Proving Riemann Integrability of Bounded Function w/ Countable Discontinuities

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Homework Help Overview

The discussion revolves around the Riemann integrability of a bounded function defined on the interval [a,b] that has a countable number of discontinuities. Participants are exploring whether an inductive proof can be applied in this context, particularly in contrast to the typical use of Lebesgue integrals for such cases.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are questioning the validity of using induction to prove Riemann integrability for functions with countable discontinuities, with some suggesting that induction on finite cases does not extend to infinite cases.

Discussion Status

The discussion is ongoing, with participants raising concerns about the applicability of induction in this scenario. Some have provided counterexamples to challenge the original poster's assumptions, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

There is a mention of the characteristic function of the irrationals as a counterexample, which has a countable number of discontinuities but is not Riemann integrable. This highlights the complexity of the topic and the need for careful consideration of definitions and properties related to integrability.

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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.
 
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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.
 
Treadstone 71 said:
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.
 
Last edited:
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]
 
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.
 
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')
 
What's wrong with it?
 
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.)
 

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