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Real Analysis
Notes selection vector ¢. Then |RS(f, ,¢) – RSI|< 1, RS(f,,¢) – RS(f,,) = (f(x ) – f(x ))((x ) – (x ))
M N i 2
and
RS(f, p, ¢) – RSI = RS(f, , ) – RSI + (f(x ) – f(x ))(( ˆ x ) – ( ˆ x )).
M N 1 2
By choosing M very large compared to N we can arrange that |f(x ) – f(x )||( ˆ x ) – ( ˆ x )| > 2.
M N 1 2
Then
1 > |RS(f, , ’) – RSI| |RS(f, , ’) – RS(f, ,)| – |RS(f, , ) – RSI| > 2 – 1 = 1.
The definition of Riemann-Stieltjes integrability is contradicted. Hence f is bounded on (, [a, b])
if f is Riemann-Stieltjes integrable with respect to .
Notes From now on, we will usually say “f is Riemann-Stieltjes integrable” instead of
“f is Riemann-Stieltjes integrable with respect to .”
22.4 A difficulty with the Definition; The Cauchy criterion for
Riemann-Stieltjes integrability
b
In order to tell whether f is Riemann-Stieltjes integrable we have to know ò f(x) d (x). The idea
a
of a Cauchy sequence leads to the following Theorem, which gives an equivalent definition.
Theorem: Cauchy criterion for Riemann-Stieltjes Integrability
A function defined on [a, b] is Riemann-Stieltjes integrable over [a, b] with respect to , defined
on [a, b], if and only if for all > 0 there exists > 0 such that for all partitions and ¢ of [a, b],
and for all selection vectors and ¢ associated with and ¢, respectively,
mesh() < and mesh(¢) < |RS(f, , , ) – RS(f, , ¢, ¢)| < .
Proof: First we suppose that f is Riemann-Stieltjes integrable over [a, b] with respect to . Then,
using /2 in the definition of Riemann-Stieltjes integrability, we obtain > 0 and RSI such that
for all partitions of [a, b],
mesh() < |RS() – RSI| < /2
Now we suppose that and ¢ are partitions of [a, b] and that
mesh() < and mesh(¢) < .
Then for all selection vectors and ¢ associated with and ¢, respectively,
|RS(f, , , ) – RS(f, , ¢, ¢)| |RS(, ) – RSI| + |RSI – RS(¢, ¢)| < /2 + /2 = .
This completes half the proof.
Next we suppose that the Cauchy condition, given in the Theorem, is satisfied. We have to find
b
a candidate for ò f(x) d (x). We first construct a sequence of partitions of [a, b]. We let denote
a n
-
æ b aö
=
the partition that divides [a, b] into n equal parts n has points x : a i n ø ÷ . Finally we
+
ç
ni
è
define selection vectors by
n
-
b a n
:= a + i , i = 1, ..., n and define := å f( )( (x ) - (x )),
-
ni n ni ni n,i 1
n i 1
=
a Riemann-Stieltjes sum ( = RS(f, , , )). Now, given > 0, we use /2 in the Cauchy
n n n
criterion, and obtain > 0 such that
mesh() < and mesh(¢) < |RS(f, , , ) – RS(f, , ¢, ¢)| < /2.
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