Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a function that is continuous everywhere except on a countable dense subset, but is bounded?

Bounded in the sense that supremum of $f$ is a finite number.

share|improve this question
    
It needs to be integrable, Riemann or Lebesgue. –  Rajesh D Aug 3 at 15:55
    
OP: Which criterion do you know to determine some function is Riemann integrable? –  Did Aug 3 at 16:22
    
@Did countable set of points of discontinuity –  Rajesh D Aug 3 at 16:25
    
And this gives you no idea to solve the exercise? –  Did Aug 3 at 16:36

2 Answers 2

The function $f(p/q)=1/q$ and zero at irrationals seems to fit the bill.

share|improve this answer
    
thanks @user72694 –  Rajesh D Aug 3 at 15:52
    
Forgot to include another condition in the question : Is it Riemann integrable? if not, then atleast lebesgue integrable? –  Rajesh D Aug 3 at 15:54
    
It's actually Riemann integrable believe it or not. –  user72694 Aug 3 at 15:57

Yes. The typical example is K.J. Thomae's function $f$ from 1875 mentioned in another answer: Define $f(x)=0$ if $x$ is irrational. If $x$ is rational, write $x=p/q$ where $q>0$ and $p,q$ are relatively prime, and set $f(x)=1/q$. One easily checks that the function is bounded (having range contained in $[0,1]$), continuous at all irrationals, and discontinuous at all rationals.

More generally, given any real-valued function (with domain an interval), its discontinuities form an $F_\sigma$ set (a countable union of closed sets), and any $F_\sigma$ set can be obtained this way. This is a result of W.H. Young from 1903. The example above corresponds to taking as $F_\sigma$ set the set of rationals. Note that the rationals are countable union of singletons, and every singleton is closed.

share|improve this answer
    
Is it Riemann integrable? if not, then atleast lebesgue integrable? –  Rajesh D Aug 3 at 15:55
    
Thomae's function? Yes: If a function is bounded and has only countably many discontinuities then it is Riemann integrable. You can even relax the latter requirement to "the set of discontinuities has measure zero", which all countable sets satisfy. –  Andres Caicedo Aug 3 at 15:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.