这一节开始讲Riemann积分了,整个Riemann积分这一章的处理是比较新的,用相对好理解的piecewise constant function来定义上积分和下积分,二者相等即为可积,和以往比较经典的达布和其实是一回事,但接受起来就更容易。第一节主要说分法partition。
Exercise 11.1.1
If
then the statement is obvious, since
is bounded and connected and is an interval for trivial reasons.
Now suppose
, then (b) implies (a) is clear, by the definition of an interval. Now suppose (a) holds, then
, since
is non-empty and bounded, we can say
and
both exist in
. Now choose any
such that
By the definition of supremum and infimum, we can find
, s.t.
Since
is connected,
, thus
. So we have the relation
, also we can easily have
, then
is equal to
plus possibly two points
and
, anyway
is an interval.
Exercise 11.1.2
By Lemma 11.1.4,
and
are bounded and connected, we have
, s.t.
Thus
is bounded.
Let
, then as
, we have
, similarly we have
, so
, this means
is connected.
We now can conclude, using Lemma 11.1.4 again, that
is a bounded interval.
Exercise 11.1.3
First, if
is not of the form
or
for any
, then we can have some
s.t.
, as
is an interval, it’s connected, thus we have
Now assume all intervals
is not of the form
or
for any
, then we have
. Then we can conclude
so we have
, this is a contradiction since
form a partition of
.
Thus there must exists at least one of
in the form of
or
. By the definition of partition, such
can only appear once if
form a partition of
.
Exercise 11.1.4
is a partition of
:
Choose
, then there’s exactly one
and
s.t.
, thus
, and any other element belongs to
can’t contain
, otherwise
or
can’t be unique. This means
is a partition of
.
is finer than
or
:
For every
, by definition we shall be able to find
and
s.t.
This means
and
, thus we have
is finer than
or
.