Proving Riemann Integrability of Bounded Function w/ Countable Discontinuities

[Treadstone 71]

Is it possible to show by induction that f:[a,b]->R, a bounded function, is Riemann integrable if f has a countable number of discontinuities? I'm told this is usually done with Lebesgue integrals, but I don't see why an inductive proof of this using Riemann/Darboux integrals can't work.

 

[matt grime]

What are you inducting on? It can't be the unmber of discontinuities. showing something for a finite number of things in no way says anything about the infinite case.

 

[NateTG]

Is it possible to show by induction that f:[a,b]->R, a bounded function, is Riemann integrable if f has a countable number of discontinuities? I'm told this is usually done with Lebesgue integrals, but I don't see why an inductive proof of this using Riemann/Darboux integrals can't work.

No.

Consider, for example, the characteristic function of the irrationals.
f(x) is 1 if x is irrational, and 0 if x is rational.

This has a countable number of discontinuities -- one at every rational number -- but is most certainly not Riemann integrable as the lower sum will always be 0, and the upper sum will always be 1.

 

[matt grime]

A 'rational' number of discontinuities? No, it has an uncountable number of discontinuities. It is discontinuous at every point at of the interval [0,1]

 

[Treadstone 71]

Well, f is integrable if it has 1 discontinuity, and whenever it has n discontinuities and is integrable, it implies that having n+1 discontinuities is integrable.

 

[matt grime]

and that tells you nothing (as given) about countably (but not finitely) many discontinuities. (Proof... well, according to this logic, the set of natural numbers is finite by 'induction')

 

[Treadstone 71]

What's wrong with it?

 

[matt grime]

Because countable includes infinite, and induction on finite things tells you nothing about infinite things. as i said according to your logic all countable sets (ie the set of natural numbers) are finite. proof: a set with n elements is finite, adding one more it remains finite, hence all countable sets are finite, but the natural numbers are a countable set that are not finite. (induction does not apply.)