Let
denote the vector space of all step functions
on the closed geometric set
Then the assignment
of
into
has the following properties:
- (Linearity)
is a vector space, and
and
for all
and
for all real numbers
- If
is a linear combination of
indicator functions of geometric sets that are subsets of
then
- (Positivity) If
for all
then
- (Order-preserving) If
and
are step functions on
for which
for all
then
Suppose
is constant on the elements of a partition
and
is constant on the elements of a partition
Let
be the partition of the geometric set
whose elements are the sets
Then both
and
are constant on the elements
of
so that
is also constant on these elements.
Therefore,
is a step function, and
and this proves the first assertion of part (1).
The proof of the other half of part (1), as well as parts (2), (3), and (4),
are totally analogous to the proofs of the corresponding parts of
[link] , and we omit the arguments here.
Now for the other necessary consistency condition:
let
be a closed geometric set in the plane.
- If
is a sequence of step functions that converges uniformly to a function
on
then the sequence
is a convergent sequence of real numbers.
- If
and
are two sequences of step functions on
that converge uniformly to the same function
then
-
If
is a real-valued function on a closed geometric set
in the plane,
then
is
integrable on S if it is the uniform limit of a sequence
of step functions on
We define the
integral of an integrable function
on
by
where
is a sequence of step functions on
that converges uniformly to
Let
be a closed geometric set in the plane,
and let
denote the set of integrable functions on
Then:
-
is a vector space of functions.
- If
and
and one of them is bounded, then
- Every step function is in
- If
is a continuous real-valued function on
then
is in
That is, every continuous real-valued function on
is integrable on
- Prove
[link] .
Note that this theorem is the analog of
[link] , but that some things are missing.
- Show that integrable functions on
are not necessarily bounded; not even step functions have to be bounded.
- Show that, if
and
is a function on
for which
for all
in the interior
of
then
That is, integrable functions on
can do whatever they like on the boundary.
Let
be a closed geometric set.
The assignment
on
satisfies the following properties.
- (Linearity)
is a vector space, and
for all
and
- (Positivity) If
for all
then
- (Order-preserving) If
and
for all
then
- If
then so is
and
- If
is the uniform limit of functions
each of which is in
then
and
- Let
be a sequence of functions in
and suppose that for each
there is a number
for which
for all
and such that the infinite series
converges.
Then the infinite series
converges uniformly to an integrable function,
and
- If
and
is a partition of
then
for all
and