1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thomae Function

  1. Mar 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Show the Thomae's function f : [0,1] → ℝ which is defined by [itex]f(x) = \begin{cases} \frac{1}{n}, & \text{if $x = \frac{m}{n}$, where $m, n \in \mathbb{N}$ and are relatively prime} \\ 0, & \text{otherwise} \end{cases}[/itex] is Riemann integrable.

    2. Relevant equations
    Thm: If fn : [a,b] → ℝ is Riemann integrable for each n, and fn → f uniformly on [a,b], then f is Riemann integrable.

    Trying to do this without invoking any sort of measure theory/Lebesgue integration.

    3. The attempt at a solution
    I've been stumped on this one for awhile now. Basically I've been trying to construct a sequence of Riemann integrable functions from [0,1] to ℝ that are continuous (and therefore Riemann integrable) that converge uniformly to Thomae's function. Unfortunately, I haven't been able to construct such a sequence of functions. The function that I'm trying to "tweak" right now is:

    gn : [0,1] → ℝ defined by [itex]g_n = x \cos^{2n}(p! \pi x)[/itex].

    I need to show that gn is continuous for each n (easy). I need to show that gn converges uniformly to the function g : [0,1] → ℝ given by

    [tex] g(x) = \begin{cases} x, & \text{if $x=k/p!$, where $0 \leq k \leq p!$} \\ 0, & \text{otherwise} \end{cases} [/tex]

    Here lies a bit of a mental barrier for me. I know that I haven't showed that g(x) = f(x), but I'm having trouble showing that gn → g uniformly since the value of g depends pointwise on x. I argue that if x is of the form k/p!, where 0 ≤ k ≤ p!, then gn(x) → g(x) uniformly (i can show this). Then I argue that this implies that if y is "otherwise," then gn(y) → g(y) uniformly as well since g(y) ≤ g(x), where x is of the form mentioned above.

    Anyways, I'm having a hard time believing my own argument. Does this work? If it doesn,t is it even possible to construct a sequence of continuous functions converging uniformly to Thomae's function?
  2. jcsd
  3. Mar 1, 2012 #2


    User Avatar
    Science Advisor

    Do you know of any results relating the set of points of continuity to Riemann

  4. Mar 1, 2012 #3
    No I don't. I need to show this using uniform convergence (if that is possible).
  5. Mar 1, 2012 #4


    User Avatar
    Science Advisor

    A suggestion:

    Try defining a sequence that mimicks/approaches the Thomae function.

    Start with an enumeration of the rationals, and assume they are given

    in reduced/lowest form. Then define f_n(x). You can

    start by defining f_1(x)=1/m for x=a_1, and ? otherwise . Does that help?

    I'm sorry, I must go for a few hours; if I'm not back tonight, I will be back tomorrow.
    Last edited: Mar 1, 2012
  6. Mar 1, 2012 #5
    Sorry I'm a bit confused when you say an enumeration of the rationals. Arn't the only rationals we're concerned about between 0 and 1? Can you elaborate a bit more on how you're defining f_n and what is a_1?
  7. Mar 1, 2012 #6


    User Avatar
    Science Advisor

    Yes, sorry, I meant the rationals in [0,1]. Since the rationals are countable,

    we can arrange them as a sequence {rn} , which assigns to n the

    n-th rational, and we assume we have these rationals in lowest terms, i.e., if

    rn=an/bn, then :


    ( this reduction is always possible, e.g., by the

    fund. thm of arithmetic ). Now, I'm trying to suggest how to construct the sequence : we

    want a sequence {f_n} of

    functions that converges to the Thomae function uniformly. We can then define the first

    term f1 of the sequence like this:

    f1(r1)=f(a1/b1):=1/b1, and

    f1(x)=0 for x in [0,1]\{r_1}; f_2(a_1)=1/b_1 ; f_2(a_2)=1/b_2 ; f:[0,1]\{r_1\/r_2}=?

    Can you see how to continue defining f2(r_n), and then fn?

    Does that help?
    Last edited: Mar 2, 2012
  8. Mar 2, 2012 #7
    Yes. That makes perfect sense. Thank you. Currently I'm working on showing fn → f uniformly.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook