Schinazi R.B. From Calculus to Analysis

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180 6 Riemann Integration

Using the induction hypothesis, we have

n+1



k=1

−a

k−1

) =(a

−a

) +(a

n+1

−a

) =a

n+1

−a

This proves Lemma 6.1.

Example 6.2 Constant functions on [a,b] are Riemann integrable.

Let f(x)=c for all x in [a, b].LetP ={x

,...,x

} be a partition of [a,b].

Since f is constant, it is clear that for all i =1,...,n,wehave



f, [x

i−1

]





f, [x

i−1

]



=c.

Hence,

U(f,P)=L(f, P ) =



i=1

c(x

−x

i−1

) =c(x

−x

) =c(b −a)

by Lemma 6.1. In particular, for any >0,

0 =U(f,P)−L(f, P ) < .

The constant function f is integrable.

Example 6.3 Let f be deﬁned on [0, 1] by f(x)=1ifx is rational and f(x)=0if

f is irrational. Then f is not Riemann integrable on [0, 1].

Let P ={x

,...,x

} be a partition of [0, 1].Fori =1,...,n, we have ratio-

nals and irrationals in [x

i−1

](recall that the rationals and the irrationals are dense

in the reals). Hence, f takes the values 0 and 1 in the set [x

i−1

]. In particular,



f, [x

i−1

]



=0 and M



f, [x

i−1

]



for all i =1,...,n. Therefore,

U(f,P)=



i=1

1(x

−x

i−1

) =x

−x

=1 −0 =1

by Lemma 6.1. On the other hand,

L(f, P ) =



i=1

0(x

−x

i−1

) =0

for every partition P . Hence, for every partition P ,

U(f,P)−L(f, P ) =1.

In particular, for  =1/2, there is no P such that U(f,P)−L(f, P ) < . Therefore,

f is not Riemann integrable.

Note that in both of the preceding examples, U(f,P)and L(f, P ) do not depend

on P . This is due to the very particular examples we took. In general, they will

depend on P .

6.1 Construction of the Integral 181

Monotone functions are integrable

Let f be deﬁned, bounded, and monotone on [a,b]. Then the function f is

Riemann integrable.

Let >0. We can pick n so that

b −a



f(b)−f(a)



<

(why?). Then we deﬁne the partition P ={x

,...,x

} of [a,b] by setting

=a +i

b −a

for i =0, 1,...,n.

Assume that f is increasing on [a,b] (the decreasing case is analogous and is left

to the exercises). Then



f, [x

i−1

]



=f(x

i−1

) and M



f, [x

i−1

]



=f(x

Hence,

U(f,P)=



i=1

f(x

)(x

−x

i−1

) and L(f, P ) =



i=1

f(x

i−1

)(x

−x

i−1

Using that x

−x

i−1

=(b −a)/n for every i =1,...,n,wehave

U(f,P)=

b −a



i=1

f(x

) and L(f, P ) =

b −a



i=1

f(x

i−1

By Lemma 6.1,

U(f,P)−L(f, P ) =

b −a



i=1



f(x

) −f(x

i−1

)



b −a



f(x

) −f(x

)



That is,

U(f,P)−L(f, P ) =

b −a



f(b)−f(a)



But this is less than  by our choice of n. This proves that f is Riemann integrable.

Another large class of Riemann integrable functions are the continuous functions.

However, in order to prove this, we will need the notion of uniform continuity that

we now introduce.

Uniformly continuous functions

A function f is said to be uniformly continuous on the set D if for every

>0, there is a δ>0 such that if x and y are in D and if |x −y| <δ, then

|f(x)−f(y)|<.

182 6 Riemann Integration

Recall that a function f is said to be continuous on D if for every a ∈ D and

for every >0, there is a δ>0 such that if x is in D and if |x − a| <δ, then

|f(x)− f(a)|<. There are some subtle but crucial differences between the two

deﬁnitions. First, continuity concerns points (f is continuous at every a in D), and

uniform continuity concerns sets (f is uniformly continuous on the whole set D).

Second, if f is continuous at a, for a given , the corresponding δ may depend on a

and on . For a different point b in D, the corresponding δ will typically be different.

It may or may not be possible to pick the same δ (for a given ) for all the points of

asetD. If this is possible, the function is said to be uniformly continuous.

Continuous functions are Riemann integrable

Let f be continuous on [a,b]. Then f is Riemann integrable on [a,b].

A continuous function on a closed and bounded interval [a,b] is uniformly con-

tinuous on [a,b]. This will be proved in Sect. 6.3 whose focus is uniform continuity.

Let >0. Since f is uniformly continuous on [a,b], there is δ>0 such that if x

and y are in [a,b] and |x −y|<δ, then



f(x)−f(y)





b −a

By the Archimedean property we may ﬁnd n so that

b −a

<δ.

Now that n is ﬁxed, let P ={x

,...,x

} be a partition of [a,b] deﬁned by

=a +i

b −a

for i =0, 1,...,n.

For any i =1,...,n, f is continuous on [x

i−1

], and the extreme value theorem

states that f attains its maximum and minimum on this interval. Thus, there are c

and d

in [x

i−1

] such that



f, [x

i−1

]



=f(c

) and M



f, [x

i−1

]



=f(d

Hence,

U(f,P)=



i=1

f(d

)(x

−x

i−1

) and L(f, P ) =



i=1

f(c

)(x

−x

i−1

Using that for every i =1,...,n,wehavex

−x

i−1

=(b −a)/n, it follows that

U(f,P)=

b −a



i=1

f(d

) and L(f, P ) =

b −a



i=1

f(c

Therefore,

U(f,P)−L(f, P ) =

b −a



i=1



f(d

) −f(c

)



6.1 Construction of the Integral 183

Note that since for every i =1,...,n, c

and d

are in [x

i−1

],wehave

−d

|≤|x

i−1

−x

b −a

<δ

by our choice of n. Hence,

0 ≤f(d

) −f(c



b −a

and

U(f,P)−L(f, P ) =

b −a



i=1



f(d

) −f(c

)



≤

b −a



i=1



b −a

=.

This proves that f is Riemann integrable.

Up to this point we have not deﬁned the integral of a function. We now do so.

The Riemann integral

Let f be bounded on [a,b]. Then the set of upper Darboux sums has the

greatest lower bound I(U,f,[a,b]), and the set of lower Darboux sums has

the least upper bound I(L,f,[a,b]).Iff is integrable, then I(U,f,[a,b]) =

I(L,f,[a,b]), and this real number is denoted by



Moreover, for every partition P of [a,b],wehave

L(f, P ) ≤



f ≤U(f,P).

There are several claims to be proved in the deﬁnition above. We will prove them

at the end of this section.

Example 6.4 Let f be a constant c. Then



f =c(b −a).

We already know that a constant function is Riemann integrable. Moreover, for

every partition P of [a,b],wehaveL(f, P ) = U(f,P)= c(b −a). Since



f is

between the lower and upper sums, we must have



f =c(b −a).

Fundamental Theorem of Calculus

Let f be Riemann integrable on [a,b], and let F be continuous on [a,b] and

differentiable on (a, b). Assume that F



(x) =f(x)for all x in (a, b). Then



f =F(b)−F(a).

184 6 Riemann Integration

We now prove the FTC. Let P ={x

,...,x

} be a partition of [a,b].Using

that x

=a and x

=b and Lemma 6.1, we get

F(b)−F(a)=F(x

) −F(x

) =



i=1



F(x

) −F(x

i−1

)



Since F is continuous on [x

i−1

]and differentiable on (x

i−1

),fori =1,...,n,

the Lagrange mean value theorem applies: there is y

in (x

i−1

) such that

F(x

) −F(x

i−1

) =F



)(x

−x

i−1

) =f(y

)(x

−x

i−1

In particular,

F(b)−F(a)=



i=1

f(y

)(x

−x

i−1

Observe that for every i =1,...,n,wehave



f, [x

i−1

]



≤f(y

) ≤M



f, [x

i−1

]



Hence,



i=1



f, [x

i−1

]



−x

i−1

) ≤



i=1

f(y

)(x

−x

i−1

)

≤



i=1



f, [x

i−1

]



−x

i−1

That is, for every partition P ,wehave

L(f, P ) ≤F(b)−F(a)≤U(f,P).

Therefore, F(b)−F(a)is an upper bound of the set of lower Darboux sums. Since

f is integrable, we know that



f is the least upper bound of the set of lower

Darboux sums. Thus,



f ≤F(b)−F(a).

On the other hand, F(b)−F(a) is also a lower bound of the set of upper Darboux

sums. Since



f is also the greatest lower bound of that set, we have

F(b)−F(a)≤



Thus,



f =F(b)−F(a), and this completes the proof of the FTC.

Example 6.5 Let a =−1. Show that f deﬁned by f(x)=x

on [1, 2] is integrable

and ﬁnd its integral.

Note that f(x)=exp(a ln x) is continuous on [1, 2] (as a composition of contin-

uous functions) and is therefore integrable on [1, 2].LetF be deﬁned by

F(x)=

a +1

a+1

6.1 Construction of the Integral 185

Then F is continuous on [1, 2] and differentiable on (1, 2) with F



=f . Hence, the

FTC applies, and we have



f =

a +1



a+1

−1



The FTC is very useful when one can ﬁnd an antiderivative F for a given func-

tion f . Unfortunately, there are very few functions for which one can ﬁnd an an-

tiderivative. The examples from Calculus are, most of the time, manufactured so

thattheFTCcanbeapplied....

We now turn our attention to the claims made to deﬁne the integral of an inte-

grable function.

There are four such claims:

1. The set of upper Darboux sums has a greatest lower bound I(U,f,[a,b]).

2. The set of lower Darboux sums has a least upper bound I(L,f,[a,b]).

3. If f is integrable, then I(U,f,[a,b]) = I(L,f,[a,b]).

4. If f is integrable, then for every partition P ,wehave

L(f, P ) ≤



f ≤U(f,P).

The fourth claim is an easy consequence of the fact that



f is a lower bound of

upper Darboux sums and an upper bound of lower Darboux sums.

The ﬁrst three claims will be proved at once thanks to the following lemma. We

designate the least upper bound of a set A by sup A and the greatest lower bound by

infA, if they exist!

Lemma 6.2 Let A and B be two nonempty subsets of reals with the property that for

all x in A and y in B, we have x ≤y. Then supA and infB exist and supA ≤infB.

If, in addition, for every >0, there are x in A and y in B such that

y −x<,

then sup A =inf B.

We ﬁrst apply Lemma 6.2, and we then prove it. In our application of this lemma,

A is the set of lower Darboux sums, and B is the set of upper Darboux sums. We will

show that any lower Darboux sum is less than any upper Darboux sum. According

to Lemma 6.2, this implies that A has a least upper bound (denoted by sup A) and

B has a greatest lower bound denoted by inf B. Moreover, by the deﬁnition of an

integrable function we know that for every >0, there is a partition P such that

U(f,P)−L(f, P ) < .

Therefore, the second part of Lemma 6.2 applies as well, and we can conclude that

supA =infB. T

hat is, I(U,f,[a,b]) =I(L,f,[a,b]).

The material in the rest of this section may be difﬁcult for a beginner. It is not essential for the

sequel and may be omitted.

186 6 Riemann Integration

Hence, to apply Lemma 6.2, we only need to check that any lower Darboux sum

is less than any upper Darboux sum. We start with the so-called reﬁnement lemma.

Lemma 6.3 Let P and Q be partitions of [a,b] with P ⊂Q. Then

L(f, P ) ≤L(f, Q) ≤U(f,Q)≤U(f,P).

The partition Q is said to be a reﬁnement of partition P because in addition to

the points of P , Q has other points as well.

We will prove that L(f, P ) ≤ L(f, Q). The proof that U(f,P) ≥ U(f,Q) is

similar and is omitted. We have already observed that L(f, Q) ≤U(f,Q).

We do a proof by induction. Our induction hypothesis is: if Q has k more points

than P , then L(f, P ) ≤ L(f, Q). Assume ﬁrst that k = 1. That is, Q has just one

more point than P .LetP ={x

,...,x

}, and let y be the additional point of Q.

Assume that y is in (x

j−1

) for some j in {1,...,n}. Note that

L(f, P ) =



i=1



f, [x

i−1

]



−x

i−1

)

and

L(f, Q) =

j−1



i=1



f, [x

i−1

]



−x

i−1

)



f, [x

j−1

,y]



(y −x

j−1

) +m



f, [y,x

]



−y)



i=j +1



f, [x

i−1

]



−x

i−1

Hence,

L(f, Q) −L(f, P ) =m



f, [x

j−1

,y]



(y −x

j−1

) +m



f, [y,x

]



−y)

−m



f, [x

j−1

]



−x

j−1

By writing x

−x

j−1

−y +y −x

j−1

we get

L(f, Q) −L(f, P ) =



f, [x

j−1

,y]



−m



f, [x

j−1

]



(y −x

j−1

)



f, [y,x

]



−m



f, [x

j−1

]



−y).

It is easy to see that if C ⊂D, then infC ≥inf D (see the exercises). Since



f(x),x∈[x

j−1

,y]



⊂



f(x),x∈[x

j−1

]



we have



f, [x

j−1

,y]



≥m



f, [x

j−1

]



Similarly,



f, [y,x

]



≥m



f, [x

j−1

]



6.1 Construction of the Integral 187

Hence, L(f, Q) −L(f, P ) is the sum of two positive terms and is therefore positive.

This proves the induction hypothesis for k =1.

Assume now that the induction hypothesis holds for k.LetQ have k + 1more

points than P .LetR be the partition obtained by deleting one of the additional

points of Q. In particular, Q has one more point than R, and so

L(f, R) ≤ L(f, Q).

On the other hand, R has k more points than P , and by the induction hypothesis we

have

L(f, P ) ≤L(f, R).

Putting together the two inequalities above, we get

L(f, P ) ≤L(f, Q),

and the lemma is proved by induction.

Lemma 6.3 is the main ingredient to prove the following:

Lemma 6.4 Let P and Q be any two partitions of [a,b]. Then

L(f, P ) ≤U(f,Q).

Deﬁne R as the union of P and Q. Hence, R is a reﬁnement of both P and Q.

Using Lemma 6.3 to get the ﬁrst and third inequalities below, we get

L(f, P ) ≤L(f, R) ≤ U(f,R)≤U(f,Q).

This completes the proof of Lemma 6.4.

With Lemma 6.4 in hand, Lemma 6.2 implies that



U,f,[a,b

]





L,f,[a,



Our last task in this section is the proof of Lemma 6.2.

Recall that we are assuming that A and B are two nonempty subsets of reals with

the property that for all x in A and y in B,wehavex ≤y.Fixx

in A and y

in B.

Then, for all y in B,wehavey ≥ x

. That is, B is bounded below and nonempty.

Therefore, by the fundamental property of the reals, B has a greatest lower bound

infB. Similarly, A is bounded above by x

and is not empty. Hence, it has a least

upper bound sup A. Since y

is an upper bound of A,wehavey

≥sup A. But this

is true for any y

in B. Hence, sup A is a lower bound of B. It must be smaller than

the greatest lower bound inf A. Thus, supA ≤ inf B. That proves the ﬁrst part of

Lemma 6.2.

Assume in addition that for every >0, we have x in A and y in B such that

0 ≤y −x<.

By the deﬁnition of sup A and inf B we have

x ≤sup A and y ≥inf B.

188 6 Riemann Integration

Hence,

y −x ≥infB −supA ≥0,

and for every >0, we have

0 ≤infB −supA<.

Therefore, inf B −supA is necessarily 0 (why?), and we are done.

Exercises

1. Let A be a set of reals bounded below and above. Let B be a nonempty subset

of A. Then,

supA ≥supB and infA ≤inf B.

2. Pick a partition of [0, 1] and compute an upper and a lower Darboux sum for

the function f(x)=x.

3. Redo Example 6.1 with the partition P ={0, 1/4, 1/2, 3/4, 1}.

4. Let f be an integrable function on [a,b]. Assume that there is a real I such that

for every partition P ,

L(f, P ) ≤I ≤U(f,P).

Show that I



f .

5. Draw a triangle on the plane. Use the FTC to compute the area of the triangle.

Check your computation with a geometric formula.

6. Justify the existence and compute the following integrals.

(a)



1+x

(b)



1+x

(c)



−2

x exp(−x

(d)



√

7. Let f be continuous on (0, 1]

(a) Show that f is integrable on [1/n,1] for every natural n.

(b) Give an example of f for which



1/n

f converges as n goes to inﬁnity.



1/n

f does not converge.

8. Let f be a bounded function on [a,b].

(a) Show that if f is constant on [a,b], then U(f,P) = L(f, P ) for every

partition P .

(b) Show that if there is a partition P of [a,b] such that U(f,P) = L(f, P ),

then f is constant on [a,b].

9. Show that if f is decreasing on [a,b], then f is integrable.

10. Show that if f is integrable on [a,b] and a<c<b, then f is integrable on

[a,c].

11. Let a<c<b. Assume that f is integrable on [a,c] and on [c, b].

(a) Let >0. Show that there are partitions P and Q of [a,c] and [c, b] such

that

U(f,P)−L(f, P ) < /2 and U(f,Q)−L(f, Q) < /

6.2 Properties of the Integral 189

(b) Let R be the union of P and Q. Show that

U(f,R)=U(f,P)+U(f,Q) and L(f, R) =L(f, P ) +L(f, Q).

12. Let f be continuous on (a, b] and bounded on [a,b]. That is, there is K such

that |f(x)|<Kfor all x in [a,b]. In this problem we show that f is integrable.

(a) Sketch the graph of a function deﬁned on [a,b] but continuous only on

(a, b].

(b) Let >0. Show that there is a natural n such that 2K/n < /2.

U(f,P)−L(f, P ) < /2.

(d) Let Q be the partition of [a,b] obtained by adding a to the partition P .

Show that



U(f,P)−

U(f,Q)



<K/

n and



L(f, P ) −L(f, Q)



<K/n.

(e) Show that U(f,Q)−L(f, Q) < . Conclude that f is integrable on [a,b].

13. Let f be bounded on [a,b] and continuous except at one point c ∈(a, b).

(a) Show that f is integrable on [a,c] and on [c, b]. Use Exercise 12.

(b) Show that f is integrable on [a,b]. Use Exercise 11.

ﬁnitely many points, then the function is integrable.

14. Assume that f is 0 on [a,b] except at c where a<c<b.

(a) Use Exercise 13(c) to show that f is integrable.

(b) Assume that f(c)>0. Show that for every natural n,wehave

0 ≤



f ≤f(c)

b −a

(Recall that for any partition P ,wehaveL(f, P ) ≤



f ≤U(f,P).)



f =0.

15. Show that if a function is 0 except at ﬁnitely many points, then the function is

integrable, and its integral is 0. (Use Exercise 14.)

16. (a) Assume that f is integrable on [a,b] and that f =g except at ﬁnitely many

points. Show that g is integrable. (Use Exercise 15.)

(b) Show that



g =



f .

6.2 Properties of the Integral

We start by stating some useful properties of greatest lower bounds and least upper

bounds. Recall that if f is a bounded function on [a,b], then the set



f(x);x ∈[a,b]



has a least upper bound M(f,[a,b]) and a greatest lower bound m(f, [a,b]). When

there is no ambiguity, we will use the shorter notation M(f) and m(f ).