I is Riemann integrable, the Riemann sum converges to F (x) as the mesh of the partition tends to zero. This gives G(x) - G(a) = F(x) as claimed. 0

36 Corollary Standard integrallormulas, such as

are valid.

Proof Every integral formula is actually a derivative formula, and the Antiderivative Theorem converts derivative formulas to integral formulas. 0

In particular, the logarithm function is defined as the integral,

Ix 1

logx = - dt. 1 t

Since the integrand 1/ t is well defined and continuous when t > 0, log x is well defined and differentiable for x > 0. Its derivative is 1/ x. By the way, as is standard in post calculus vocabulary, log x refers to the natural logarithm, not to the base 10 logarithm. See also Exercise 16.

An antiderivative of I has G' (x) = I (x) everywhere, and differs from the indefinite integral F(x) by a constant. But what if we assume instead that H'(x) = I(x) almost everywhere? Should this not also imply H(x) differs from F (x) by a constant? Surprisingly, the answer is "no."

37 Theorem There exists a continuous junction H : [0, 1] -+ lR whose derivative exists and equals zero almost everywhere, but which is not constant.


Proof. The counter-example is the Devil's staircase function, also called the Cantor function. It is defined as

H(x)

1/2 if x E [1/3,2/3]

1/4 if x E [1/9,2/9]

3/4 if x E [7/9,8/9]

See Figures 77, 78.

I .............. ~ ___ ~ :2

I .... ~_ 4

1 2 I 99:3

2 7 8 :3 () ()

Figure 77 The devil's staircase.

On each discarded interval in the Cantor set construction, H is constant. Thus H is differentiable with derivative zero at all points of [0, 1] \ C, and since C is a zero set, this implies H' (x) = 0 for almost every x.

To show that H is continuous we use base 2 and base 3 arithmetic. If x E [0, 1] has base 3 expansion x = (.XIX2 ..• )3 then the base 2 expansion of y = H(x) is (.YIY2 •.. h where

Yi = {~ Xi

2

if 3k < i such that Xk = 1

if Xi = 1 and jJk < i such that Xk = 1

if Xi = 0 or Xi = 2 and jJk < i such that Xk = 1.

You should ask: is this a valid definition of H (x)? If two different base 3 expansions, (.XIX2 ... h and (.x;x~ ... h represent the same number x, do the two base 2 expressions for H(x) represent the same number y? Two


Figure 78 The devil's undergraph.

base 3 expansions of x represent the same number x if and only if x is an endpoint of C; one of its base 3 expansions ends in 0' s, the other in 2' s. For example,

(.XlX2 ... xe02h = x = (.XlX2 ... xe10h.

If for some (smallest) k :5 e, Xk = 1 then

(Xl X2 Xk-l-)

(·YlY2 ... h = '22'" -2- 10 2

unambiguously. If none of Xk with k :5 e equals 1 then the two base 2 expansions corresponding to H (x) are

.-- ... -01 (Xl X2 XC-I-) 2 2 2 2

(.~ ~2 ... Xl; I 10)2'

These two base 2 expansions represent the same number y. The same reasoning applies to the only other ambiguous case,

(.XlX2 ... Xc 12h = X = (,XIX2 ... xe20h.

The point Y = H (x) is well defined. Continuity is now easy to check. Let E > 0 be given. Choose k such

that Ij2k :5 E. If Ix - x'I < Ij3k then there are base 3 expansions of x, x'


whose first k symbols agree. Therefore the first k symbols of H (x) and H (x') agree, which implies that

IH(x) - H(x')1 .:::: 2k~1 < E. D

A yet more pathological example is a strictly monotone, continuous function 1 whose derivative is almost everywhere zero. Its graph is a sort of Devil's ski slope, almost everywhere level but also every~here downhill. To cons~ct l, start with H and extend it to a function H : lR ---+ lR by setting H(x + n) = H(x) + n for all n E Z and all x E [0,1]. Then set

00 if(3kx) lex) = L 4k .

k=O

The values of if (3 k x) for x E [0, 1] are .:::: 3k , which is much smaller than the denominator 4k. Thus the series converges and lex) is well defined. According to the Weierstrass M -test, proved in the next chapter, 1 is continuous. Since if (3k x) strictly increases for any pair of points at distance > 1j3k apart, and this fact is preserved when we take sums, 1 strictly increases. The proof that l'(x) = 0 almost everywhere requires more and deeper theory.

Next, we justify two common methods of integration.

38 Integration by substitution If fER and g : [c, d] ---+ [a, b] is a continuously differentiable bijection with g' > 0 (g is a C l diffeomorphism) then lb f(y)dy = ld f(g(x»g'(x)dx.

Proof The first integral exists by assumption. By Corollary 31, the composite fog E R, and since g' is continuous, the second integral exists by Corollary 25. To show that the two integrals are equal we resort again to Riemann sums. Let P partition the interval [c, d] as

c = Xo < Xl < ... < Xn = d

and choose tk E [Xk-l, Xk] such that

The Mean Value Theorem ensures that such a tk exists. Since g is a diffeomorphism we have a partition Q of the interval [a, b]

a = Yo < YI < ... < Yn = b


where Yk = g(Xk), and II P II --+ 0 implies that II Q II --+ O. Set Sk = g(tk). This gives two equal Riemann sums

n n

k=! k=!

which converge to the integrals f: fey) dy and t f(g(t)g'(t) dt as II P II --+ O. Since the limits of equals are equal, the integrals are equal. 0

Actually, it is sufficient to assume that g' E R.

39 Integration by Parts Iff, g : [a, b] --+ rrf. are differentiable and f', g' E

Rthen

ib f(x)g'(x) dx = f(b)g(b) - f(a)g(a) _ib f'(x)g(x) dx.

Proof Differentiability implies continuity implies integrability, so f, g E

R. Since the product of Riemann integrable functions is Riemann integrable, f' g, f g' E R, and both integrals exist. By the Leibniz Rule, (fg)'(x) = f(x)g'(x) + f'(x)g(x) everywhere. That is, fg is an antiderivative of f' g + f g'. The Antiderivative Theorem states that f g differs from the indefinite integral of f' g + f g' by a constant. That is, for all t E [a,b],

f(t)g(t) - f(a)g(a) = it f'(x)g(x) + f(x)g'(x) dx

= it f'(x)g(x) dx + it f(x)g'(x) dx.

Setting t = b gives the result. o

Improper Integrals

Assume that f : [a, b) --+ rrf. is Riemann integrable when restricted to any closed subinterval [a, c] C [a, b). You may imagine that f(x) has some unpleasant behavior as x --+ b, such as limsupx---+b If(x)1 = 00 and/or b = 00. See Figure 79.

If the limit of f: f(x) dx exists (and is a real number) as c --+ b then it is natural to define it as the improper Riemann integral

Ib

f(x) dx = lim iC

f(x) dx. a c---+b a

The same idea works of course on an interval to the left of a. In order that the two sided improper integral exists for a function f : (a, b) --+ rrf. it is

Section 3 Series

Figure 79 The improper integral converges if and only if the total undergraph area is finite.

179

natural to fix some point m E (a, b) and require that both improper integrals lam f (x) dx and f: f (x) dx exist. Their sum is the improper integral

1: f (x) dx. With some ingenuity you can devise a function f : ffi. -+ ffi. whose improper integral f~oo f (x) dx exists despite the fact that it is unbounded at both ±oo. See Exercise 71.

3 Series

A series is a formal sum L ak where the terms ak are real numbers. The nth

partial sum of the series is

The series converges to A if An -+ A as n -+ 00, and we write

00

A series that does not converge diverges. The basic question to ask about a series is: does it converge or diverge?


For example, if A is a constant and IAI < 1 then the geometric series

00

LAk = l+A+···+An + ... k=O

converges to 1/(1 - A). For its partial sums are

1 _An+1

An = 1 + A + A 2 + ... + An = --I-A

and A n+ 1 --+ 0 as n --+ 00. On the other hand, if I A I ~ 1, the series L A k

diverges. Let L an be a series. The Cauchy Convergence Criterion from Chapter 1

applied to its sequence of partial sums yields the CCC for series

L ak converges if and only if

'VE > 0 3N such that m, n ~ N :::} n

Lak < E.

k=m

One immediate consequence of the CCC is that no finite number of terms affects convergence of a series. Rather, it is the tail of the series, the terms ak with k large, that determines convergence or divergence. Likewise, whether the series leads off with a term of index k = 0 or k = 1, etc. is irrelevant.

A second consequence of the CCC is that if ak does not converge to zero as k --+ 00 then L ak does not converge. For Cauchyness of the partial sum sequence (An) implies that an = An - A n- 1 becomes small when n --+ 00. If I A I ~ 1 the geometric series L A k diverges since its terms do not converge to zero. The harmonic series

gives an example that a series can diverge even though its terms do tend to zero. See below.

Series theory has a large number of convergence tests. All boil down to the following result.

40 Comparison Test If a series L bk dominates a series L ak in the sense that Jor all sufficiently large k, lak I ::: bb then convergence oJL bk implies convergence oj L ak·

Section 3 Series 181

Proof Given E > 0, convergence of L bk implies there is a large N such that for all m, n ::::: N, L~=m bk < E. Thus

n n n

k=m k=m k=m

and convergence of L ak follows from the CCC. o

Example The series L sin(k)/2k converges since it is dominated by the geometric series L I 12k.

A series L ak converges absolutely if L lak I converges. The comparison test shows that absolute convergence implies convergence. A series that converges but not absolutely converges conditionally: L ak converges and L lak I diverges. See below.

Series and integrals are both infinite sums. You can imagine a series as an improper integral in which the integration variable is an integer,

More precisely, given a series L ak, define f : [0, (0) -+ ffi. by setting

f(x) = ak if k - 1 < x ::: k.

See Figure 80. Then

00 100

Lak = f(x)dx. k=O 0

The series converges if and only if the improper integral does. The natural extension of this picture is the

41 Integral Test Suppose that Jooo f (x) dx is a given improper integral and L ak is a given series.

(a) If lakl ::: f(x) for all sufficiently large k and all x E (k - 1, k] then convergence of the improper integral implies convergence of the series.

(b) Iflf(x)1 ::: akforall sufficiently large k and all x E [k, k + 1) then divergence of the improper integral implies divergence of the series.


() 2 4 5 k-l k k+l

o 2 4 5 k k+1 k+2

Figure 80 The pictorial proof of the integral test.

Proof See Figure 80. (a) For some large No and all N ::: No we have

t lakl:s iN f(x) dx :s 100

f(x) dx, k=No+l No 0

which is a finite real number. An increasing, bounded sequence converges to a limit, so the tail of the series L lak I converges, and the whole series L lakl converges. Absolute convergence implies convergence.

The proof of (b) is left as Exercise 73. 0

Example The p-series, L 1/ kP converges when p > 1 and diverges when p:sl. Case 1. p > 1. By the fundamental theorem of calculus and differentiation rules

lb 1 b1- p - 1 1 -dx = -+--

I x P 1- p p - 1 as b -+ 00. The improper integral converges and dominates the p-series, which implies convergence of the series by the integral test. Case 2. p :s 1. The p-series dominates the improper integral

lb 1 {10g b -dx = b1- p -l

1 x P I-p

if p = 1

ifp<l.

As b -+ 00, these quantities blow up, and the integral test implies divergence of the series. When p = 1 we have the harmonic series, which we have just shown to diverge.


The exponential growth rate of the series L ak is

ex =limsup~. k---?oo

42 Root Test Let ex be the exponential growth rate of a series L ak· If ex < 1 the series converges, if ex > 1 the series diverges, and if ex = 1 the root test is inconclusive.

Proof If ex < 1, fix a constant fJ,

ex<fJ<1.

Then for all large k, lad/ k :::: fJ; i.e.,

which gives convergence of L ak by comparison to the geometric series

LfJk•

If ex > 1, choose fJ, 1 < fJ < ex. Then lakl ?: fJk for infinitely many k. Since the terms ak do not converge to 0, the series diverges.

To show that the root test is inconclusive when ex = 1, we must find two series, one convergent and the other divergent, both having exponential growth rate ex = 1. The examples are p-series. We have

( 1 )l/k -plog(k) -plog(x) -pix

log kp = k '" x '" -1 - '" ° byL'Hospital'srule as k = x --+ 00. Therefore ex = limk---?oo(ll kP)l/k = 1. Since the square series L 1 I k2 converges and the harmonic series L 1 I k diverges the root test is inconclusive when ex = 1. 0

43 Ratio Test Let the ratio between successive terms of the series L ak be rk = lak+1/akl, and set

liminf rk = A k---?oo

lim sup rk = p. k---?oo

If P < 1 the series converges, if A > 1 the series diverges, and otherwise the ratio test is inconclusive.


Proof If p < 1, choose,8, p < ,8 < 1. For all k ::: some K, I ak+I/ak I < ,8; i.e.,

lakl ~ ,8k-K laKI = C,8k

where C ,8-K laK I is a constant. Convergence of L ak follows from comparison with the geometric series L C,8k . If A > 1, choose ,8, 1 < ,8 < A. Then lak I ::: ,8k / C for all large k, and L ak diverges because its terms do not converge to O. Again the p-series all have ratio limit p = A = 1 and demonstrate the inconclusiveness of the ratio test when p = 1 or A = 1. 0

Although it is usually easier to apply the ratio test than the root test, the latter has a strictly wider scope. See Exercises 56, 60.

Conditional Convergence

If (ak) is a decreasing sequence in lR that converges to 0 then its alternating series

converges. For,

A2n = (al - a2) + (a3 - a4) + ... (a2n-l - a2n).

and ak-l - ak is the length of the the interval h = (ab ak-l). The intervals Ik are disjoint, so the sum of their lengths is at most the length of (0, ao), namely ao. See Figure 81.

• o

Figure 81 The pictorial proof of alternating convergence.

The sequence (A2n ) is increasing and bounded, so limn-+oo A 2n exists. The partial sum A 2n+1 differs from A2n by a2n+l, a quantity that converges to 0 as n ~ 00, so

and the alternating series converges. When ak = 1/ k we have the alternating harmonic series,

00 (_I)k+l 1 1 1 L k =1-2:+3"-4+··· k=l

which we have just shown is convergent.


Series of Functions

A series of functions is of the form

where each term fk (a, b) -+ lR is a function. For example, in a power series

the functions are monomials Ckxk. (The coefficients Ck are constants and x

is a real variable.) If you think of A = x as a variable, then the geometric series is a power series whose coefficients are 1, L xk. Another example of a series of functions is a Fourier series

L ak sin(kx) + bk cos(kx).

44 Radius of Convergence Theorem If L Ckxk is a power series then there is a unique R, 0 ::: R ::: 00, its radius of convergence, such that the series converges whenever Ix I < R, and diverges whenever Ix I > R. Moreover R is given by the formula

1 R - -----== -limsup~'

k-+oo

Proof Apply the root test to the series L Ckxk. Then

. k~1 kl . k;-;-:::-;- Ixl hmsupy ICk X " I = Ixlhmsupv ICkl =-. k-+oo k-+oo R

If Ix I < R the root test gives convergence. If Ix I > R it gives divergence. D

There are power series with any given radius of convergence, 0 ::: R ::: 00. The series L kk xk has R = O. The series L xk / (}k has R = () for o < () < 00. The series L Xk / k! has R = 00. Eventually, we show that a function defined by a power series is analytic: it has all derivatives at all points and it can be expanded as a Taylor series at each point inside its radius of convergence, not merely at x = O. See Section 6 in Chapter 4.


Exercises

1. Assume that f : lR ---+ lR satisfies If(t) - f(x)1 ::: It - xl2 for all t, x. Prove that f is constant.

2. A function f : (a, b) ---+ lR satisfies a HOlder condition of order a if a > 0, and for some constant H and all u, x E (a, b),

If(u) - f(x)1 ::: H lu - xla .

The function is said to be a-HOlder, with a-HOlder constant H. (The terms "Lipschitz function of order a" and "a-Lipschitz function" are sometimes used with the same meaning.)

(a) Prove that an a-Holder function defined on (a, b) is uniformly continuous and infer that it extends uniquely to a continuous function defined on [a, b]. Is the extended function a-HOlder?

(b) What does a-HOlder continuity mean when a = I? (c) Prove that a-Holder continuity when a > 1 implies that f is

constant. 3. Assume that f : (a, b) ---+ lR is differentiable.

(a) If f'(x) > 0 for all x, prove that f is strictly monotone increasing.

(b) If f' (x) 2: 0 for all x, what can you prove? 4. Prove that.Jn+T - In ---+ 0 as n ---+ 00.

5. Assume that f : lR ---+ lR is continuous, and for all x i= 0, f'(x) exists. If lim f'(x) = L exists, does it follow that f'(O) exists?

x---+o Prove or disprove.

6. If a differentiable function f : (a, b) ---+ lR assumes a maximum or a minimum at some B E (a, b), prove that f'(B) = O. Why is the assertion false when [a, b] replaces (a, b)?

7. In L'Hospital's Rule, replace the interval (a, b) with the half-line [a, (0) and interpret "x tends to b" as "x ---+ 00." Show that if fig tends to % and f' / g' tends to L then fig also tends to L also. Prove that this continues to hold when L = 00 in the sense that if f' / g' ---+ 00 then fig ---+ 00.

8. In L'Hospital's Rule, replace the assumption that fig tends to 0/0 with the assumption that it tends to 00/00. If f' / g' tends to L, prove that fig tends to L also. [Hint: Think of a rear guard instead of an advance guard.] [Query: Is there a way to deduce the 00/00 case from the % case? NaIvely taking reciprocals does not work.]

9. (a) Draw the graph of a continuous function defined on [0, 1] that is differentiable on the interval (0, 1) but not at the endpoints.

Exercises

(b) Can you find a formula for such a function? (c) Does the Mean Value Theorem apply to such a function?

10. Let f : (a, b) -+ JR be given. (a) If f" (x) exists, prove that

r f(x - h) - 2f(x) + f(x + h) - f"( ) h~ h2 - X .

187

(b) Find an example that this limit can exist even when f" (x) fails to exist.

11. Assume that f : (-1, 1) -+ JR and l' (0) exists. If an, fin -+ 0 as n -+ 00, define the difference quotient

Dn = f(fin) - f(an) . fin - an

(a) Prove that lim Dn = f' (0) under each of the following con-n~oo

ditions (i) an < 0 < fin.

(ii) 0 < an < fin and fin ::: M. fin - an

(iii) l' (x) exists and is continuous for all x E (-1, 1). (b) Set f(x) = x 2 sin(l/x) for x =I 0 and f(O) = O. Observe that

f is differentiable everywhere in (-1, 1) and l' (0) = O. Find an, fin that tend to 0 in such a way that Dn converges to a limit unequal to 1'(0).

12. Assume that f and g are rth order differentiable functions (a, b) -+ R r ::: 1. Prove the rth order Leibniz product rule for the function

f 'g,

(f. g)(r) (x) = t (r)f(k)(X) . g(r-k) (x). k=O k

where G) = r!/(k!(r - k)!) is the binomial coefficient, r choose k. [Hint: Induction.]

13. Assume that f : JR -+ JR is differentiable. (a) If there is an L < 1 such that for each x E JR, l' (x) < L, prove

that there exists a unique point x such that f(x) = x. [x is a fixed point for f.]

(b) Show by example that (a) fails if L = 1. 14. Define e : JR -+ JR by

e(x) = !oe- 11x if x > 0 if x ::: 0


(a) Prove that e is smooth; that is, e has derivatives of all orders at all points x. [Hint: L'Hospital and induction. Feel free to use the standard differentiation formulas about eX from calculus.]

(b) Is e analytic? (c) Show that the bump function

f3(x) = e2e(1 - x) . e(x + 1)

is smooth, identically zero outside the interval (-1, 1), positive inside the interval (-1, 1), and takes value 1 at x = O. (e2 is the square of the base of the natural logarithms, while e(x) is the function just defined. Apologies to the abused notation.)

(d) For Ixl < 1 show that

f3(x) = e-2x2 /(x2-1).

Bump functions have wide use in smooth function theory and differential topology. The graph of f3 looks like a bump. See Figure 82.

0.5

-1 --D.S o 0.5

Figure 82 The graph of the bump function f3.

**15. Let L be any closed set in R Prove that there is a smooth function f : lR. --+ [0, 1] such that f(x) = 0 if and only if x E L. To put it another way, every closed set in lR. is the zero locus of some smooth function. [Hint: Use Exercise 14(c).]

16. log x is defined to be It l/t dt for x > O. Using only the mathematics explained in this chapter,

Exercises 189

(a) Prove that log is a smooth function. (b) Prove that log(xy) = log x + log y for all x, y > O. [Hint:

Fix y and define f (x) = log(xy) - log x - log y. Show that f(x) == 0.]

(c) Prove that log is strictly monotone increasing and its range is all ofR

17. Define f(x) = x 2 if x < 0 and f(x) = x + x 2 if x ~ O. Differentiation gives f" (x) == 2. This is bogus. Why?

18. Recall that the K -oscillation set of an arbitrary function f : [a, b] ---+

lR is

DK = {x E [a, b] : oscxCf) ~ K}.

(a) Prove that DK is closed. (b) Infer that the discontinuity set of f is a countable union of

closed sets. (This is called an Fa-set.) (c) Infer from (b) that the set of continuity points is a countable

intersection of open sets. (This is called a G8-Set.) *19. Baire's Theorem (page 243) asserts that if a complete metric space

is the countable union of closed subsets then at least one of them has non-empty interior. Use Baire's Theorem to show that the set of irrational numbers is not the countable union of closed subsets of R

20. Use Exercises 18 and 19 to show thatthere is no function f : lR ---+ lR which is discontinuous at every irrational number and continuous at every rational number.

**21. Find a subset S of the middle-thirds Cantor set which is never the discontinuity set of a function f : lR ---+ R Infer that some zero sets are never discontinuity sets of Riemann integrable functions. [Hint: How many subsets of C are there? How many can be countable unions of closed sets?]

* *22. Suppose that fn : [a, b] ---+ lR is a sequence of continuous functions that converges pointwise to a limit function f : [a, b] ---+ lR. Such an f is said to be of Baire class 1. (Pointwise convergence is discussed in the next chapter. It means what it says: for each x, fn (x) converges to f(x) as n ---+ 00. Continuous functions are considered to be of Baire class 0, and in general a Baire class k function is the pointwise limit of a sequence of Baire class k - 1 functions. Strictly speaking, it should not be ofBaire class k -1 itself, but for simplicity I include continuous functions among Baire class 1 functions. It is an interesting fact that for every k there are Baire class k functions not of Baire class k - 1. You might consult A Primer of Real Functions by Ralph Boas.)


Prove that the K-oscillation set of f is nowhere dense, as follows. To arrive at a contradiction, suppose that DK is dense in some interval (a, f3) c [a, b]. By Exercise 18, DK is closed, so it contains (a, f3). Cover lR by countably many intervals (ae, be) oflength < K and set

He = fpre(ae, be).

(a) Why does UeHe = [a, b]? (b) Show that no He contains a subinterval of (a, f3). (c) Why are

1 Femn = {x E [a, b] : ae +

closed? (d) Show that

m

He = U E£mN.

m,NEN

1 < be - -}

m

(e) Use (a) and Baire's Theorem (page 243) to deduce that some E£mN contains a subinterval of (a, f3).

(1) Why does (e) contradict (b) and complete the proof that DK is nowhere dense?

23. Combine Exercises 18,22, and Baire's Theorem to show that a Baire class 1 function has a dense set of continuity points.

24. Suppose that g : [a, b] ---+ lR is differentiable. (a) Prove that g' is of Baire class 1. [Hint: Extend g to a differen

tiable function defined on a larger interval and consider

fn(x) = g(x + lin) - g(x) lin

for x E [a, b]. Is fn (x) continuous? Does fn (x) converge pointwise to g'(x) as n ---+ oo?]

(b) Infer from Exercise 23 that a derivative can not be everywhere discontinuous. It must be continuous on a dense subset of its domain of definition.

25. Consider the characteristic functions f(x) and g(x) of the intervals [l, 4] and [2, 5]. The derivatives f', g' exist almost everywhere. The integration by parts formula says that

13

f(x)g'(x) dx = f(3)g(3) - f(O)g(O) -13

f'(x)g(x) dx.

Exercises 191

But both integrals are zero, while f(3)g(3) - f(O)g(O) = 1. Where is the error?

26. Let Q be a set with a transitive relation ::S. It satisfies the conditions that for all WI, W2, W3 E Q ,WI ::S WI, and if WI ::S W2 ::S W3 then WI ::S W3. A function f : Q -+ JR converges to a limit L with respect to Q if, given any E > 0 there is an Wo E Q such that Wo ::S W

implies If(w) - LI < E. We write limn few) = L to indicate this convergence. Observe that

• When fen) = an and N is given its standard order relation :S, limn- Hx) an means the same thing as limN fen).

• When JR is given its standard order relation :S, limHoo f(t) means the same thing as limlR f(t).

• Fix an x E JR and give JR the new relation tl ::S t2 when It2 - xl :s It I - xl. Then limHx f(t) means the same thing as lim(lR,:o) f(t).

(a) Prove that limits are unique: iflimn f = Ll and limn f = L2 then Ll = L 2.

(b) Prove that existence of limn f and limn g imply that

lim(f + cg) = lim f + climg n n n lim(f . g) = lim f . lim g n n n lim (f / g) = lim fllim g n n n

where c is a constant and, in the quotient rule, limn g =1= 0, g =1= O.

(c) Let Q consist of all partition pairs (P, T); define (P, T) ::S (Pi, T') when pi is finer than P, mesh pi :s mesh P. Observe that ::S is transitive and that limn R(f, P, T) = I means the same as limmesh p---+o R (f, P, T) in the definition of the Riemann integral.

(d) Review the proof of Theorem 16 and use (b) to justify the fact that linearity of Riemann sums with respect to the integrand,

R(f + cg, P, T) = R(f, P, T) + cR(g, P, T),

actually does imply linearity of the integral with respect to the integrand.

(e) Formulate this limit definition for functions from Q to a general metric space in place of JR.


27. Redefine Riemann and Darboux integrability using dyadic partitions. (a) Prove that the integrals are unaffected. (b) Infer that Riemann's integrability criterion can be restated in

terms of dyadic partitions. (c) Repeat the analysis using partitions into subintervals of length

(b - a)/n. 28. In many calculus books, the definition of the integral is given as

~ b-a lim ~ f(x:)--

n---+oo n k=l

where xt is the midpoint of the interval

[a + (k - l)(b - a)/n, a + k(b - a)/n].

See Stewart's Calculus with Early Transcendentals, for example. (a) If f is continuous, show that the calculus book limit exists and

equals the Riemann integral of f. [Hint: This is a one-liner.] (b) Show by example that the calculus style limit can exist for

functions which are not Riemann integrable. (c) Infer that the calculus style definition of the integral is inade

quate for real analysis. 29. Suppose that Z c lR. Prove that the following are equivalent.

(i) Z is a zero set. (ii) For each E > ° there is a countable covering of Z by closed

intervals [ai, b;] with total length L bi - ai < E.

(iii) For each E > ° there is a countable covering of Z by sets Si with total diameter L diam Si < E.

30. Prove that the interval [0, 1] is not a zero set. [Hint: Be careful; this is not entirely trivial.]

31. The standard middle-quarters Cantor set is formed by removing the middle quarter from [0, 1], then removing the middle quarter from each of the remaining two intervals, then removing the middle quarter from each of the remaining four intervals, and so on.

(a) Prove that it is a zero set. (b) Formulate the natural definition of the middle f3 -Cantor set. (c) Is it also a zero set? Prove or disprove.

*32. Define a Cantor set by removing from [0, 1] the middle interval of length 1/4. From the remaining two intervals Fl remove the middle intervals oflength 1/16. From the remaining four intervals F2 remove the middle intervals oflength 1/64, and so on. At the nth step in the construction F n consists of 2n subintervals of F n

-1•

Exercises 193

(a) Prove that F = nFn is a Cantor set but not a zero set. It is often referred to as a fat Cantor set.

(b) Infer that being a zero set is not a topological property: if two sets are homeomorphic and one is a zero set, then the other need not be a zero set.

[Hint: To get a sense of this fat Cantor set, calculate the total length of the intervals which comprise its complement. See Figure 49 and Exercise 36.]

33. Consider the characteristic function of the dyadic rational numbers, f(x) = 1 if x = kl2n for some k E Z and n E N, and f(x) = ° otherwise.

(a) What is its set of discontinuities? (b) At which points is its oscillation:::: K? (c) Is it integrable? Explain, both by the Riemann-Lebesgue The

orem and directly from the definition. (d) Consider the dyadic ruler function g(x) = 1/2n if x = kl2n

and g(x) = ° otherwise. Graph it and answer the questions posed in (a), (b), (c).

34. (a) Prove that the characteristic function f of the middle-thirds Cantor set C is Riemann integrable but the characteristic function g of the fat Cantor set F (Exercise 32) is not.

(b) Why is there a homeomorphism h : [0, 1] ---+ [0, 1] sending C onto F?

(c) Infer that the composite of Riemann integrable functions need not be Riemann integrable. How is this example related to Corollaries 26, 30 of the Riemann-Lebesgue Theorem? See also Exercise 36.

*35. Assume that ljr : [a, b] ---+ JR is continuously differentiable. A critical point of ljr is an x such that ljr' (x) = 0. A critical value is a number y such that for at least one critical point, y = ljr(x).

(a) Prove that the set of critical values is a zero set. (This is the Morse-Sard Theorem in dimension one.)

(b) Generalize this to continuously differentiable functions JR ---+ JR.

*36. Let F C [0, 1] be the fat Cantor set from Exercise 32, and define

ljr(x) = foX dist(t, F) dt

where dist(t, F) refers to the minimum distance from t to F.


(a) Why is 1/1 a continuously differentiable homeomorphism from [0, 1] onto [0, L] where L = 1/I(1)?

(b) What is the set of critical points of 1/1? (See Exercise 35.) (c) Why is 1/I(F) a Cantor set of zero measure? (d) Let f be the characteristic function of 1/1 (F). Why is f Riemann

integrable but f 0 1/1 not? (e) What is the relation of (d) to Exercise 34?

37. Generalizing Exercise 30 in Chapter 1, we say that f : (a, b) ---+ JR. has a jump discontinuity (or a discontinuity of the first kind) at c E (a, b) if

f(c-) = lim f(x) and f(c+) = lim f(x) x-).c- x---+c+

exist, but are either unequal or are unequal to f (c). (The three quantities exist and are equal if and only if f is continuous at c.) An oscillating discontinuity (or a discontinuity of the second kind) is any non-jump discontinuity.

(a) Show that f : JR. ---+ JR. has at most countably many jump discontinuities.

(b) Show that

{

. 1

f(x) = ~m~ if x > 0

if x ~ 0

has an oscillating discontinuity at x = o. (c) Show that the characteristic function of the rationals, XIQi, has

an oscillating discontinuity at every point. **38. Recall that peS) = 2s is the power set of S, the collection of all

subsets of S, and n is the set of Riemann integrable functions f : [a, b] ---+ R

(a) Prove that the cardinality of n is the same as the cardinality of P(JR.), which is greater than the cardinality of R

(b) Call two functions in n integrally equivalent if they differ only on a zero set. Prove that the collection of integral equivalence classes of n has the same cardinality as JR., namely 21'1.

(c) Is it better to count Riemann integrable functions or integral equivalence classes of Riemann integrable functions?

(d) Show that f, g E n are integrally equivalent if and only if the integral of If - g I is zero.

39. Suppose that 1/1 : [c, d] ---+ [a, b] is continuous and for every zero set Z C [a, b], 1/Ipre(z) is a zero set in [c, d].

Exercises 195

(a) If / is Riemann integrable, prove that / 0 1jf is Riemann integrable.

(b) Derive Corollary 30 from (a). 40. Let 1jf(x) = x sin l/x for ° < x ::::: 1 and 1jf(0) = 0.

(a) If / : [-1, 1] ---+ lR is Riemann integrable, prove that / 0 1jf is Riemann integrable.

(b) What happens for 1jf(x) = JX sin l/x? *41. Assume that 1jf : [e, d] ---+ [a, b] is continuously differentiable.

(a) Ifthe critical points of 1jf form a zero set in [e, d] and / is Riemann integrable on [a, b] prove that / 01jf is Riemann integrable on [e, d].

(b) Conversely, prove that if / 0 1jf is Riemann integrable for each Riemann integrable / on [a, b], then the critical points of 1jf form a zero set. [Hint: Think in terms of Exercise 35.]

(c) Prove (a) and (b) under the weaker assumption that 1jf is continuously differentiable except at finitely many points of [e, d].

(d) Derive part (a) of Exercise 36 from (c). (e) Weaken the assumption further to 1jf being continuously dif

ferentiable on an open subset of [e, d] whose complement is a zero set.

The following assertion, to be proved in Chapter 6, is related to the preceding exercises. If / : [a, b] ---+ lR satisfies a Lipschitz condition or is monotone then the set of points at which f' (x) fails to exist is a zero set. Thus: "a Lipschitz function is differentiable almost everywhere," which is Rademacher's Theorem in dimension one, and a "monotone function is almost everywhere differentiable," which is the last theorem in Lebesgue's book, Le~ons sur l'integration et la recherche des/onetions primitives. See Theorem 39 and Corollary 41 in Chapter 6.

42. Set

lex) = {a. rr SlO

x

if x ::::: ° if x > ° and g(x) = {~

Prove that / has an anti derivative but g does not.

if x ::::: ° if x > 0.

43. Show that any two anti derivatives of a function differ by a constant. [Hint: This is a one-liner.]

44. (a) Define the oscillation for a function from one metric space to another, / : M ---+ N.


(b) Is it true that I is continuous at a point if and only if its oscillation is zero there? Prove or disprove.

(c) Fix a number K > O. Is the set of points at which the oscillation of I is ::: K closed in M? Prove or disprove.

45. (a) Prove thatthe integral of the Zeno' s staircase function described on page 161 is 2/3.

(b) What about the Devil's staircase? 46. In the proof of Corollary 26 of the Riemann-Lebesgue Theorem, it

is asserted that when <p is continuous the discontinuity set of <p 0 I is contained in the discontinuity set of I.

(a) Prove this. (b) Give an example where the inclusion is not an equality. (c) Find a sufficient condition on 0, II(x)1 ::: m for all x E [a, b]. Prove that the reciprocal of I, l/I(x), also belongs to 'R. If I E 'R, II(x)1 > 0, but no m > 0 is an underbound for III, prove that the reciprocal of I is not Riemann integrable.

48. Corollary 26 to the Riemann-Lebesgue Theorem asserts that if I E 'R and <p is continuous, then <p 0 I E 'R. Show that piecewise continuity can not replace continuity. [Hint: Take I to be a ruler function and <p to be a characteristic function.]

**49. Assume that I : [a, b] -+ [c, d] is a Riemann integrable bijection. Is the inverse bijection also Riemann integrable? Prove or disprove.

50. If I, g are Riemann integrable on [a, b], and I(x) < g(x) for all x E [a, b], prove that t I (x) dx < J: g(x) dx. (Note the strict inequality. )

51. Let I : [a, b] -+ lR be given. Prove or give counter-examples to the following assertions.

(a) I E 'R :::} III E 'R. (b) I I I E 'R :::} I E 'R. (c) I E 'Rand II(x)l::: c > o for all x :::} 1/1 E'R. (d) I E 'R :::} 12 E 'R. ( e) 12 E 'R :::} I E 'R. (f) 13 E 'R :::} I E 'R.

Exercises 197

(g) f2 E Rand f (x) :::: ° for all x =} fER. [Here F and f3 refer to the functions f (x) . f (x) and f (x) . f (x) . f(x), not the iterates.]

52. Given f, g E R, prove that max(j, g), min(j, g) E R, where max(j, g)(x) = max(j(x), g(x)) and min(j, g)(x) = min(j(x) , g(x)).

53. Assume that f, g : [0, 1] ~ ffi. are Riemann integrable and f(x) = g(x) except on the middle-thirds Cantor set C.

(a) Prove that f and g have the same integral. (b) Is the same true if f(x) = g(x) except for x E Q? (c) How is this related to the fact that the characteristic function of

Q is not Riemann integrable? 54. Prove that if an :::: ° and L an converges, then L Cia;') / n converges. 55. (a) If L an converges and (bn) is monotonic and bounded, prove

that L anbn converges. (b) If the monotonicity condition is dropped, or replaced by the

assumption that limn---+ oo bn = 0, find a counter-example to convergence of L anbn.

56. Find an example of a series of positive terms that converges despite the fact that lim sUPn---+oo lan+J/an I = 00. Infer that p can not replace A in the divergence half of the ratio test.

57. Prove that if the terms of a sequence decrease monotonically, a] ::::

a2 :::: ... , and converge to ° then the series L ak converges if and only if the associated dyadic series

al + 2a2 + 4a4 + 8as + ... = L2ka2k

converges. (I call this the block test because it groups the terms of the series in blocks oflength 2k- 1.)

58. Prove that L 1/ k 10g(k)P converges when p > 1 and diverges when p :s 1. Here k = 2, 3, .... [Hint: Integral test or block test.]

59. Concoct a series L ak such that (-l)kak > 0, ak ~ 0, but the series diverges.

60. (a) Show that if a series has ratio lim sup p then it has exponential growth rate p. Infer that the ratio test is subordinate to the root test.

(b) Concoct a series such that the root test is conclusive but the ratio test is not. Infer that the root test has strictly wider scope than the ratio test.


61. Show that there is no simple comparison test for conditionally convergent series:

(a) Find two series L ak and L bk such that L bk converges conditionally, ak/bk -+ 1 as k -+ 00, and L ak diverges.

(b) Why is this impossible if the series L bk is absolutely convergent?

62. An infinite product is an expression TI Ck where Ck > O. The nth

partial product is Cn = Cl ... Cn. If Cn converges to a limit C -:f=. 0 the product converges to C. Write Ck = 1 + ak. If each ak :::: 0 or each ak .:s 0 prove that L ak converges if and only if TI Ck converges. [Hint: Take log's.]

63. Show that conditional convergence of the series L ak and the product TI (l + ak) can be unrelated to each other:

(a) Set ak = (_I)k / -Jk. The series L ak converges but the corresponding product TI (l + ak) diverges.

(b) Let ek = 0 when k is odd and ek = 1 when k is even. Set bk = ek/ k + (_I)k / -Jk. The series L bk diverges while the corresponding product TIk>2(1 + bk ) converges.

64. Consider a series L ak and rearrange its terms by some bijection f3 : N -+ N, forming a new series L ap(k). The rearranged series converges if and only if the partial sums ap(l) + ... + ap(n) converge to a limit as n -+ 00.

(a) Prove that every rearrangement of a convergent series of nonnegative terms converges - and converges to the same sum as the original series.

(b) Do the same for absolutely convergent series. *65. Let L ak be given.

(a) If L ak converges conditionally, prove that rearrangement totally alters its convergence in the sense that some rearrangements L bk of L ak diverge to +00, others diverge to -00, and still others converge to any given real number.

(b) Infer that a series is absolutely convergent if and only if every rearrangement converges. (The fact that rearrangement radically alters conditional convergence shows that although finite addition is commutative, infinite addition (series) is not.)

**66. Suppose that L ak converges conditionally. If L bk is a rearrangement of L ak, let Y be the set of subsequential limits of (Bn) where Bn is the nth partial sum of L bk . That is, y E Y if and only if some Bn£ -+ y as e -+ 00.

Exercises 199

(a) Prove that Y is closed and connected. (b) If Y is compact and non-empty, prove that L bk converges to

Y in the sense that dH(Yn , Y) -+ 0 as n -+ 00, where dH is the Hausdorff metric on the space of compact subsets of IR and Yn

is the closure of {Bm : m ?: n}. See Exercise 2.124. (c) Prove that each closed and connected subset of IR is the set of

subsequential limits of some rearrangement of L ak. The article, "The Remarkable Theorem of Levy and Steinitz" by Peter Rosenthal in the American Math Monthly of April 1987 deals with some of these issues.

***67. Let V be a Banach space - a vector space with a norm such that V is complete with respect to the metric induced by the norm. (For example, IRm is a Banach space.) If L Vk is a convergent series of vectors in V such that L II Vk II diverges, what is the generalization of Exercise 66? In particular, is Y convex?

*68. Absolutely convergent series can be multiplied in a natural way, the result being their Cauchy product,

where Ck = aObk + a1bk- 1 + ... + akbO. (a) Prove that L Ck converges absolutely. (b) Formulate some algebraic laws for such products (commutativ-

ity' distributivity, and so on.) Prove two of them. [Hint for (a): Write the products aibj in an 00 x 00 matrix array M, and let An, Bn, en be the nth partial sums of L ai, L bj , L Ck. You are asked to prove that (lim An)(lim Bn) = lim en. The product of the limits is the limit ofthe products. The product AnBn is the sum of all the aibj in the n x n comer submatrix of M and Cn is the sum of its anti-diagonal. Now estimate AnBn - en. Alternately, assume that an, bn ?: 0 and draw a rectangle R with edges A, B. Observe that R is the union of rectangles Rij with edges ai, b j .]

**69. With reference to Exercise 68, (a) Reduce the hypothesis that both series Lai and Lbj are ab

solutely convergent to merely one being absolutely convergent and the other convergent. (Exercises 68 and 69(a) are known as Mertens'Theorem.)

(b) Find an example to show that the Cauchy product of two conditionally convergent series may diverge.


**70. The Riemann ~-function is defined to be ~(s) = L~l n-S where s > 1. It is the sum of the p-series when p = s. Establish Euler's product formula,

00 1 ~(s) = n 1 -s

k=l - Pk

where Pk is the kth prime number. Thus, PI = 2, P2 = 3, and so on. Prove that the infinite product converges. [Hint: Each factor in the infinite product is the sum of a geometric series 1 + P';;s + (p';;S)2 + .... Replace each factor by its geometric series and write out the nth partial product. Apply Mertens' Theorem, collect terms, and recall that any integer has a unique prime factorization.]

71. Invent a continuous function I : lR ---+ lR whose improper integral is zero, but which is unbounded as x ---+ -00 and x ---+ 00. [Hint: I is far from monotone.]

72. Assume that I : lR ---+ lR and that restricted to any closed interval, I is bounded.

(a) Formulate the concepts of conditional and absolute convergence of the improper Riemann integral of I.

(b) Find an example that distinguishes them. 73. Let I : [0, (0) ---+ [0, (0) and L ak be given. Assume that for all

sufficiently large k and all x E [k, k + 1), I(x) ::: ak. Prove that divergence of the improper integral Jo

oo I (x) dx implies divergence ofLak.

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