Consider if we chose the following definition for the definite integral of a step function:
Which of the following properties (all valid for the actual definition) would remain valid under this new definition:
-
.
-
.
-
.
-
.
- If
for all
, then
.
- True.
Proof. Letbe a partition of
and
be a partition of
such that
is constant on the open subintervals of
and
. Then, let
, where
, so
is a partition of
and
is constant on the open subintervals of
. Then,
- False.
Counterexample: Letfor all
. Then,
while
Hence, this property does not hold. (More generally, since
.)
- False. Again, this is because
. A specific counterexample is given.
Counterexample: Letfor all
and let
. Then,
while,
- True.
Proof. Letbe a partition of
such that
on the
th open subinterval of
. Then,
is a partition of
and
on
. So,
Thus, indeed we have
- True.
Proof. Sinceimplies
, the result follows immediately
Wouldn’t it be P_1 = {x_0,…,x_m} partition of [a,b] ?