Suppose we defined the integral of a step function as:
Which of the following properties of the integral would still hold:
- If for all , then .
Proof. Let be a partition of and be 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,
Proof. Let be partition of such that is constant on the open subintervals of . (We know such a partition exists since we can take the common refinement of the partitions of on which and individually are constant.) Assume if for . Then,
Proof. Let be a partition of such that is constant on the open subintervals of . Assume if for . Then if so using our alternative definition of the integral we have,
- False. Since . In particular, a counterexample is given by letting for all and let . Then,
A counterexample is given by considering and on the interval . Then, on the interval, but
Hi! I appreciate your efforts. But (b) doesn’t seem convincing enough, I would suggest to use induction to prove that the s+t is constant on every open subinterval of union of Pt and Ps (partition of [a,b] according to t and s respectively). I don’t think it is a good choice to assume this is true by default. By the way, thanks a lot!
that e) is surprising. positive area, negative integral.
Yeah, it’s negative, but keep in mind that we aren’t using the original definition of integral, but a modified one, so we can’t say we are dealing with “real” integrals here.