Riemann Integrability of Thomae's Function

[c1fn]

Homework Statement

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.

Homework Equations

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

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

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:

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

I need to show that g_n is continuous for each n (easy). I need to show that g_n 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 g_n → 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 g_n(x) → g(x) uniformly (i can show this). Then I argue that this implies that if y is "otherwise," then g_n(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?

 

[Bacle2]

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

integrability?

 

[c1fn]

No I don't. I need to show this using uniform convergence (if that is possible).

 

[Bacle2]

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.

 

[c1fn]

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?

 

[Bacle2]

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

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

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

r_n=a_n/b_n, then :

gcd(a_n,b_n)=1

( 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 f_n?

Does that help?

 

[c1fn]

Yes. That makes perfect sense. Thank you. Currently I'm working on showing f_n → f uniformly.