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 .
Proof. Let be 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,
Counterexample: Let for all . Then,
Hence, this property does not hold. (More generally, since .)
- False. Again, this is because . A specific counterexample is given.
Counterexample: Let for all and let . Then,
Proof. Let be a partition of such that on the th open subinterval of . Then,
is a partition of and on . So,
Thus, indeed we have
Proof. Since implies , the result follows immediately