Proving that a function is Riemann integrable

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"Let (r_n) be any list of rational numbers in [0,1]. Let (a_n) be a sequence such that 1>a_1>a_2>...>a_n>... converging to 0. Let f be defined such that

f=0 if x is irrational
=a_n if x is equal to r_x

Prove that f is Riemann integrable."

We are doing integrals from Darboux' approach, so no tagged partitions or whatnot. I have to somehow show that the upper sum L(f,P) and lower sum L(f,P), for some partition P, differ by less than a given e>0.

L(f,P) = 0 for any partition, so |U(f,P) - L(f,P)|=U(f,P). I have no idea how to proceed, because if p is a partition of n equal parts, U(f,P) \leq \sum_{i=0}^{n}\frac{1}{n} which diverges. I am unable to refine the inequality.
 
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1.You need to take advantage of the fact that the a_n converge to zero.
2.You don't know that L(f,P) is going to be 0 since the a_n can be less than zero.

Hint:
If you know that for n>N, a_n < 1/k can you produce an upper bound for U(f,P) as the norm of the partition goes to zero?
 
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a_n is a strictly decreasing sequence converging to 0, a_n is therefore never less than 0.

I think I have an idea on how to proceed, regardless.
 
Treadstone 71 said:
a_n is a strictly decreasing sequence converging to 0, a_n is therefore never less than 0.

<opens mouth, switches feet>
Oh, I missed that. It makes things a bit more convenient.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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