Find a specific partition to make $U(f,P) < 2/100$

Question

" Suppose $f : [0,1] \to \mathbb{R}$ is given by $f(1/n) = 1/n$ when $n \in \mathbb{N}$ and $f(x) = 0$ for all other $x \in [0,1]$. Show that for some partition $P$ of $[0,1]$, we have $U(f,P) < \frac{2}{100}$" . My attempt is to subdivide the interval $[0,1]$ into subintervals with equal length $d$. Then I will make this number $d$ agree with the inequality $U(f,P) < 2/100$. But it turns out not to work well. How can I approach to solve this? (Let $M_k = sup \{ f(x) : x \in [x_{k-1}, x_k ] \}$ the upper sum $U(f,P) = \sum_{k=1}^{n} M_k (x_k - x_{k-1})$ and $P = \{ 0 = x_0 < x_1 < ... < x_n = 1 \}$)

Answer 1

If subdividing into equal intervals does not work, you need to subdivide into unequal intervals. This is a classic technique, which shows that the upper sum for any countable set like the rationals is zero. Order the set, then put an interval of length $k$ around the first, $\frac k4$ around the second, $\frac k9$ around the third, and $\frac k{n^2}$ around the $n^{th}$. The sum of the intervals is $\frac {k\pi^2}6$, so by choosing $k$ small enough you can get the sum as small as you want.

For your problem, equal intervals will work fine as long as you choose $d$ large enough. The bottom subinterval will have infinitely many points in it, but will only contribute $d^2$ to the sum. There will be only finitely many intervals above that have a $\frac 1n$ in them and each will contribute $\frac 1{nd}$ to the sum. As long as $d$ is large enough you will be there. Given a $d$ you need to bound this sum and you will be there.