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and discontinuous at each rational point, except 0.
We will consider the integrability of this function on the interval
[0, 1] by seeking an upper sum and a lower sum which differ by
less than
0.
(i) Identify a lower step function for which the lower sum
0.
(ii) Choose a positive integer
m
such that 1/
m
. Show that
there cannot be more than
m
(
m
1) points
x
[0, 1] such
that
f
(
x
)
1/
m
.
(iii) Suppose that there are exactly
N
such points, which we
denote in order of magnitude by
c
,
c
,
c
,...,
c
.
Find a valueof
such that an upper step function defined on
thesubdivision
0
c
c
c
c
...
c
1
by
S
(
x
)
1
when
c
x
c
,
S
(
x
)
1/
m
otherwise,
gives an upper sum
, where points are suppressed in the
subdivision where there is overlap.
(iv) Deduce that the function
is
Riemann integrable on [0, 1].
Integration and continuity
1/
x
when
x
37
(i) Why is thefunction
f
defined by
f
(
x
)
0 and
f
(0)
0 not integrable on [0, 1]?
(ii) Givean exampleto show that a function which is not
continuous at one point of a closed interval may none the less
be integrable on that interval.
38 A real function
f
is continuous on [
a
,
b
]. How do you know that
f
is bounded and that therefore
f
has an upper and a lower integral?
39 (
Cauchy
, 1823) Thereal function
f
is continuous on [
a
,
b
] and
a
x
b
.
(i) Must thefunction
f
bebounded and attain its bounds on
[
x
x
x
...
x
,
x
]?
(ii) Let
m
inf
f
(
x
)
x
x
x
,
and
M
sup
f
(
x
)
x
x
x
.
Explain why
m
(
x
x
) is a lower sum for
f
and why
M
(
x
x
) is an upper sum for
f
.
