Results 1 to 8 of 8

Math Help - question on Integrals

  1. #1
    Member
    Joined
    Dec 2008
    Posts
    154

    question on Integrals

    Define f(x)=x if x is rational and f(x)=0 if x is irrational.
    Is f(x) Riemann Integrable on [0,1]?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,658
    Thanks
    1615
    Awards
    1
    Quote Originally Posted by Kat-M View Post
    Define f(x)=x if x is rational and f(x)=0 if x is irrational.
    Is f(x) Riemann Integrable on [0,1]?
    Given any subdivision, \mathcal{D}, of the interval [0,1] what is the upper sum and the lower sum with respect to \mathcal{D}, ?

    Can that difference be made ‘small’?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2008
    Posts
    154

    subdivisions

    the lower sum should be zero since in any subdivision, there is an irrational number. and if each subdivision is so small, then upper sum could be really small. so i think the integral is 0
    but if you do the samething to Dirichlets funtion, shouldnt the integral be 0 also? but Dirichlets funtion is not Riemann integrable.
    i dont see the difference in these functions.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    it is not continuous on (0,1], so it is not Riemann integrable.
    There is a theorem which finally solve the problem to judge which limitary function is integrable: the set of uncontinuous points forms a "null set",that is, we can use countable numbers of open interval to cover a real numbers' set completely and the sum of the "length" of these interval ( length(a,b)=b-a)can be arbitrarily close to 0.
    A obvious corollary is that if the uncontinuous points set forms an interval (a,b), then the function must not be Riemann integrable.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    Quote Originally Posted by Kat-M View Post
    and if each subdivision is so small, then upper sum could be really small.
    really? in [\frac{1}{2},1], any upper sum would be greater than \frac{1}{4}!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Dec 2008
    Posts
    154

    upper sum

    i see. so the lower and upper sum are not the same. so it is not Riemann integrable.

    What if the function is deined as f(x)=\frac{1}{m} if x=\frac{n}{m} where n,m \in N and f(x)=0 is x is irrational?

    would it be integrable? since we dont know that f(x)> \frac{1}{2}on [\frac{1}{2},1], we cant extimate the upper sum.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    Quote Originally Posted by Kat-M View Post
    i see. so the lower and upper sum are not the same. so it is not Riemann integrable.

    What if the function is deined as f(x)=\frac{1}{m} if x=\frac{n}{m} where n,m \in N and f(x)=0 is x is irrational?

    would it be integrable? since we dont know that f(x)> \frac{1}{2}on [\frac{1}{2},1], we cant extimate the upper sum.
    for every irrational number x, we can use a sequence of rational numbers to approach it. In these sequence, the denominator would approach to \infty. So the result will have limit 0. So it will be continuous at x. So the uncontinous points set would be countable.
    Let we list these points by a_1,a_2,...., then the sum of the length of the sequence (a_1-\frac{1}{2^k},a_1+\frac{1}{2^k}),(a_2-\frac{1}{2^{k+1}},a_2+\frac{1}{2^{k+1}})... would be \frac{1}{2^{k-2}}, which will have limit 0 when k\rightarrow \infty. So it would be a "null set". So the function is Riemann integrable
    Last edited by ynj; August 17th 2009 at 06:07 PM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    ynj
    ynj is offline
    Senior Member
    Joined
    Jul 2009
    Posts
    254
    I am sorry I do not follow your idea but use another theorem..but this theorem is really convenience when judging the integrability
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exam question - Integrals
    Posted in the Calculus Forum
    Replies: 14
    Last Post: February 1st 2011, 09:28 AM
  2. Quick question about integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 25th 2009, 07:58 PM
  3. T/F Question on integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 20th 2009, 01:55 PM
  4. Question about Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 21st 2008, 01:52 PM
  5. integrals question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 16th 2007, 05:53 AM

Search Tags


/mathhelpforum @mathhelpforum