A Proof of the Squeeze Theorem for Integrals Using Cauchy
Sequences
Spiros Konstantogiannis [email protected]
Abstract. We use the sequential criterion for Riemann integrability to give a proof of the squeeze theorem for integrals using Cauchy sequences. As an auxiliary lemma, we prove a criterion for a sequence to be a Cauchy sequence.
Keywords: Riemann integral; sequential criterion; Cauchy sequence; squeeze theorem.
1 Introduction
Sequential criterion for Riemann integrability
A function
f
: ,
[ ]
a b
®
¡
is Riemann integrable on[ ]
a b
,
if and only if for every sequence( )
P
&
nof tagged partitions of
[ ]
a b
,
which is such thatlim
n0
n®¥
P
=
&
, the sequence(
(
;
)
)
n
S f P
&
ofRiemann sums is a Cauchy sequence, and then
lim
(
;
)
b n n a
S f P
f
®¥=
ò
&
. ProofThe proof is given in [1].
Auxiliary lemma 1 (a consequence of the definition of limit)
If the sequences
( )
x
n and( )
y
n both converge andlim
nlim
nn®¥
x
<
n®¥y
+
a
, for somea
Î
¡
, thenfor every
e
>
0
there existsN
Î
¥
such thatx
n<
y
n+ +
a
e
for alln N
>
.Proof
Since both sequences
( )
x
n and( )
y
n converge, then, by linearity, the sequence(
x
n-
y
n-
a
)
also converges and
lim
(
n n)
lim
nlim
nn®¥
x
-
y
-
a
=
n®¥x
-
n®¥y
-
a
(1).Further, by assumption,
lim
nlim
nn®¥
x
<
n®¥y
+
a
, thus nlim
®¥x
n-
lim
n®¥y
n- <
a
0
, and by means of (1),(
)
lim
n n0
n®¥
x
-
y
-
a
<
(2).Next, setting
lim
(
n n)
Besides, since
lim
(
n n)
n®¥
x
-
y
-
a
=
l
, then, by definition, for everye
>
0
there existsN
Î
¥
suchthat
x
n-
y
n- - <
a l
e
for alln N
>
, from which it follows thatx
n-
y
n- - <
a l
e
, and thenn n
x
-
y
- < +
a l
e
(4) for alln N
>
.Further, adding
e
to both sides of (3) yieldsl
+ <
e e
and combining with (4) then yieldsn n
x
-
y
- <
a
e
, and thusx
n<
y
n+ +
a
e
for alln N
>
.We have thus shown that for every
e
>
0
there existsN
Î
¥
such thatx
n<
y
n+ +
a
e
for alln N
>
, which is what we wanted to prove.Auxiliary lemma 2 (a criterion for a sequence to be Cauchy)
A sequence
( )
x
n is a Cauchy sequence if and only if for everye
>
0
there existsN
Î
¥
andl
eÎ
¡
(that depends one
), such thatx
n- <
l
ee
for alln N
>
.Proof
Note that, since
l
e depends one
, thenl
e is not a fixed number, and thus it cannot be derived thatl
e is the limit of( )
x
n .(i) Let
( )
x
n be a Cauchy sequence.Then, by definition, for every
e
>
0
there existsN
Î
¥
such thatx
m-
x
n<
e
for allm n N
,
>
.The number
N
generally depends one
.Then, choosing
m N
=
+ >
1
N
, we have thatx
N+1-
x
n<
e
, and since(
)
1 1 1
N n N n n N
x
+-
x
= -
x
+-
x
=
x
-
x
+ , the last inequality readsx
n-
x
N+1<
e
for alln N
>
.Further, since
N
depends one
, thenN
+
1
also depends one
, and thusx
N+1 depends one
too.
Then setting
x
N+1º
l
e, we have that for everye
>
0
there existsN
Î
¥
such thatx
n- <
l
ee
for all
n N
>
.(ii) We assume that for every
e
>
0
there existsN
Î
¥
andl
eÎ
¡
such thatx
n- <
l
ee
(1)for all
n N
>
.We will show that the sequence
( )
x
n is a Cauchy sequence.Further, by the triangle inequality,
x
m- + -
l
el
ex
n³
(
x
m-
l
e) (
+
l
e-
x
n)
=
x
m-
x
n , thusm n m n
x
-
x
£
x
- + -
l
el
ex
, and combining with (4) yieldsx
m-
x
n<
2
e
.The last inequality holds for every
e
>
0
, thus it also holds for everye
¢ =
2
e
>
0
, because ase
takes any positive value,
2
e
also takes any positive value.We have thus shown that for every
e
¢ >
0
there existsN
Î
¥
such thatx
m-
x
n<
e
¢
for all,
m n N
>
, and then, by definition,( )
x
n is a Cauchy sequence.Finally, combining (i) and (ii) completes the proof.
2 Squeeze theorem
A function
f
: ,
[ ]
a b
®
¡
is Riemann integrable on[ ]
a b
,
if and only if for everye
>
0
there existfunctions
g h
e,
e: ,
[ ]
a b
®
¡
such thatg x
e( )
£
f x
( )
£
h x
e( )
for allx
Î
[ ]
a b
,
, and such that,
g h
e e are both Riemann integrable on[ ]
a b
,
and(
)
b a
h
e-
g
e<
e
ò
.Proof
(i) Let
f
be Riemann integrable on[ ]
a b
,
.For every
e
>
0
, we setg
eº
f
andh
eº
f
, and then we haveg x
e( )
=
f x
( )
=
h x
e( )
, thusalso
g x
e( )
£
f x
( )
£
h x
e( )
, for allx
Î
[ ]
a b
,
, and since, by assumption,f
is Riemannintegrable on
[ ]
a b
,
, theng h
e,
e are both Riemann integrable on[ ]
a b
,
, and also(
)
0 0
b b
a a
h
e-
g
e=
=
ò
ò
, becauseS
( )
0;
P
&
=
0
for any tagged partitionP
&
of[ ]
a b
,
, thus taking anysequence
( )
P
&
n of tagged partitions of[ ]
a b
,
which is such thatlim
n0
n®¥
P
=
&
, we have that( )
0;
n0
S
P
&
=
, and thenlim
( )
0;
n0
n®¥
S
P
=
&
, and thus, by the sequential criterion for Riemannintegrability,
0 0
b a=
ò
. Since(
)
0
b ah
e-
g
e=
ò
, then(
)
b ah
e-
g
e<
e
ò
for everye
>
0
.We have thus shown that for every
e
>
0
there exist functionsg h
e,
e: ,
[ ]
a b
®
¡
such that( )
( )
( )
g x
e£
f x
£
h x
e for allx
Î
[ ]
a b
,
, and such thatg h
e,
e are both Riemann integrable on[ ]
a b
,
and(
)
b a
h
e-
g
e<
e
ò
.(ii) We assume that for every
e
>
0
there exist functionsg h
e,
e: ,
[ ]
a b
®
¡
such that( )
( )
( )
g x
e£
f x
£
h x
e for allx
Î
[ ]
a b
,
, and such thatg h
e,
e are both Riemann integrable on[ ]
a b
,
and(
)
b a
h
e-
g
e<
e
ò
.Next, we consider any sequence
( )
P
&
n of tagged partitions of[ ]
a b
,
which is such thatlim
n0
n®¥
P
=
&
.Then, since
g h
e,
e are integrable on[ ]
a b
,
, the sequential criterion for Riemann integrability impliesthat
lim
(
;
)
b n n ag
eS g P
e ®¥=
ò
&
(1) and blim
(
;
)
n n a
h
eS h P
e ®¥=
ò
&
(2).Since both integrals
b a
g
eò
and b ah
eò
exist, then, by linearity,(
)
b b b
a a a
h
e-
g
e=
h
e-
g
eò
ò
ò
, and by (1)and (2),
(
)
lim
(
;
)
lim
(
;
)
b n n n n a
h
eg
eS h P
eS g P
e ®¥ ®¥-
=
-ò
&
&
, and then the relation b(
)
a
h
e-
g
e<
e
ò
reads(
)
(
)
lim
;
nlim
;
n n®¥S h P
e-
n®¥S g P
e<
e
&
&
, thuslim
(
;
)
lim
(
;
)
n n
n®¥
S h P
e<
n®¥S g P
e+
e
&
&
.Then, by the auxiliary lemma 1, there exists
N
Î
¥
such thatS h P
(
e;
&
n) (
<
S g P
e;
&
n)
+ +
e e
, thus(
;
n) (
;
n)
2
S h P
e&
<
S g P
e&
+
e
(3) for alln N
>
.Besides, since, by assumption,
g x
e( )
£
f x
( )
£
h x
e( )
for allx
Î
[ ]
a b
,
, then for every taggedpartition
P
&
of[ ]
a b
,
, it holds thatS g P
(
e;
&
) ( ) (
£
S f P
;
&
£
S h P
e;
&
)
.Let us prove it.
If
{ }
0 m k kx
= and{ }
1 m k kt
= are, respectively, the partition points and tags ofP
&
, then, by definition,(
;
)
( )(
1)
m k k kS g P
e&
=
å
g t
ex
-
x
- ,( )
;
( )(
1)
m k k kS f P
&
=
å
f t
x
-
x
- ,Further, since
g x
e( )
£
f x
( )
£
h x
e( )
for allx
Î
[ ]
a b
,
and sincet
kÎ
[
x
k-1,
x
k] [ ]
Í
a b
,
for1, 2,...,
k
=
m
, theng t
e( )
k£
f t
( )
k£
h t
e( )
k , and multiplying all sides byx
k-
x
k-1>
0
(x
k>
x
k-1so that the interval
[
x
k-1,
x
k]
is well defined) yields( )(
k k k 1)
( )(
k k k 1)
( )(
k k k 1)
g t
ex
-
x
-£
f t
x
-
x
-£
h t
ex
-
x
- , fork
=
1, 2,...,
m
, and adding them
inequalities then yields
( )(
1)
( )(
1)
( )(
1)
1 1 1 m m m k k k k k k k k k k k k
g t
ex
x
-f t
x
x
-h t
ex
x
-= = =-
£
-
£
-å
å
å
, thus(
;
) ( ) (
;
;
)
S g P
e&
£
S f P
&
£
S h P
e&
, which is what we wanted to prove.Applying the previous relation, we have
S g P
(
e;
&
n) (
£
S f P
;
&
n) (
£
S h P
e;
&
n)
for alln
Î
¥
, thus alsofor all
n N
>
, and combining with (3) then yieldsS g P
(
e;
&
n) (
£
S f P
;
&
n) (
<
S g P
e;
&
n)
+
2
e
(4) for alln N
>
.We note that we cannot take the limit
e
®
0
, because asN
depends one
, thenN
may tend toinfinity as
e
tends to 0, and then the relation (4) does not actually hold.To proceed, we will use that the sequence
(
S g P
(
e;
&
n)
)
converges tob a
g
eò
(see (1)).Then, by definition, for every
e
¢ >
0
, thus also fore
¢ = >
e
0
, there existsN
1Î
¥
such that(
;
)
b n
a
S g P
e&
-
ò
g
e<
e
(5) for alln N
>
1.Next, we set
N
2º
max
{
N N
,
1}
.Since
N N
,
1Î
¥
, thenmax
{
N N
,
1}
Î
¥
, and thusN
2Î
¥
.Further, since
max
{
N N
,
1}
³
N N
,
1, thenN
2³
N N
,
1, and thus ifn N
>
2, thenn N N
>
,
1, andthus both (4) and (5) hold for all
n N
>
2.By (5), we have
(
;
)
b n aS g P
eg
ee
e
- <
&
-
ò
<
, and addingb a
g
eò
to all sides yields(
;
)
b b n a ag
e- <
e
S g P
e<
g
e+
e
ò
&
ò
, thus b(
;
)
n ag
e- <
e
S g P
eò
&
(6) and(
;
)
b n aS g P
e&
<
ò
g
e+
e
(7).Next, adding
2
e
to both sides of (7) yields(
;
)
2
3
b n
a
Further, since
e
>
0
, then- > -
e
3
e
, and addingb a
g
eò
to both sides yields3
b b a a
g
e- >
e
g
e-
e
ò
ò
, i.e.3
b b a ag
e-
e
<
g
e-
e
ò
ò
(9).Next, combining (6) and (9) yields
3
(
;
)
b b n a a
g
e-
e
<
g
e- <
e
S g P
eò
ò
&
, thus b3
(
;
)
n ag
e-
e
<
S g P
eò
&
,and combining with (4) then yields
3
(
;
) (
;
) (
;
)
2
b
n n n
a
g
e-
e
<
S g P
e£
S f P
<
S g P
e+
e
ò
&
&
&
, thus(
) (
)
3
;
;
2
b n n ag
e-
e
<
S f P
<
S g P
e+
e
ò
&
&
, and combining with (8) yields(
)
3
;
3
b b n a ag
e-
e
<
S f P
<
g
e+
e
ò
&
ò
, and adding ba
g
e-
ò
to all sides then yields(
)
3
;
3
b n aS f P
g
ee
e
- <
&
-
ò
<
, thus(
;
)
3
b n aS f P
&
-
ò
g
e<
e
(10).The inequality (10) holds for every
e
>
0
and for everyn N
>
2.Since the inequality (10) holds for every
e
>
0
, then it also holds for everye
¢¢ =
3
e
>
0
, because ase
takes any positive value,3
e
also takes any positive value.Also, since
e e
=
¢¢
3
, then the functiong
e that, by assumption, depends one
, depends one
¢¢
.We have thus shown that for every
e
¢¢ >
0
there existsN
2Î
¥
such that(
;
)
b n
a
S f P
&
-
ò
g
e¢¢<
e
¢¢
for all
n N
>
2, where the integralb a
g
e¢¢ò
is a real number that, generally, depends one
¢¢
(becausethe function
g
e¢¢ depends one
¢¢
).Then, by the auxiliary lemma 2, the sequence
(
S f P
(
;
&
n)
)
is a Cauchy sequence.We have thus shown that for any sequence
( )
P
&
n of tagged partition of[ ]
a b
,
which is such thatlim
n0
n®¥
P
=
&
, the sequence(
(
;
)
)
n
S f P
&
is a Cauchy sequence.3 References
1. Spiros Konstantogiannis, A Sequential Criterion for Riemann Integrability Resulting from Riemann’s Definition of the Integral. academia.edu, 2020, available at
https://www.academia.edu/43876143/A_Sequential_Criterion_for_Riemann_Integrability_Resulting_from_Riema nns_Definition_of_the_Integral
and
https://q4quantum.wordpress.com/2020/08/17/a-sequential-criterion-for-riemann-integrability-resulting-from-riemanns-definition-of-the-integral/
2. Robert G. Bartle and Donald R. Sherbert, Introduction to Real Analysis. John Wiley & Sons, Inc., Fourth Edition, 2011.