Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Could someone give me an idea for a proof of this

  1. Mar 30, 2004 #1
    Let f ( x ) = {0, if x = 1/n for some natural number n
    or 1, otherwise

    Note: Natural number would refer to the set of positive integers Z+ that is 1,2,3,...

    Prove that this function is integrable on [0,1] and it's integral is 1

    Certainly there are an infinite number of dicontinuities and nearly all of the function lies in the domain of [0,1]. But is the set of Inverse Naturals (1/n) (postive integers) bigger than the set of irrationals?

    Someone recommended using the Robustness of the Reimann Integral

    Let g and f be two functions defined on [a,b], and suppose
    that the set of numbers in [a,b] at which the functions do not
    take the same value (at which they "differ") is finite.
  2. jcsd
  3. Mar 30, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you prove it's integrable over [a, 1] for some positive a?
  4. Mar 30, 2004 #3


    User Avatar
    Science Advisor
    Homework Helper

    What definitions are you using? You should have no problem showing that the upper and lower sum converge.
  5. Mar 30, 2004 #4

    what do you mean? AS for the upper and lower sums - so then

    U(f,P) = Sum i =1 to infinity 1 = infinity

    and L (f,P) = 0

    they dont seem to converge...?
  6. Mar 30, 2004 #5


    User Avatar
    Science Advisor

    Oh, dear. I don't want to hurt your feelings but you are way, way off.
    For one thing, your integral is from x= 0 to x= 1 so the sum is NOT from 1 to infinity. The sum in U(f,P) is over the intervals in the partion- for any finite partition, it is a finite sum. Also the summand is not 1:it is 1 times the length of the interval in the partition. U(f,P) is NOT infinity.

    Also, L(f,P) is not 0. Since f(x) is 1 for all x except 1/2, 1/3, 1/4, etc. it clearly is equal to 1 for all x> 1/2: L(f,P) will be 1(1/2)+ something.

    Oh, by the way:
    "But is the set of Inverse Naturals (1/n) (postive integers) bigger than the set of irrationals?"

    Not even close: there is an obvious one-to-one correspondence between the naturals and "inverse naturals" (n<-> 1/n) so the set {1/n} is countable.
  7. Mar 30, 2004 #6
    so how can i prove taht the set of naturals is countable or finite, thus proving that the function has a finite number of discontinuities?

    is there a theorem ??
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook