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c1fn

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## 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?