Riemann Integrability of Thomae's Function

In summary: I'll let you know when I get that done.In summary, the Thomae's function f : [0,1] → ℝ is Riemann integrable and can be proven using the theorem that states if fn : [a,b] → ℝ is Riemann integrable for each n, and fn → f uniformly on [a,b], then f is Riemann integrable. This can be done without invoking any sort of measure theory/Lebesgue integration. The attempt at a solution involves constructing a sequence of Riemann integrable functions that are continuous and converge uniformly to Thomae's function. This can be done by starting with an enumeration of the rationals in [0,1] and defining each term of
  • #1
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 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.


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?
 
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  • #2
Do you know of any results relating the set of points of continuity to Riemann

integrability?
 
  • #3
No I don't. I need to show this using uniform convergence (if that is possible).
 
  • #4
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.
 
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  • #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?
 
  • #6
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 :

gcd(an,bn)=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 fn?

Does that help?
 
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  • #7
Yes. That makes perfect sense. Thank you. Currently I'm working on showing fn → f uniformly.
 

FAQ: Riemann Integrability of Thomae's Function

1. What is Thomae's Function?

Thomae's Function, also known as the Riemann function, is a mathematical function defined by German mathematician Carl Johan Thomae in 1875. It is a discontinuous function that takes on rational values at rational points and irrational values at irrational points.

2. Is Thomae's Function Riemann integrable?

Yes, Thomae's Function is Riemann integrable. This means that it satisfies the Riemann criterion for integrability, which states that a function must be bounded and have a set of points of discontinuity with measure zero in order to be Riemann integrable.

3. What is the Riemann integrability of Thomae's Function?

The Riemann integrability of Thomae's Function is 0. This means that the integral of the function over any interval is 0, regardless of the width of the interval. This is because the function is discontinuous at every point, making it impossible to calculate the area under the curve using traditional methods.

4. How is the Riemann integral of Thomae's Function calculated?

The Riemann integral of Thomae's Function is calculated using the Riemann sum, which is a method of approximating the area under the curve by dividing the interval into smaller subintervals and calculating the sum of the areas of rectangles that approximate the curve. However, since the function is discontinuous, the Riemann sum will always be 0, resulting in an integral of 0.

5. Why is the Riemann integrability of Thomae's Function important?

The Riemann integrability of Thomae's Function is important in the study of real analysis and calculus. It serves as an example of a function that is bounded and satisfies the Riemann criterion for integrability, but is still not integrable in the traditional sense. This helps to illustrate the limitations of the Riemann integral and the need for more advanced methods, such as the Lebesgue integral, to integrate certain types of functions.

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