Using Riemann criterion for integrability to prove integrability of the function $F$

Question

$F(1/n) = 1$ for $n=1,2,3..$

Else $F(x) = 0$

I considered the uniform partition $[0,1/N,...1/2, 1]$

But then the upper Riemann sum is always equal to one and the lower one always equal to 0.. what partition should I consider?

Answer 1

When $\epsilon < 1$ take partition points $0$, $1/k - \epsilon/2^{k+3}, 1/k + \epsilon/2^{k+3}$ for $k = 2, \ldots, N$, $1 - \epsilon/8$, and $1$.

Then $L(P,f) = 0$ and $U(P,f) = \frac{1}{N} - \frac{\epsilon}{2^{N+3}} + \epsilon\sum_{k=1}^N\frac{1}{2^{k+2}} < \epsilon$ for sufficiently large $N$.

When $\epsilon \geqslant 1$ take $P= (0,1)$.