Integration Riemann

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8-1-1965
Riemann-Stieltjes integration
Lawrence Woodard
Atlanta University
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il
N IiN~
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~
R:ENANN-STIELTJES INTEGRATION
A THESIS
SUBMITTED TO THE FACULTY OF ATLANTA UNIVERSITY
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS
FOR THE DEGREE OF MASTER OF SCIENCE
BY
LAWRENCE W000ARD
DEPARTMENT OF MATHEMATICS
ATLANTA, GEORGIA
AUGUST 1965
i.
s~
il
~
oh
h~o h
A CKN OWLEDGENENT S
The autrior wishes to acknowledge the assistance and
suggestions given by Dr. L. K. Williams and Dr. C. B.
Dansby in the preparation of this thesis.
He is also most grateful to his wife, Alma, for the
aid she gave in proofreading and typing the first draft.
i
il
~ I
~ I~
TABLE OF CONTENTS
Page
A CKN OWLE~DGENENTS
Cter
I.
II.
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INTRODUCTION
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PRESENTATION OF DATA ON RIEN~NN~STIELTJES
INTEGRATI ON . . . . . . . . . . . . . .
3
Construction of Riernann Integral
Construction of Riemann—StieltjeS
Integral
Special Examples of Riemann—Stieltjes
Integrals
Theorems of Riemann—Stieltjes Integration
Existence of Riemann-Stieltjes Integral
III.
PRESENTATION OF DATA ON LEBESQUE-STIELTJES
INTEGRAL. . . . . . . . . . . . . . . .
B IBLIOGRAPHY~
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2).~.
31
4,4
4
~ ~
Chapter I
Introduction
The Riemann—Stieltjes integral, as its name suggests,
is similar to the Riemann integral which is used in or
dinary elementary calculus.
It is somewhat more general
than the Rieinann integral, and its added generality makes
it more applicable and useful in the field of physics as
well as mathematics.
We assert that it is more general than the Riemann
integral for the reason that
g(~
g(x~_~) not only can
denote the length of a subinterval of a real line, but can
designate a measure of a magnitude associated with the
length xk
-
x~_~ of the real line.
Therefore, in the field
of physics, f or example, one may consider g(x) to be the
total “mass” or Ucharge?? on the interval
-
~
then g(x~)
g(x~_~) would mark the mass or charge associated with
-
xk_l which is the length of the k1~’ subinterval of the
interval
~
where
~
is on the real line.
But, the Riemann integral has limitations and restric
tions in its application basically because
ti
x~ xk
-
xk_1
can be no more than a length of a subinterval of a real
line.
This being true, only problems can be solved where
1
2
the points xk where k
=
1, 2,
3
....
n, are found on a real
line.
As the above observation has shown, the Riemann—Stielt—
jes integral is applicable in solving a number of problems
where the Rie]nann integral fails to do so.
Being aware of this fact, the writer’s purpose is to
show the construction of the Riemann and the Riemann
Stieltjes integrals and their relationship to one another.
U
lIkI&~h&~hJ ~ ~Ith!I
JkI~~L
~
Chapter II
Presentation of Data on Rieinann
Stieltjos Integration
Construction zf Riemann Integral
The first objective of the writer is to show how the
Riemann and th~e Riemann-StieltjeS integrals are constructed.
In constructing the Riemann integral, we will assume that
f is a real valued function defined on a closed interval
J
=
f~,bJ which is on the real number line.
assume that f is bounded on J
=
5,jj7.
We will also
By bounded, we mean
that there are two real numbers ~ and B such that f (x)~
~ for all x in
~
and f(x)~ for all x in 5,bJ.
Our figure wculd look like the drawing below.
Y
Fig.I
—
Some of the terms used throughout this ~aper are terms
which are infrequently used at certain levels of mathematical
3
LI.
development.
When the writer finds the need to use one of
these terms, we will define it.
is Partition.
The first of these terms
A partition P of an interval J
5,bJ is a
finite collection of non-overlapping intervals whose sum is
J
=
~
Usually, we describe a partition P by specifying
a finite set of real numbers (x0, x1, x2,
....,
x~) such
that
ax0~x1~X2~..
. .
~ =b
and such that the subinterval occurring in the partition P
are the intervals ~kl’
x~7,
k1, 2,
....,
n.
ly, we refer to the end points xk, k0, 1, 2,
More proper
)n as the
partition points.
The next step in the construction of the Riemann in
tegral is to partition J9~,bJby a partition P.
Fig. II
—
I
I
ccXe X1
We wil
I
I
I
X~•
x~ Xk
I
x~, x.~ ~
X
now consider a point ~k such that xk_1~
~x for k~., 2,
3,
•..,
n.
Now consider the k~ subin
terval
Fig. III
Xiç
111k
d
~ OW
OW 0
S
The area under the graph f (x) between xk_1 and xk
can be approximated by the product of f(ck)LX~-’~k...l]’
If
this product is taken for each subinberval k, for k1, 2,
3,
n, the sum of these products gives an approximation
•..,
of the area under the curve between x=a and x=b.
This approxi
mation is not a very close approximation as can be seen from
figure
IV.
Same of the area under the curve is not represented
in the product (as indicated by
~
).
Some of the area
represented in the product is not under the curve, but above
(indicated by~
/
).
4’
/
/
/
Fig.IV
I
I
I
-
°.~
~~
•
x~-~
~,
~= ~
The next term that will be defined at this point is
If P and Q. are partitions of
refinement of a partition P.
we say that
Q,
is a refinement of P or that
contained in some subinterval in P.
Q,
is
The refinement concept
used here is a modern concept and is used in place of the
traditional limit concept used in integral calculus.
It can
be assumed that the intermediate points are partition points,
so the partition (x0, ~
,
x1
,
figure IV, we will denote as Q.
tion Q~of Q
,
....,
X~,, ~ found in
If we considered a
the k4~subinterval would be
parti
6
Fig. V
x
3~i~
k-i
~
It is quite clear that the sum of the two products f(Fk’)
(—
~ f([k~1)(xk_ 1k~ is a much. closer approximation
to the area under the graph than the product f(ck)(Xk_Xk_l).
If there exists a real number I, which is the exact area
under the graph in
number
e
5,~7,
such that for every positive
there is a partition Q~ of J such that if Q is
any refinement of Q~€
and S(P;f) is any Riemann sum correspond
ing to Q, where S(Q;f) = £f~k
[g(X~) - g(X~_1)J
,
then
S(Q;f) —ifr€.
In this case the number I is uniquely determined and is
denoted by
S(Q;f) =
= fbfd(x)
and is called the Riemann integral of f over
~
Construction of Riemann-Stieltjes Integral
In constructing the Riemann-Stieltjes integral, we
assume f and g are real valued functions defined on a closed
interval J5,bJ of the real number line.
assume that both f and g are bounded on J.
We shall also
I
~
IJ
II
~ IIFII~II)~
p
Fig. VI
I
I
I
x
- ___________
b
Now consider a partition P of J
=
~,b7 such that
Fig. VII
Li
•
•
-
-
-
•
x~•,
~
Now consider a point ~ such that xk_1~Tk~ Xk where k
2,
....,
n.
=
1,
This is a partition of P and is denoted by
P=(x0,~,, x~, •..xk_l, ~k’ x~).
If we considered the k
subinterval, it would be as indicated in the diagram
Fig. VIII
and the product f (~k) [g(~
-
g(xk_1)J
.
As can be seen,
this product is not the area under either graph between its
bounds, as was the case when constructing the Riemann inte
gral.
When constructing the Riemann integral,4x~(x~-x~_~)
was the length of the
ktb
subinterva].
[xk..1&xkj.
But, when
considering the Rieiiiann-Stieltjes integral, ~g(x~)g(x~)-g(x~_].)
8
denotes a measure of a magnitude associated with the length,
(xk-xk_l) of the
subinterval.
Thus, if g~x) is the
total “mass” or “charge” on the interval
~
then
denotes the mass or charge on the kt~? subinterval,
[xk_1~ xk].
We denote S(P;f,g)Zf(f~)
~
and call it the Riemann—StieltieS sum of f with respect to
g corresponding to P
=
(x0,~
,
x1
,
....,
Xn_l,
~,
X~).
We say that f is integrable with respect to g on J if there
exists a real number I such that for every positive number
E there is a partition P~ of J such that if P is any refine
ment of P( and S(•P;f,g) is any Riexnann-Stieltjes sum
corresponding to P, then
S(P;f,g)
—
ii~
E.
In this case the number I is uniquely determined and is dea~oted
by
S(P;f,g)1
~fd~
it is called the Riemann—Stieltjes integral of f with respect
to g over J5,bJ.
grator.
We call f the integrand, and g the inte
In the special case g(x)x, if f is integrable with
respect to g, we usually say that f is Riemar.n integrable.
Special Examples of Riemann-Stieltjes Integrals
The seven examples which will follow are special cases.
This is done in order to keep the calculations simple.
The
more typical examples are found by combining the ones given
.
(1) The first of these examples is one which has already
been discussed.
If g(x)
x, then the integral reduces to
J
lb
ll~M:j
1
9
the ordinary Riemann integral of elementary calculus.
can easily be
en by looking at
Y
This
following figure IX.
Fig. IX
~
T ~
In figure IX the graph of g(x) is a straight line which
satisfies the condition g(x)x.
terval, the product f(Y~k)
Considering the kt~ subin—
[g(xk) _g(xk_j)J
This is true for k=l, 2, ....,n.
=
~
{~k-xk_l]
/~
f(~)
~k~k-l’~
Therefore,
fdx when g(~x.
There
fore, the Riemann—Stieltjes integral equals the Riemann in
tegral when g •(x)x.
(2) a.
The second special example is; if g is constant
on the interval
5,~7,
then any function f is integrable
with respect to g, and the value of the Riemann-Stieltjes
integral is C.(b).
su.binterval J~
J/J1
More generally if g is constant on the
of J, then any function f which vanishes on
is integrable with respect to g and the value of the
integral is C, where J/Jt
denotes the complement of J rela
tive to J.
(proof of 2a)
We will first consider the first part of the example
by assuming g is constant on the interval f5,bJ.
would look like this.
Our figure
V
Fig • I
—
‘C~
The point is an intermediate point and is defined as
where ~=l, 2, •....,n.
The values of g(x) being constant
g(x~_1)=g(x~) for k1, 2, ....,n.
I
J~fdg(x)
This being the case
=
~~(~k) {~(xk)
=
~
-
{~o~
=0
(proof of 2b)
We will consider part (2b) by assuming g is constant
on J
~
=
on. J/J1
,
of
~
where cLXtb.
and f is a function which vanished
Let x be partition points in
and x be partition points in
each interval
~
~
but n- subinterval in
and (c,~7.
I
Fig. XI
—-
•
—,1~
a~,=X~
xI’ X.~
If we considered the J
I’
-
j(~fdg
I
7’ ~
•
~
of
Ji5,~7
and let x
‘~f(~) (~(x~,)_g(x~,_1i~
=~
~
~k’~
11
If we considered J/J1 =(c,~7 and let Xkft
I~’~J~fd~
~
=
E
=
f(~ti)
(0)
1k” ~Xk ~
)7r
~(x~n)_g(x~tj.~
[~(xkn)_~(xk!Ll)~
=
0
Since both in~egra1s, I’ and I”, are zero, their sum is zero,
and
1=11
+
I”
(3)
~fdg
=
0
In the next example, let. g be defined on J~,bJ
by
g(x)0,
x=a
1,
a~X~.b
Then f is integrable with respect to g if and only if f is
continuous at a and the value of the integral is f(s).
First, we will assume that f is continuous on J
1~,bJ, and g(x) be defined as stated.
/
The figure is
I
Fig. XII
—~
I
-~
x~.
a~x~ x,
~c’~
—
-
-
-
--
-.
X,,~L
The pointç is an intermediate point of the kt~’ subinterval and is defined to be ~ !k~ xk.
for each k, where k—i, 2, ....,n.
first term of the sum
f(a)
~i-o}
=
f(a).
0 since for g~(’x~~)
2..
=
xk_l
This being the case, the
f( ~k) ~g(xk_g(xk_l)~ would be
The value of
-
Let ~k
~
g(x~_1)0, for k~-2
Therefore, I ,~f(x)dg(x)
=
f(s).
~Ef(rk)
~(xk)_g(xk.l)J
1
J
dill il
~
12
(Lb) For this example, let c be an interior point of
the interval
~
and let g be defined by
g(x)0, a~x~c
1, c~-x~b
Consider ,i(fdg
J~dg+j(~dg
It
+
I” where f is a function
and is integr~ble with respect to g if and only if it is
continuous at c from the right.
If f satisfies this condi
tion, the valae of the integral is f(c).
First we will con
sider a function f(x) continuous on J and g(x) as stated
above.
_____~__~
Fig. XIII
I
Now consider a set
If we would consider
g(x~)O for k1, 2,
)J
-g(x
=
c~
in
partition points
product
~E
g(xkhi_1~J
=
0.
....,
5,~7.
n.
Therefore, ~
Now consider (c,~7 and a set of
[xi’ I x”~
(c,~7J
.
The first term of the
f(rk)~g(xk)_g(xk_1)J would be f(c)~(xkfl)
The value of g(x~”)1, and the value of g(x~”_~)
.
The product of
1, and
~
Lx
of partition points [x’(x’ L~~~7]
f(f~,) ~x~1)_g(x~J.i)] in L~’~7~
~
fCc) h(xkn)
~ki~) [g (xkn)
-
-
g(x~n
Therefore, on the closed interval
g(x~tt_1~]
~ )~
Ea,~7,
=
f(c)
L1-o}
D’Grj Ii
I
I’
=
+
I”
=
0.
f(c)
Therefore, on the closed interva1,/~,~7 II+I~f(c)
(5)
Modifying the preceding example, let h be defined
by
h(x)
0,
a~x’~-c
=
ii kk.~~h~ibq
lli~ hi
1
II
~ ~d III
13
=
1,
c~x~b
Then h is con~inuous at c from the right and a function f
is integrable with respect to h if and only if f is continuous
at c from the left.
is f(c).
In this case the value of the integral
We will assume that f is continuous on ~5,bJ and
h. is continuous at a from the left.
The function h will be
defined as stated below.
h (x)
I
Fig. xiv
Consider 11’
=
0, a~x~-c
~
+
I”, where P
J(~dh4I11 1fdh.
5,~7,
considered the closed interval
~2Z
n~
~
(~-
subinterval f (a) [h(x)
f(c).
If we considered
we would have~
I”f(c) for the interval
oJ
=
0.
If we considered the
rh(Xk)
~ ~1k~
~ [i
-
ii
~r5~~7
=
f(~)
-
0.
[1
0]
-
h(xk_l)~
Then I
in
=
I’
+
~
(6) The next example is: Let C1
of J
~f(~k1) {il(xk?)
h (xk_l)j
-
If we
C2
be interior points
and let g be defined by
g(x)
°(~,
a~x~c
°~,
c~x4c
0(j,
c4x~b.
,
If f is continuous at the points c1, c2, then f is integrable
lL~.
with respect to g and~fdgf(c1 )(~-~ )
+
We will first assume that f is continuous at the points Ci,
and f is integrable with respect to g on
5,~7.
Our figure
would be
Fig. XV
a..
Consider I
j~ fdg
Ii+ 12+ 13, where 11% fdg 12=
=
,/~~fdg, I3t1~fdg.
g(xkt_liJ
If we considered
2
i =1(~eag
0.
=
=
-
g(x~tij)}
Therefore, I
%‘~dg
,~7,
=
~
=
f(c,)
The
~k”)f ~
(~-~ ).
case
I
f(c1)
=
The
c~z_.
~,
~~1k’~
cJ, the
,
f( k”~
~.
-
If we considered the
{g(x~n)_
f(~ktt)f~(xk11) _g(x~It_1)}
Therefore, I ~fdg
).
This being the
.
j~(x1~.j-
Therefore,
the first term would be £ (fku)
g(x~n_1)]= f(c~)(~3-~3
~fdg
0.
If we considered the sum on (.c1
first termwould be f(c1){o(~ _~j.
sum on Cc
f(3~,)
on the interval 5,cJwe would have
=
(~(xkn)
x
6
=
f(c~)
~
=
~
)
+
f(c~)
(—~).
(7) This example is: Let the function f be Direchiet’s
discontinuous function defined b7
•f(x)
and let g (x)
=
x.
=
1,
if x is rational
0,
if x is irrational
Consider these functions on I
=
II
d~[b~I~dII
‘5
If a partition P consists of n equal subintervals, then by
selecting k o~ the intermediate points in the sum s(P;f,g)
to be rational and the remaining to be irrational, S(P;f,g)
=
k/n.
It follows that f is not Riemann integrable.
Cx)
The first step will be to assume that g
1
~,
j7,
=
x on
and f (x) will be as defined above.
Y
Fig. XVI
x
We will get a partition P of I
~47
=
and P is a
partition such that for every positive number~ there is a
partition P~
of I and P is a refinement of PL
•
We will
now consider the subinterval to be divided into n equal
subintervals, and k is the end point of the ~
subinterval.
We will consider the subinterval between o and k to be the
subintervals in which we will choose
~ to be rational.
The subintervals between k and 1 will be the ones in which
3~wii]. be irrational.
This is the l~çiS
subinterval in
L~7.
Fig. XVII
)c~ç
16
The product
(xk— xk~1)=
f:fk)
{g(xk)_g(xk_1)l~
=
(1) (xk-xkl)
=
since 1 is the length of the interval and n
,
is the number of subintervals S(P;f,g)= ~
i
The
~.
f(1~)
~(Xk)
-
subinterval in~4.7 would look
like
Fig. XVIII
Xj:
ya-I
Theorems of Riemarin—Stieltjes Integration
The first theorem which will be discussed is Cauchy
Criterion for Integrability.for Riemann—Stieltjes Integrals.
(1) Cauchy Criterion for Integrability.--The function
f is integrable with respect to g over 3
if for each positive real number E
Q.
=
5,bj if and only
there is a partition
of J such that if P and Q are refinements of ~ and if
S(P;f,g) are any corresponding Riemann-Stieltjes sums, then
(S(P;f,g)
Proof:
—
s(c~;f,g)j~. ~
If f is integrable, there is a partition PE
that if P,Q are refinements of P~
,
then any corresponding
Riemann—Stieltjes sums satisfy fS(P;f,g)
(s(~;f,g)
obtain
-
i/~-
(s(P;f,g)
¼..
-
.
such
—
I(.~.. ~&/2 and
By using the Triangle Inequality, we
S(~;f,g,)(~ E
Conversely, suppose the criterion is satisfied.
To
show that f is integrable with respect to g, we need to
produce the value of its integral and use the definition
that/~fdg
=
~ICf~c)f~xk_ xk_l)3if g(x)x.
Let
Q,
then
17
be a partition of J such that if P and
then
~S(?;f,g)
-
S(Q;f,g)(~~
I
.
Q,
are refinements of
Inductively, we choose
Q~ to be a refinement of Q,~..i such that if P and Q are re
finements of Q~, then
S(P;f,g)
-
S(Q;f,g)j.~-
~
Considers sequence (s(Q~;f,g) of real numbers obtained
in this way.
Since Q~ is a refinement of Q3n when n~m, this
sequence of sums is a Cauchy sequence of real numbers,
regardless of how the intermediate points are chosen.
By
Cauchy Convergence Criterion for Riemann Integrals, the
sequence converges to some real number L.
Hence, if E~O,
there is an integer N such that 2/Nd-6 and
jS(Q;f,g)
L~
-
~/2.
~.
If P is a refinement of ~N’ then it follows from the construc
tion of Q~ that
~S(P;f,g)
S(Q;f,g)~\
-
Hence, for any refinement P of
.~.
l/NZ. e,2•
and any corresponding
Riemann—Stieltjes sum, we have
S(P;f,g)
-
This shows that £ is integrable with respect to g over J and
that the vali~e of this integral is L.
Q.E.D.
(2) Theorem.--If f1
to g on J
f~ are integrable with respect
,
ar~dcc~~ are real numbers, then ~f1
+/f~ is in
tegrable witb respect to g on J and
+
f~)dg
Proof: (a) Let~>O and P1
Q~J~dg+ ~
(x0, x1,
....,
~ and P~
=
18
(Yo~ ~i’
~‘‘
Ym~ be partitions of J
Q. is a refinetnent of both P1
=
ü~~7
such that if
and P~, then for any correspond
ing Riemann—Stie].tjes sums, we have
IL
Let P~
—
S(Q;f,g)~~E,
—
be a Dartition of J which is a refir~ement of both
and P~ (for example, all the par~bition points in P1
and P~
are combined to form Pc).
If Q. is a partition of
J such. that P ~ Q, then both of the relations above still
hold.
When the same intermediate points are Lsed, we
evidently have
S(Q;o(f1 +~f~ ,g)d~S(Q;f1,g)+ ~S(Q;ç~,g).
It follows from this and the preceding inequalities, that
JQ~Ij ~
S(Q;~(f1+8i~,g)f
~
S(Q;f~)]
+~~- S(Q;f~,g~+1~1)~
This proves that c’(I~ +Ai~is the integral of c~f1 +Af~
with respect to g, and
Q.E.D.
.(3)
Theorem.-—Suppose that aLc~b and that f is inte—
grable with respect to g over both of the sub~ntervals
5,~7
the
~ Then
interval ~ and
and
f is integrable with respect to g on
Ja fdg = ~‘~fdg +,,/~dg
Proof: If~7 0, lot P~ be a partition of
P’ is a refinement of P~
I
€
,
~
then the inequality fs(P;r,g)
holds for any Rieruann-Stieltjes sum.
corresponding partition of
such that if
LZ,~7.
Let P~
—
be a
If P~ is the partition of
5,bJ formed by using the partition points ~n both P~
19
and P~
,
and if P Is a refinement of P~
~P;f,g)
S(P ;f,g)
+
,
then
s(Pnt;f,g),
where P’, P” denote the partitions of
~ L~,bJ
induced
by P and where the corresponding intermediate points are
used.
I
TherefDre, we have
J~~dg+,/(dg - S(P;f,g)~
~Jfdg ~(Pt;f,g)(
÷ f~~g
-
s(Ptt;f,g)(L~~
It follows that f is integrable with respect to g over
5,~7
and
7~L
rb
~fdg =/fa~
/fdg.
+
(Lb) Integration by Parts.--A function f is integrable
with respect to g over
5,~7
if and only if g is integrable
with respect to f over ~,bJ.
/[fdg
Proof:
to g.
+
J7’~df
In this case,
f (b)g(b)
=
-
f ~a)g(a).
We shall suppose that f is integrable wIth respect
LetE) O~ and let P~
be a partition of
that if Q. Is a refinement of P~
/5~7
such
and S(Q;f,g) is any
corresponding Riemann—Stieltjes sum, then
(s(Q;r,g)
.-
/[~ag
~ ~
Now let P be a refinement of P~ and consider a Riemann—
Stieltjes sum S(P;g,f) given by
S(P;g,f)
whore xk_l ~ ~ ~
tion of
5,~7
=
~
g(~~)~(x~)
(yo..iYii~
Let Q
)4r~) be the parti
obtained by using both the
tion points; hence ~
the terms f (y4~)g (y~~), k
rearrange to obtain
and xk as parti
and y~_~ ~‘k•
=
0, 1,
...,
-.
Add and subtract
n, to S(P;g,f) and
j~k ~
IS_n:
:~~I:III,nöIIkIhE~~
20
S(P;g,f) = f(b)g(b)
-~r(\)
f(a)g(a)
-
g(y~)
-
where the intermediate points
points x~ .
Thus we have
S(P;g,f) = f(b)g(b)
where the partition Q~
P~
are selected to be the
~
—
f(a)g(a)
y1~
y~)
In of the fact that 1S(~;f,g)
.
(s(P;g,f)
S(Q,;f,g),
—
is a refinement of
fdg
-
~(b)g(b) -~f(a)g(a)
-
provided P is a refinen~nt of P~ .
integrable with respect to f over
f
-
This prcves that g is
~
if = f(b)g(b)
dg
(~
and
f(a)g(a).
-
(5) Integrability Theorem.--If f is continuous on J
and g is monotone increasing, then f is integrable with
3.
respect to g over
Proof: Since f is uniformly continuous, giveno0
a real number
fx
y16(e)
—
,
0 such that if x,y belong to J and
then
~f(x)
—
~
Let P~ =
be a partition such that sup{~_Xfr1I..
=
(yD,)~c~
Vr.i)
I(e)
(Xb)XI).•
X~)
and let Q
be a refinement of P€ ; we shall estin~te
the difference S(P~;f,g)
P~
there is
-
S(Q;f,g).
Since every point in
appears in Q, we can express these Rieniann—Stieltjes sums
in the form
S(P ;f,g) =
g(Y~)
-
~ ~ g(/~ In order to write ~(~E ;f,g) in terms of the partition points
in
~,
we must permit repetitions for the intermediate points
and we do not require ¶~ to be contained in
r/k_1’ YkJ.
U Iha:u
1~~liIiI. iI1bJt~ ii ii
110
0±
Lxh_l, x~
However, both!k and~k belong to some interval
and, according to the choice of P~
,
we therefore have
-
If we write t3e difference of the two Rieiiann—StieltjeS
sums and employing the preceding estimate, we have
~ ~
~
=
~(~‘k)~
-
g(y~)
=
Therefore, if P and
ments of P~
Q
-
&
-
g(y~~~)
~(b)
)
f(~’k)J (~(y
-
-
~r )
-
are partitions of J which are refine
and if S(P;f,g) and S(Q~;f,g) are anycorrespond
ing Riemann-Stieltjes sums, then
~S(P;f,g)
S(~;f,g)~~
—
fs(P;f~g)
±
—
(S(P;f,g)
—
s(P
;f,g)
S(Q;f,g)j~. 2 C—~(b)
-
g(a)J
From the Cauchy Criterion for Riemann-Stieltjes integral,
we conclude that 1° is integrable with respect to g.
Existence of Riemann-Stieltjes Integral
It should be clearly understood t~t I ~f dg may
or may not exist, depending on what functions f(x) and
g(x) are used.
It is only when the limit of
~
içz~
-
g(x~_15} exist that the integral is defined.
example let
Example A
a
f(x)
=
0,
g(x)
b=2
=
0,
O.~x≤~l
For
22
1,
V
lL-x~2
Fig. XIX
I
I
-
t
I
0
Consider I
P
+
I”, where I’
=
~dg and I”
547,
If we considered the closed interval
~g(x~~)
g(x~,_i)} =
~
Eo
f(~,)
considered the first term f(~~t?)[g(x~
-
f( ~k”).
g(x~nj)~
in
Li’~7
~(1k~)
=
limit.
[1
f(f~,)
=
-
Since ~k” is defined to be xk”_l .~-Jk~’ ~-Ak”, the
co2
(1)
and I
ir we
g(x~tI_1)}of the
-
value f(~kt!)O, if X~L1Ik~T, and f(fktt)
The ~
~
ol ~o.
-
)
~dg.
I’
+
~(xkt?)
0.
~
=
-
g(x~tt~)~
Therefore, I
=
(1)
=
I’
1 if Xkt?_1±c~%kll.
=
I”
+
=
Li
=
-
0 in one case,
1 in another, and does not approach a
Hence the Riemann-Stieltjes integral does not exist.
~
We will let
5,~7
and g(x) have the same
definition a~ in Example A and let f(x) be identically 1,
In this example, we will show that by changing f(x)
identically, the Rien2ann-Stieltjes integral exists.
Fig. XX
60
I
I
=
1
lJkL.1~Ii~ hI 01
d&fr.”
Consider I
=
~tIlIIIHiih0IfrIliIh II111Th1111
I’
+
I”, where I’ =~dg and I~= fiag.
If we considered the closed interval
~54.7’ ~
fg (xk.It)
0
I
=
-
g(xi~
0 regardless of how
terval
~
~l-O~
-
~k11
=
Jk
C
f(~ii.)
-
is chosen.
The first term
of the product
(1)
(1)
_~)~ =
-
1, no matter where
g(x~lt_~)~
is chosen.
=
~E(i)
Therefore, I”
gardless of w~aere
~1
=
0; therefore,
Now consider the in—
~~fk”~
{~(Xkfl)
=
~(xkIt)
-
g(x~n_1i~would equal
is chosen.
-
l~
g(x~rI_1)~
=
Thea ~
0 regardless of where
1, and I
I’
+
I~’
=
1 re
is chosen, and the Riemann—StieltjeS in
tegral exist inder the conditions given.
In general, we are
able to say that if f(x) and g(x) are bounded on an interval
5,bJ, and do not have a common point of discontinuity in
L’i,~7,
the Riemann—StieltjeS integral exists in
5,j~7.
If f(x) and g(x) have a common point of discontinuity in
the integral fails to exist in
5,~7.
Ildik
~Uidr.~IflI~~dIU
Chapter III
The Presentation of Dabs on
ebesgue—Stielties Integral
Before defining the Lebesgue-Stieltjes integral, the
writer will define a number of key words and symbols with
which the reader should be familiar.
—
If q is a positive integer, we shall say that each
ordered q
t.uple (x”~
—
“point in q
A I
-
The first of these is:
—
,
)
~
of real numbers is a
dimensional space”, or a point in Rq~
If I is a closed interval ~ ~
~
bW, for
i1, 2, ....,q, we define the volume 41 of the interval I
to be (t~
T~
[iJ
-
—
a ~
~
(t~
)
a~
—
(i~~
•. •
—
Let us suppose that f(x) is defined and finite
valued on an interval I
~i,bJ.
=
We subdivide I by means
of a finite number of points Q~i, where a
and form the sum
~
f(~j÷i
)
-
f(oC~
)
•
The supremum of
this sum for all subdivisions of I is called the total
variation, and it is denoted by T~
Bounded Variation
-
If T~
[iJ
~7.
is finite we say that f(x) is
of bounded variation (abbreviated “f (x) is B V”) on I.
1*
-
The sum of intervals
as i
—
If g(x) iseefined and finite on 1*, and the
=
1, 2, ....n.
subinterval I of 1* is the sum of non-overlapping intervals
2~
~tiliiiIiL~iiifl~lS~i Oil
25
~—
I ,...In, then~I
Su!mnable
I
~c_
~≤_.(-l)
If f(x) is defined on the closed interval I, and
—
its upper and lower integrals are finite and equal, f(x)
is said to be sunmiable over I.
Minimum decomposition
Let g(x) be defined on the space H
-
and of B V on every interval 1*.
Then there exist functions
P(x), N (x) with the following properties.
(a) P(x) and N(x) are defined, finite and positively
is on the whole space;
(b) P(x)
-
(c) Tg
N(x)
=
g(x)
[i) =~i +~~I for every intervalI;
-rr
(d) If
then
(x), Y(x) have properties (a), (b), (c),
4~-
I ~ ~I and
interval I.
4y’I ~ for every
A pair of functions P(x), N(x) with
properties (a), (b), (c), and
(a)
is. called a minimum decomposition of g(x).
Semi-continuous (lower, upper).--Let f(x) be defined
on a set E, and let x, of B be an accumulation point of B.
We define f(x) to be lower semi-continuous at x0 if urn inf
f(x) ~-..f(x
),
and we define f(x) to be upper semi-continuous
at x0 if urn sup f(x) ~ f(x).
Accumulation Point.-—A point A is an accumulation
point of a set S, if every deleted neighborhood of A con
tains points of S.
Note that the point A may or may not be
a men~ber of S.
LI_-Function.--A function u(x), defined on a closed
interval I, will be called a
-function if (a) u(x) is
on, d~I
26
lower semi-continuous on I, and (b) u(x) is bounded below;
that is there is a finite constant M such that u(x)~ N
for all x in I.
L
—
function.——A Function 1(x) defined on a closed
interval I, will be called a L
function if (a) u(x) is
—
lower semi-continuous on I, and (b) 1(x) is bounded above;
that is, there is a finite constant N such that u(x)
N
~-
for all x in I.
Intepral of V-function.--If u(x) is a ~J-function on
a closed interval I, J(x) is defined to be sup
ft(x)
a
x
for all continuous functions ~f(x)~ u(x).
Integral of L
on I,
—
function.--If 1(x) is an L
1(x)d(x) is defined to be inf
-
function
[~D(x) for~L1 con
tinuous functions ~‘j0(x) ~-l(x).
Remark 1.
If u(x) is a LI —function, it is bounded
below; so there really do exist continuous functions ~-f’(x)
~-U(x).
Likewise, if 1(x) is an L
a continuous function
Reniark 2.
(f~L~~
—
function, there is
1(x).
If u(x) is a bounded ti-function, say Iu(x)~
~ M.~°then for every continuous function~f(x)
we have
(~ (x) ~
N,
£
u(x)
hence
[(x)dx~ M’~I
It
Therefore, the integral of u is finite, and does not exceed
N’~I.
Like~4se, every bounded L
-
function has a finite in
te gra 1.
Upper Laniell Integral.--Let f(x) be any function
defined on the closed interval I, then its upper Daniell
,j~p,
Ilk
III~II I:. ~dL~Ol
27
integral is
—
dx2
inf fu(x) dx
for all 1J-f~nctions u(x) ~ f(x).
Lower Daniell Integral.--Let f(x) be any function de
fined on the olosed interval I, then its lower Daniell in
tegral is
/f(x) d(x)
for all L
-
sup
/1(x)
functions 1(x) ~f(x).
Remark 1.
If f(x) is bounded on I say
it lies between the continuous functions
—
If(x)I~
N and N.
N,
So by
this definition its lower and upper integrals are finite.
Riemann Integral.--If f(x) is defined on the closed
interval I, and its upper and lower integrals are finite
and equal, f(x) is said to be sumnmable over I, and its in
tegral is to be the common value of its upper and lower
integrals
a x ~Jc(x)~x~/(x) d x.
The writer feels that at this point the definition of
Lebesgue—Stieltjes integral can be understood with the aid
of the definitions given above.
It can be derived by using
the following theorem.
Theoreni.--Let g(x) be defined and finite on the space
and of EV on every closed interval,
be a minimum decomposition of g(x).
,
Let P(x), N(x)
If f(x) is defined on
it is summable with respect to g, or g-sunimable, over
R if it is sunimable with respect to both P and N; and in
this case we define
~(x)dg(x)~~f(X)dP(X)
/f(x)dN(x).
-
There ~s an apparent ambiguity in this definition,
since the functions P(x) and N(x) are not uniquely determined
by g(x).
However, our next theorem will show that this
causes no trouble, and that any minimum decomposition of
g(x) is in a sense the best possible way of representing
g(x) as a difference of positively monotonic functions.
Theorexa.--If g(x) is of BV on every interval I, and
P(x), N(x) are a minimum decomposition of g(x), and 7T(x),
“
(x) are positively monotonic functions such that 4i~1I
I
for every interval I,
then every function f(x) which is summable with respect to
both and v is also suinmable with respect to P and N and
therefore with respect to g, and
ffd !T_
Proof: Defien
-
,~
/~d1
-
h(x)
-
7r(x)
~I we also have ~h’
J4dP
P(x).
-
=
-
[fdN
-
Since 4~~-I
4n~
~
-
4(gI
=
From (d) in
the definition of minimum decomposition, h(x) is positive
ly monotomic, since ~kI
=
.A~p I
-
f(x) is sumraable with respect to
Tr
0.
≥.
and
Therefore if
f is sun~mable
with respect to P, h, and N, and
/fdi7-=
=
-
a
~dY
/~dP
+
Ifdh,
frdN
+
Jfdh.
=
JfdP_ /faN+ ftdh_ ffdh.
/faP
-
(fdN
=
ftdg
~.E.D.
hr ~1Wh
L ~
29
The au-thor feels it is important to define a number
of terms which are used in the examples for the existence
and non—existence of the Lebesgue-Stieltjes Integral.
Sunmaable functions.--are defined to be the limit
“almost everywhere” of sequences of step functions, but
there are sequences of step functions which converge al
most everywhere, but whose limits are not su.xnmable.
We
define every function which is the limit almost everywhere
of step functions a Measurable function.
A measurable set is defined to be a set whose charac
teristic function is measurable where the characteristI~c~
function
K(x) of a set e is defined to be equal to 1 where
x belongs tc e, and the characteristic function K(x) of a
set e is defined to be
o where x does not belong to the
set e.
For a measurable set e, the ii~asure of e
is
defined to be the value of the integral pf t~.-ie characteris
tic function K(x) if ~x) is summable, and the value of the
integral of the characteristic function K(x) is
where
K(x) is not summable.
The next step of the author will be to illustrate
when the Lebesgue—Stieltjes integral exist and when it
doesn’t exist.
Example I, will illustrate its non—existence
If e is a non-measurable subset of E
f(x)
=
1,
~547,
and
if x is in e
if x is not in e
This being the case
I
Lilt ii~~HiLiLIi~~i1 ILL
LI I~U
IL bLLIEiII
30
m(E)
m(e)
+
m(Ce)
=
1’
~jf(x)]~f
mess
•o Lmeasxi f(~)q.
Since e is non-measurable, f is not Lebesgue integrable on
Bi47
and
I
II.
on E
=
=
/f(x) dg(x) does not exist.
We will define f(x)
~47,
g(x)
since x which is continuous
=
x2 is monotone increasing on
=
~47.
Then
i
=
/sin xdx2
=
/2x sin xdx
=
21x
sin xdx
/1=
Let u
=
f~U=
J udv
uv sin xdx
x, dv
=
dxv
sin xdx
cosx
=
x (-cos x)J~
+
,os xdx
_xcosx+sinxJ0
-
cos 1
sin 1
+
-
L-
oJ
sin
sinl- cosi
=
sin 57 18’
5L~o2L1.
Since I
=
Jf(x) dg(x)
that the integral exists.
-
cos
57 18’
=
.8L~i5i-.
.30127
=
.30127, it is
quite obvious
*I~I
J~i,i
~
Bibliography
Books
Bartle, Robert G. The Elements of Real Analysis.
John Wiley and Sons, Inc., 196L~.
McSh.ane, Edward J. Integration.
University Press, l9L~I~.
Princeton
Olmstead, John. N. H. Real Variables.
Century-Crofts, Inc., l9~9.
New York:
Princeton
New York: Appleton
Riesz, Frifyes and Sz.-Nagy, Bela. Functional Analysis.
New York: Frederick Ungar Publishing Company, l9~S.
31