The set of
where
is not continuous is contained in
the union of the sets where
and
are not continuous. These last two
sets are of measure 0 by Theorem 3-8; so theee first set is also of measure
0. But then
is integrable by Theorem 3-8.
If
has content 0, then it is bounded by Problem 3-9 (a); so it is a subset
of an closed rectangle
. Since
has content 0, one has
for some open rectangles
the sum of whose volumes can be made as small
as desired. But then the boundary of
is contained in the closure of
,
which is contained in the union of the closures of the
(since this
union is closed). But then the boundary of
must be of content 0, and
so
is Jordan measurable by Theorem 3-9. Further, by Problem 3-5, one
has
which can be made as small as desired; so
.
Let
be the set of rational numbers in
. Then the boundary of
is
, which is not of measure 0. So
does not exist
by Theorem 3-9.
Using the hint, let
be a partition of
where
is a closed rectangle
containing
. Then let
be a rectangle of
of positive volume.
Then
is not of measure 0 by Problem 3-8, and so
.
But then there is a point of
outside of
; so
.
Since this is true of all
, one has
. Since this
holds for all partitions
of
, it follows that
if the integral exists.
Following the hint, let
be a positive integer and
.
Let
. Let
be a partition of
such that
.
Then if
is a rectangle of
which intersects
, we have
. So
.
By replacing the closed rectangles
with slightly larger open rectangles,
one gets an open rectangular cover of
with sets, the sum of whose volumes
is at most
. So
has content 0. Now apply Theorem 3-4 to
conclude that
has measure 0.
The set of
where
is not continuous is
which is not
of measure 0. If the set where
is not continuous is not of measure 0,
then
is not integrable by Theorem 3-8. On the other hand, if it
is of measure 0, then taking the union of this set with the set of measure 0
consisting of the points where
and
differ gives a set of
measure 0 which contains the set of points where
is not continuous. So this
set is also of measure 0, which is a contradiction.
This is an immediate consequence of Problem 3-12 and Theorem 3-8.
Suppose
is Jordan measurable. Then its boundary is of content 0 by
Theorem 3-9. Let
and choose a finite set
for
of open rectangles the sum of whose volumes is less than
and
such that the
form a cover of the boundary of
. Let
be a
partition of
such that every subrectangle of
is either contained
within each
or does not intersect it. This
satisfies the
condition in the statement of the problem.
Suppose for every
, there is a partition
as in the statement.
Then by replacing the rectangles with slightly larger ones, one can obtain
the same result except now one will have
in place of
and the
will be open rectangles. This shows that the boundary of
is of content 0; hence
is Jordan measurable by Theorem 3-9.
Let
be a closed rectangle containing
. Apply Problem 3-21 with
as the Jordan measurable set. Let
be the partition as in Problem 3-21.
Define
. Then
and clearly
is Jordan measurable by Theorem 3-9. Further
.