# Thread: i just want to see the proof..

1. ## i just want to see the proof..

i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..

Let $A = \{ x\in \Re | x > 0 \}$

let h(x) = $\left\{ {\begin{array}{cc} 0 &, x \in Q' \\ \, \\ \dfrac{1}{n} &, x = \dfrac{m}{n}\\ \end{array}} \right.$

(why cant i remove the right brace?)
where x is always nonnegative and m,n are natural numbers relatively prime..

show that this function is Riemann integrable on any interval $\left[ a,b \right] \subseteq A$

(This function is called Thomae's Function)..

2. Originally Posted by kalagota
i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..

Let $A = \{ x\in \Re | x > 0 \}$

let h = $\left\{ {\begin{array}{cc} 0 &, x \in Q' \\ \, \\ \dfrac{1}{n} &, x = \dfrac{m}{n}\\ \end{array}} \right.$

(why cant i remove the right brace?)
where x is always nonnegative and m,n are natural numbers relatively prime..

show that this function is integrable on any interval $\left[ a,b \right] \subseteq A$

(This function is called Thomae's Function)..
The function is 0 a.e.

RonL

3. Originally Posted by CaptainBlack
The function is 0 a.e.

RonL
huh? i don't get your response.. i think you were saying that the integral is 0, which i also believe to be true..
but, what i want to see is the proof of its Riemann integrability..

4. Given ε>0, choose n with 1/n < ε/(b–a). In the interval [a,b], there are only finitely many rational numbers with denominator ≤n. Let's say there are N of them. Surround each of these N points with an interval of length ε/N, and use the endpoints of these intervals to form a dissection of the interval [a,b]. It's not hard to check that the upper Riemann sum of the function h(x) for this dissection is less than 2ε (the function has an upper bound 1 in each of the N intervals, whose combined length is ε; and an upper bound 1/n in all the remaining intervals of the dissection, whose combined length is less than b–a). Since the lower Riemann sum for any dissection is zero, you can deduce from this that h is Riemann integrable over [a,b] (with integral zero).

5. Originally Posted by Opalg
Given ε>0, choose n with 1/n < ε/(b–a). In the interval [a,b], there are only finitely many rational numbers with denominator ≤n. Let's say there are N of them. Surround each of these N points with an interval of length ε/N, and use the endpoints of these intervals to form a dissection of the interval [a,b]. It's not hard to check that the upper Riemann sum of the function h(x) for this dissection is less than 2ε (the function has an upper bound 1 in each of the N intervals, whose combined length is ε; and an upper bound 1/n in all the remaining intervals of the dissection, whose combined length is less than b–a). Since the lower Riemann sum for any dissection is zero, you can deduce from this that h is Riemann integrable over [a,b] (with integral zero).
What you say is a special case of a more general case.

A set $S$ has Lebesque measure zero in $\mathbb{R}$ if for any $\epsilon > 0$ there exists finitely many compact intervals $I_k$ such that thier union is a cover on $S$ and the total length of the intervals is $< \epsilon.$
(Again, I am not sure if "measure zero" is the correct term).

Functions that are discontinous on a Lebesque measure zero set are always integrable.

i'd like to put this in the "Problem of the week" section but TPH might get angry at me.. anyhow, i want to see the proof of this..
Integrability is not an elementary question to ask.