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
let h(x) =
(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
(This function is called Thomae's Function)..
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 has Lebesque measure zero in if for any there exists finitely many compact intervals such that thier union is a cover on and the total length of the intervals is
(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.
Integrability is not an elementary question to ask.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..