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This gives
S
s
, which contradicts qn 20.
22 A constant f
u
nction is a step function and therefore the inequality
S
guarantees the first condition before qn 12 for
any supposed 'integral' lying between the upper and the lower integral.
The second condition also holds because it holds for step functions.
s
f
f
23 Since
f
sup
s
, there is a lower step function
s
such that
f
s
f
and, since
f
inf
S
, there is an upper step
function
S
such that
f
S
f
.
24 If the difference between the upper and lower integrals were
, then to
find an upper step function
S
and a lower step function
s
such that
S
s
would contradict the chain of inequalities before qn 22.
25 If we consider upper and lower step functions on the intervals
(0, 1/
n
), (1/
n
, 1/(
n
1)), . . ., (
,
), (
,
), (
, 1);
then we have an upper sum
0. Sincethseare
arbitrarily close, the function is integrable, and has integral 0.
1/
n
and a lower sum
26
s
(
x
)
is a lower sum and
k
is an upper sum for
k
·
f
.
k
·
k
·
k
·
, so the difference
between an upper sum and a lower sum may be made arbitrarily small
and so
k
·
f
is integrable. Clearly the integral is
k
·
F
.
f
(
x
)
S
(
x
)
k
·
s
(
x
)
k
·
f
(
x
)
k
·
S
(
x
)
k
27
s
(
x
)
f
(
x
)
S
(
x
)
S
(
x
)
f
(
x
)
s
(
x
)
S
is a lower step
function and
s
is an upper step function for
f
.So
is a lower
sum and
is an upper sum for
f
.(
)
(
)
.
Clearly the integral is
F
.
28 From qn 26,
k
·
f
is integrable so, from qn 27,
k
·
f
is integrable, with
integral
k
·
F
.
29
s
(
x
)
f
(
x
)
S
(
x
) and
s
(
x
)
g
(
x
)
S
(
x
)
s
(
x
)
s
(
x
)
f
(
x
)
g
(
x
)
S
(
x
)
S
(
x
).
Now
s
s
is a step function on theunion of thesubintrvals for
s
and
s
, in the sense of qn 20, and is therefore a lower step function for
f
g
.
Moreover,
using thedefinition of theintegral of a
step function on the union of the subintervals. So
(
s
s
)
s
s
is a lower sum
for
f
g
, and likewise
is an upper sum.
(
)
(
)
(
)
(
)
2
, which may bemade
arbitrarily small, so
f
g
is integrable.
F
and
G
F
G
,
so theintegral is
F
G
.
30
1) from qn 4. Some caution is called for in the appeal
to qn 4, sincethedissction of [0,
a
] in qn 4 is infinite, and so the
approximating functions are not step functions. However, by choosing
arbitrarily large finite dissections we can obtain upper and lower sums
x
a
/(
k
